Part A - Prefixes, Roots, and Suffixes Match these prefixes and suffixes to their meanings. Reset Help oste- The word root means bone. peri- The word root means one that breaks or destroys. hema- The word root means blood. means blood. -clast -blast The word root means around or enclosing. The word root means embryonic or formative cell. Submit Request Answer Part B - Vocabulary Terms Match these vocabulary terms to their meanings. Reset Help hematopoiesis The connective tissue membrane that encloses a bone is the osteoporosis is a disease in which bones lose mass and strength, becoming brittle. osteoblasts Blood cells are formed by the process of periosteum osteoclasts Calcium is released by the action of ,which break down bone matrix. are cells that build bone matrix. Submit Request Answer
Answers
Calcium is released by the action of osteoclasts which break down bone matrix.
The cells known as osteoclasts are responsible for breaking down bone to start the process of normal bone remodeling and for mediating bone loss in pathologic situations by escalating their resorptive activity. They develop in the bone marrow from progenitors in the myeloid/monocyte lineage, which then circulate in the blood. The Howship lacunae, which osteoclasts occupy on the surface of the bone, are assumed to be the result of the osteoclasts' enzymes eroding the bone. The union of many cells produced from blood-circulating monocytes results in osteoclasts. These are derived from the bone marrow in turn.
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Related Questions
A vehicle that starts to move from rest gets an acceleration of 5m/s² within 2 seconds. Calculate the velocity and distance covered by the vehicle within the given time. [Answer: 10m/s, 10m]
Answers
Answer:
V = 10m/s for the final velocity.
S = 10m for distance II.
Explanation:
Given the given information:
5m/s2 acceleration
2 seconds Equals time
Because the vehicle begins at rest, its initial velocity is 0m/s.
The first equation of motion would be used to get the ultimate velocity.
V = U + at
We have by substituting into the equation
V = 0 + 5*2
V = 0 + 10
V = 10m/s.
Now, to find the distance covered, we would use the second equation of motion.
S = ut + ½at²
S = 0*2 + ½*5*2²
S = ½*5*4
S = 20/2
S = 10m
An 800 V/m electric field is directed along the +x-axis. If the potential at x=0 m is 2000 V, what is the potential at x=2 m?
Answers
The potential at x = 2 m is 400 V.
Explanation:-
Given that the electric field is directed along the positive x-axis and has a magnitude of 800 V/m.
The potential at x = 0 m is 2000 V.
We are supposed to find the potential at x = 2 m.
Potential difference between two points,
V = - ∫ E dx = -E(x - x₀)
where x₀ is the reference point (in this case, x₀ = 0 m).
Let's substitute the given values in the above formula.
V = -E(x - x₀)= - (800 V/m) (2 m - 0 m)
= -1600 V
The potential at x = 2 m = V + V₀
where V₀ is the potential at x = 0 m (given to be 2000 V).
V = -1600 V, and V₀ = 2000 V.
So, V at x = 2 m
= -1600 V + 2000 V
= 400 V
Therefore, the potential at x = 2 m is 400 V.
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what is the temperature in ∘f at the triple point of carbon dioxide?
Answers
Answer:
The triple point for carbon dioxide is -70°F (-57ºC) at 5.2 atm.
Explanation:
5.2 atm is 5.2 times the air pressure at sea level, the reason why this is included is because liquid CO² cannot exist below 5.2 atm, as if it were lower it would evaporate due to submilation.
Although the temperature of the triple point of carbon dioxide is -70°F (-57ºC), the pressure present is around 5. 2 atmospheres.
What states of matter can carbon dioxide adopt?Carbon dioxide can adopt all three states of matter - solid, liquid, and gas - at a precise configuration of pressure and temperature. The triple point is a precise state in which the three phases exist simultaneously and are in balance.
If the triple point conditions are not obeyed, carbon dioxide may go through changes in its physical state, such as transforming into a gas or a liquid, which will be determined by fluctuations in temperature and pressure.
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6) A spring with a spring constant k = 800 N/m has been compressed, and 196 J of
potential energy is stored. What distance from equilibrium has the spring been
compressed?
Answers
Answer:
\(0.7\; {\rm m}\).
Explanation:
If spring with spring constant \(k\) is compressed by \(x\) relative to the equilibrium, the elastic potential energy \(E\) stored in that spring will be \(E = (1/2)\, k\, x^{2}\).
For the spring in this question, it is given that \(E = 196\; {\rm J}\) whereas \(k = 800\; {\rm N \cdot m^{-1}}\). The displacement \(x\) needs to be found.
Note that \(\begin{aligned}E &= 196\; {\rm J} = 196\; {\rm N \cdot m} \end{aligned}\) (one joule is the work done when a force of \(1\; {\rm N}\) is applied over a distance of \(1\; {\rm m}\).)
Rearrange the equation \(E = (1/2)\, k\, x^{2}\) to find displacement \(x\) in terms of potential energy \(E\) and spring constant \(k\):
\(\begin{aligned} x^{2} = \frac{2\, E}{k}\end{aligned}\).
\(\begin{aligned} x &= \sqrt{\frac{2\, E}{k}} \\ &= \sqrt{\frac{2 \times 196\; {\rm N \cdot m}}{800\; {\rm N\cdot m^{-1}}}} \\ &= \sqrt{0.49\; {\rm m^{2}}} \\ &= 0.7\; {\rm m} \end{aligned}\).
In other words, the spring has been compressed by \(x = 0.7\; {\rm m}\) from the equilibrium.
Suppose the velocity of an object moving along a line is positive. are position, displacement, and distance traveled equal? explain. my math lab
Answers
No, the object's displacement and distance traveled will be equal; however, because the object's initial position is unknown, the object's position may differ from its displacement and distance traveled.
Because the initial position is not provided, we cannot assert that the displacement or distance equals the position. We might have arrived at a different conclusion if the starting point had been zero, because the distance from zero equals the position.
What precisely do you mean by position, displacement, and distance?
The location of an object (whether a person, a ball, or a particle) at a given point in time is referred to as its position. The displacement of an object is the difference in its position from one time to the next. Distance is the total amount the object has traveled in a certain period of time.
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a small object with a charge of magnitude q creates an electric field. at a point 1.0 m to the north of the charge the field has a value of 2.0 n/c directed south. at a point 0.5 m to the west of the charge the field has a value of
Answers
We have that for the Question,it can be said that the field has a value of
E_B=12N/C due westFrom the question we are told
a small object with a charge of magnitude q creates an electric field. at a point 1.0 m to the north of the charge the field has a value of 2.0 n/c directed south. at a point 0.5 m to the west of the charge the field has a value of
Electric fieldGenerally the equation for the electric field is mathematically given as
\(E=\frac{q}{4\piE_0r^2}\)
Therefore
\(if E_A=3N/C\\\\E_B=\frac{3}{0.25}\\\\\frac{q}{4\piE_0}=3N/C\\\\\)
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why does heat flow from a body at higher temperature to body at lower temperature?
Answers
Because, body at higher temperature has higher average KE of molecules and body at lower temperature has lower average KE of molecules. When two bodies having higher temperature and lower temperature are keep in contact, the body at higher temperature shares their energy to molecules at lower temperature.
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Using what you have learned about series and parallel circuits, explain how some appliances in your kitchen receive power while others do not. (PLEASE HELPP)
Answers
To learn series and parallel circuits, to know the some appliances in your kitchen receive power while others do not.
What is series circuits?
A series circuit is an easy way for electrons to travel to one or more resistors. Anything that depends on a cell for power is a resistor. A bulb frequently serves as the resistor in a series circuit.
What is parallel circuit?
Branching is used in a parallel circuit so that the current is divided and only a portion of it passes through each branch. In a parallel circuit, the voltage, or potential difference, across each branch is the same, but the currents may differ.
While plugged in, a range of various electronic equipment and appliances, such as televisions, toasters, lamps, and more, can use electricity even when they're off.
Therefore, known the series and parallel circuits, the some appliances receive powers.
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Experiment 1: Exploring Charge with Scotch® Tape
In this experiment, you will observe the behavior of charged objects using pieces of Scotch® tape.
Materials
Scotch® Tape
Ruler
*Pen
*Flat Work Surface
Procedure
Part 1
1. Use the ruler to measure a piece of tape that is 10 cm long.
2. Tear the tape to remove the 10 cm piece from the roll.
3. Create a "handle" on one side of the piece of tape by folding down the piece of tape 1 cm from the end, leaving a 9 cm sticky piece with a 1 cm handle.
4. Stick the entire sticky surface of the tape to a table top, counter top, or another flat surface.
5. Repeat Steps 1 – 4 with a second 10 cm piece of tape. Stick the second piece of tape at least 15 cm away from the first piece on the same surface.
6. Quickly pull off both strips of tape from the surface and ensure that the pieces do not touch.
7. Carefully bring the non-sticky sides of the tape together and record observations about the behavior of the pieces in Table 1.
8. Discard the tape.
Part 2
1. Use the ruler to measure a piece of tape that is 10 cm long.
2. Tear the tape to remove the 10 cm piece from the roll.
3. Create a "handle" on one side of the piece of tape by folding down 1 cm of tape from one end.
4. Stick the entire sticky surface of the tape to a table top, counter top, or another flat surface.
5. Use a pen and write "B1" on the tape. "B" stands for bottom.
6. Repeat Steps 1 – 4 with a second 10 cm piece of tape. This time, press the second strip of tape on top of the one labeled "B1".
7. Use the pen to label the top piece with a "T1". "T" stands for top.
8. Create a second pair of pieces of tape by repeating Steps 1 – 7. This time, label the bottom piece "B2" and the top piece "T2".
9. Use the T1 handle to quickly pull off T1 strip of tape from the flat surface.
10. Use the B1 handle to peel off the bottom strip from the flat surface. Keep both B1 and T1 pieces away from each other.
11. Bring the non-sticky sides of B1 and T1 together and record observations about the behavior of the pieces in Table 1.
12. Set the pieces of tape, non-sticky side down, on the table approximately 15 cm away from each other. Do not stick them back on the table!
13. Repeat Steps 9 - 12 for B2 and T2.
14. Carefully bring the non-sticky sides of piece "T1" and "B2". Record observations about the behavior of the pieces in Table 1.
15. Set them back down, non-sticky side down.
16. Repeat Steps 14 - 15 for "T1" and "T2". Record your observations in Table 1.
17. Repeat Steps 14 - 15 for "B1" and "B2". Record your observations in Table 1.
18. Repeat Steps 14 and 15 for "T1" and the hair on your leg or arm. Record your observations in Table 1.
19. Repeat Steps 14 and 15 for "B1" and the hair on your leg or arm. Record your observations in Table 1.
Table 1: Electric Charge Observations
procedure
interacting pieces observation
Part 1 Two pieces on table Part 2 T1 / B1 T2 / B2 T1 / B2 T2 / B1 B1 / B2 T1 / Arm Hair B1 / Arm Hair ***The observation is filled.
Post-Lab Questions
1. Describe the interaction between the top and bottom strips as they relate to electric charge. Did the behavior of the pieces change when the tape was from different sets?
2. Describe the interaction between two top and two bottom pieces of tape as they relate to electric charge. Is this consistent with the existence of only two types of charge? Use your results to support your answer.
3. Did the top tape attract your arm hair? Did the bottom tape attract your arm hair? Usually arm hair is neutral; it has equal number positive and negative charges. Use this information to explain your results.
4. Which pieces of tape are positively charged? Which pieces of tape are negatively charged? Explain your reasoning.
5. Use your data to create a rule describing how like charges, opposite charges, and neutral bodies interact.
6. What do you observe about the force of attraction or repulsion when the pieces of tape are closer together and farther apart? Does this change happen gradually or quickly?
Answers
1.When the non-sticky sides of the two pieces of tape recording are brought together, they repel each other. This is due to the buildup of electric charge on the face of the tape recording when it was hulled off from the flat face.
2.The pieces didn't change when the tape recording was from different sets. When two top or two nethermost pieces of tape recording are brought together, they repel each other.
3.When a top and nethermost piece of tape recording are brought together, they attract each other. This is harmonious with the actuality of only two types of charge, positive and negative. The results support the fact that the top and nethermost pieces of tape recording had contrary charges. The top tape recording attracted the arm hair, while the bottom tape recording didn't attract the arm hair. Arm hair is generally neutral, but it can be concentrated by the electric field of the charged tape recording.
4.The top tape recording is negatively charged, and it concentrated the arm hair, which has a positive charge. This redounded in magnet between the top tape recording and the arm hair. The pieces of tape recording labeled" T1" and" B2" are appreciatively charged, while the pieces of tape recording labeled" B1" and" T2" are negatively charged. This can be determined from the compliances.
5.When the appreciatively charged tape recording was brought near to a negatively charged tape recording, they attracted each other. When two appreciatively charged videotapes or two negatively charged videotapes were brought near together, they repelled each other. Like charges repel each other, contrary charges attract each other, and neutral bodies aren't affected by electric fields.
6.The force of magnet or aversion between the pieces of tape recording increases as they get near together and decreases as they move further piecemeal. This change happens gradationally, not snappily.
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we investigated a jet landing on an aircraft carrier. in a later maneuver, the jet comes in for a landing on solid ground with a speed of 87 m/s, and its acceleration can have a maximum magnitude of 8.55 m/s2 as it comes to rest. (a) from the instant the jet touches the runway, what is the minimum time interval needed before it can come to rest? (b) can this jet land on a small tropical island airport where the runway is 0.800 km long? (c) explain your answer.
Answers
For a successful landing the aircraft will need:
(a) A time from the instant the jet touches the runway of: 10.1754 s(b) A distance of: 0.44263 km(c) Due to the jet need just 0.44263 km of distance to landing it can land on a small tropical island airport where the runway is 0.800 km longThe formula for uniformly varied rectilinear motion (UVRM) and procedure we will use to solve this exercise is:
vf = vi + (a * t)x = (vf²-vi²)/ (2a)Where:
x = distancevf = final velocityvi = initial velocitya = accelerationInformation about the problem:
vf= 0 m/svi = 87 m/sa = -8.55 m/s²t =?x= ?Applying the final velocity formula, and clearing the time we get:
vf = vi + (a * t)
t = (vf – vi) /a
t = (0 m/s – 87 m/s) /-8.55 m/s²
t = -87 m/s/ -8.55 m/s²
t = 10.1754 s
Applying the distance formula, we get:
x = (vf²-vi²)/ (2a)
x = [(0 m/s)²- (87 m/s)²]/ (2 * -8.55 m/s²)
x = (0 m²/s²- 7569 m²/s²)/ -17.1 m/s²
x = -7569 m²/s²/ -17.1 m/s²
x = 442.63 m
By converting the distance units form (m) to (km) we have
x = 442.63 m * 1 km/1000 m
x = 0.44263 km
What is acceleration?
It is a physical quantity that indicates the variation of velocity as a function of time, it is expressed in units of distance per time squared e.g.: m/sec2 ; km/h2
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A car is travelling along a road when the driver spots a hedgehog crossing ahead. The driver applies a braking force of 2500 N and manages to stop the car after 7.5 metres to avoid the hedgehog. Calculate the work done by the brakes to stop the car.
Answers
The work done by the brakes to stop the car will be 56.25 kN-m.
What is the work done?If the amount of force is applied to the body and the body gets displaced from its original place. Then it is said to work is done on the system. The work done is given as,
WD = Force x Displacement
A vehicle is going along a street when the driver detects a hedgehog crossing ahead. The driver applies a slowing down the power of 2500 N and figures out how to stop the vehicle after 7.5 meters to stay away from the hedgehog.
The work done is calculated as,
WD = 7500 x 7.5
WD = 56,250 N-m
WD = 56.25 kN-m
The work done by the brakes to stop the car will be 56.25 kN-m.
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Two resistors connected in series (R1, R2) are connected to two resistors that are connected in parallel (R3,R4). The series-parallel combination is connected to a battery. Each resistor has a resistance of 10.00 Ohms. The wires connecting the resistors and battery have negligible resistance. A current of 2.00 A runs through resistor R1. What is the voltage supplied by the voltage source?
Answers
Answer:
50volts
Explanation:
Each of the resistance is 10ohms. 2 in series and 2 in parallel. it means the 2 parallel can combines with total of 10×10÷(10+10)= 5ohms and combine in series with 10ohms which is 10+10+5= 25ohms. in series connection, we have same current flowing. therefore the voltage source will be 25ohms × 2amps = 50volts
If 3.5 waves have a distance of 42M what is the wavelength of one wave?
Please please please tell me how you exactly found the answer!!
Answers
Answer:
The answer is 12
Explanation:
Just divide 42/3.5
a car is being lifted by a crane so that it moves upwards at a constant speed. which force is doing negative work on the car as it is being lifted? group of answer choices
Answers
The correct answer is option B.
A car is being lifted by a crane so that it moves upwards at a constant speed. Force of gravity is doing negative work on the car as it is being lifted upward.
In this case, we can see a certain amount of work is being done by the crane to uplift the car upward with constant speed.
The car also covers a displacement in upward direction.
In Physics, Work is defined as the distance covered by an object under a particular force. Work is the product of force and displacement i.e. w = F.d
However, as the car is moving in upward direction, so the displacement is upward while the force of gravity due to displacement is also acting on the car which is opposite the direction of displacement i.e. downward. Therefore, gravity is doing negative work on the car.
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The complete question is:
A car is being lifted by a crane so that it moves upwards at a constant speed. which force is doing negative work on the car as it is being lifted?
a) The force of rope acting on the car
b) The force of gravity acting on the car due to earth
c) The force of car acting on the rope
d) There is no work being done
A spring with a force constant of 140 N/m is used to push a 0.26-kg block of wood against a wall, as shown in the figure
A: Find the minimum compression of the spring needed to keep the block from falling, given that the coefficient of static friction between the block and the wall is 0.40.
B: Does your answer to part A change if the mass of the block of wood is doubled?
Answers
The minimum compression of the spring needed to keep the block from falling is 0.0074 m and the minimum compression of the spring needed to keep the block from falling will increase if the mass of the block of wood is doubled.
A)The minimum compression of the spring needed to keep the block from falling is given by the expression,kΔx = mgsinθ + μmgcosθ
Here,m = 0.26 kg = 9.8 ms-2θ = 90° (since the block is pushed against a vertical wall)μ = 0.4 (coefficient of static friction between the block and the wall)k = 140 N/m
Therefore,140 Δx = 0.26 × 9.8 sin 90° + 0.4 × 0.26 × 9.8 cos 90°⇒ Δx = (0.26 × 9.8 × 0.4)/140⇒ Δx = 0.0074 m
Therefore, the minimum compression of the spring needed to keep the block from falling is 0.0074 m.
B) Doubling the mass of the block of wood will affect the minimum compression of the spring needed to keep it from falling.
Since the gravitational force acting on the block increases when the mass is doubled, the spring will need to be compressed more to balance the weight. Therefore, the minimum compression of the spring needed to keep the block from falling will increase if the mass of the block of wood is doubled.
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PLEASE ANSWER CLEARLY :)
A composite material is to be made from type E-glass fibers
embedded in a matrix of ABS plastic, with all fibers to be aligned
in the same direction. For the composite, the el
A composite material is to be made from type E-glass fibers embedded in a matrix of ABS plastic, with all fibers to be aligned in the same direction. For the composite, the elastic modulus parallel to
Answers
The elastic modulus parallel to the fibers of a composite material made of type E-glass fibers embedded in a matrix of ABS plastic, with all fibers to be aligned in the same direction can be calculated as follows:First, we need to calculate the elastic modulus of each component of the composite material.
The elastic modulus of type E-glass fibers is 72 GPa, and the elastic modulus of ABS plastic is 2.5 GPa.Next, we need to calculate the volume fraction of each component. If we assume that the composite material is made up of 60% type E-glass fibers and 40% ABS plastic, then the volume fraction of type E-glass fibers is 0.6, and the volume fraction of ABS plastic is 0.4.
Finally, we can use the rule of mixtures to calculate the elastic modulus parallel to the fibers. The rule of mixtures states that the elastic modulus of a composite material is equal to the weighted average of the elastic moduli of the individual components, where the weights are the volume fractions.
Therefore, the elastic modulus parallel to the fibers is given by:
Elastic modulus parallel to fibers = (Volume fraction of type E-glass fibers x Elastic modulus of type E-glass fibers) + (Volume fraction of ABS plastic x Elastic modulus of ABS plastic)
Elastic modulus parallel to fibers = (0.6 x 72 GPa) + (0.4 x 2.5 GPa)
Elastic modulus parallel to fibers = 43.5 GPaSo, the elastic modulus parallel to the fibers of the composite material is 43.5 GPa.
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Rita is a salon owner. She notices that her salon charged one of her clients, Linda, extra for a service that the clent did not request. What do you think Rita should do? Α. Remaln qulet about the extra money. B. Distribute the money equally among the staff. C. Call the client and Inform her that she was incorrectly charged. D. Try overcharging the next client too and check if it goes unnoticed.
Answers
Answer:
C hope it helps
call the client and inform her that she was incorrectly charged.
Answer:
The answer is C. call the client and inform her that she was incorrectky charged
Explanation:
What is the electric force between a +3 µC point charge and a –3 µC point charge if they are separated by a distance of 5.0 cm? Show your work. (µC = 1.0 × 10–6 C)
Answers
The electric force between a +3 µC point charge and a –3 µC point charge if they are separated by a distance of 5.0 cm can be calculated by using Coulomb’s law. F = kq1q2 / r²F = (9.0 × 109 N·m2/C2) × (3.0 × 10-6 C) × (-3.0 × 10-6 C) / (0.05 m)²F = -243 N
The electric force between a +3 µC point charge and a –3 µC point charge if they are separated by a distance of 5.0 cm can be calculated by using Coulomb’s law.
Coulomb’s law states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Mathematically, F = kq1q2 / r²where, F is the electric forceq1 and q2 are the magnitudes of the two charges is Coulomb’s constant (9.0 × 109 N·m2/C2)r is the separation between the two charges.
Let's use the values from the question to calculate the electric force: F = kq1q2 / r²F = (9.0 × 109 N·m2/C2) × (3.0 × 10-6 C) × (-3.0 × 10-6 C) / (0.05 m)²F = -243 N (the negative sign indicates that the force is attractive)Therefore, the electric force between the two charges is 243 N.
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what is the relationship between vapour pressure of the liquid and temperature
Answers
Answer:
As temperature increases, vapor pressure increases. As temperature decreases, vapor pressure decreases.
Explanation:
Hot molecules move fast, cold molecules move slow.
when astronomers study the light emitted from the sun they notice that the light emitted from the east limb (edge) is blueshifted while the light emitted from the west limb is redshifted. the reason for this is that
Answers
This phenomenon occurs due to the Doppler effect, which results from the Sun's rotation.
The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the source of the wave.
In the case of the Sun, it rotates from west to east, causing the light emitted from the east limb to move towards the observer, leading to blueshift.
Conversely, the light emitted from the west limb moves away from the observer, causing redshift.
Summary: The blueshift observed in the light from the east limb of the Sun and the redshift in the light from the west limb are caused by the Doppler effect due to the Sun's rotation from west to east.
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2) A 30-N box is pulled upward 6.0 m along the surface of a ramp that rises at 37° above the horizontal. How
much work does gravity do on the box during this process?
A) -140 J B) -110 J C) -1100 J D) -180 J E) 120 J
Answers
Since the box is pulled upward, work done by gravity is negative, so the answer is (A) -140 J.
What is meant by gravity?Force by which any planet or other body draws objects toward its center is called as gravity.
As ΔPE = mgh
ΔPE is the change in potential energy, m is mass of the box, g is acceleration due to gravity, and h is change in height.
Given : h = 6.0 m sin 37° = 3.6 m
Given, 37° is the angle of the ramp.
As F = ma
F is the force applied to the box (30 N) and a is the acceleration of the box along the ramp. The component of the gravitational force parallel to the ramp is : Fg|| = mg sin θ
θ is the angle of ramp, and is equal to 37° in this case. The force of friction opposing the motion is : Ff = μFg⊥
μ : coefficient of kinetic friction between box and ramp, and Fg⊥ : component of gravitational force perpendicular to ramp : Fg⊥ = mg cos θ
Fnet = Fg|| - Ff = mg sin θ - μmg cos θ
and we know, ma = mg sin θ - μmg cos θ
so, m = a / (g sin θ - μg cos θ)
m = 30 N / [(9.81 ) sin 37° - (0.4)(9.81 cos 37°] = 2.32 kg
ΔPE = mgh = (2.32)(9.81 )(3.6 ) = 81.5 J
Since the box is pulled upward, the work done by gravity is negative, so the answer is (A) -140 J.
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If a light ray is traveling from air into a glass prism, how will the direction of the light ray change after it enters the prism? A. The light ray will be refracted toward the normal. B. The light ray will be refracted away from the normal. C. The light ray will be refracted perpendicular to the normal. D. The light ray will be refracted along the normal. E. The light ray will not be refracted.
Answers
the light ray will be refracted towards the normal.
Choose the correct meaning of the underlined word: his eyes had that mischievious glint they sometimes got when he pulled a pop quiz and made all the multiple choice answers b. a enough; an adequate amount b a small flash, sparkle c in a shy way d in a sneaky way
Answers
Based on the context provided, the correct meaning of the underlined word "mischievous" is d) in a sneaky way. Hence, option d) is the correct answer.
This is because the sentence describes the person's eyes having a "glint" when he pulls a pop quiz and makes all the multiple choice answers, which suggests that he may be doing it in a playful or mischievous manner. The word "mischievous" is often used to describe behavior that is playful or teasing, but can also be seen as sneaky or sly.
The correct meaning of the underlined word "mischievous" in the given sentence is: d. in a sneaky way.
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a light aluminum ball and a heavy lead ball of the same size are allowed to roll down an incline. when they are halfway down the incline, they will have identical
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In the context of momentum, leftward momentum should equal rightward momentum. This is chapter 1.2. That ought to be 1.8 kg into two meters per second. Finally, this we equals 1.8 when two is divided by 1.2, which is 3.0 meters per second.
There are four different challenges in the example given. The first one refers to the initial ball's mass. This is assumed assuming that each person is equivalent to 1.5 telegrams and the second pool ball's mass. M2 is that This also equates to an initial speed of 2.3 kg.
The first pool ball is moving at 25 meters per second, whereas the second pool ball was initially at rest. The first ball ultimately comes to rest after colliding with the second pool ball, becoming zero, and we must determine the second ball's final speed. To do this, we will apply the principle of conservation of linear momentum, which states that the linear momentum of both balls remains constant.
The balls before to impact indicate M 1. We have a total of two from one plus M. That corresponds to the ultimate momentum and one we F plus M. This is 1.5 kg in 2 25 m per second plus zero, and we need to plug in all the known variables. For the first ball after coalition because it comes to rest Plus 2.3 kg Into VF two, it will be zero for the second ball is equal to here.
Lasting Speed of Second Pool Ball. Therefore, it works out to be 2.3 kg into VF 2. And canceling this kilogram here, it is equivalent to 37.5 kilogram meters per second.
The answer to the first problem supplied to him now in the second problem, problem number two, mass of the bullet Small, is determined to be equivalent to 16.3 meters per second for the second pool ball. The bullet's speed is five kilograms per second, and the block in which it is implanted has a mass of 0.005 kilograms.
The bullet designated as Small B has an initial speed of 500 m/s and an overall speed of both. The block and bullet that are absent. We need to locate it.
Since the blocks were released earlier and the spring was also handled, their initial momentum was zero. Eventually, they began to move in the other direction. hence where momentum is concerned, leftward momentum should equal rightward momentum. This is chapter 1.2. That ought to be 1.8 kg into two meters per second. Finally, this we equals 1.8 till two divided by 1.2 results in a speed of 3.0 meters per second.
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If the sun exploded would the solar system collaspe?
Answers
Answer:
Yes, yes it would since we need light
Explanation:
Calculate the frequency of a vibrating body if it vibrates 400 times in 20 seconds. Identify the category of sound it belongs to.(sonic sound/ ultrasound/ infra sound)
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The category of sound is a sonic sound
Answer:
F+20Hz
Explanation:
One strategy that has been implemented to address the drug epidemic in Philadelphia was the creation of something called
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The One strategy implemented to address the drug epidemic in Philadelphia is the creation of Comprehensive User Engagement Sites (CUES). These sites aim to tackle the widespread issue of drug addiction and related energy health concerns.
The CUES are safe spaces where individuals battling addiction can energy access various harm reduction services, such as clean syringes, medical support, and overdose prevention. These sites provide connections to addiction treatment programs and mental health services, helping people on their path to recovery. By offering safe and supervised spaces, CUES work to reduce public drug use, discarded syringes, and other related issues in the community. CUES also serve as educational hubs, raising awareness and providing information about the dangers of drug addiction and available resources for support. Lastly, these sites foster community engagement and collaboration, uniting various stakeholders in the fight against the drug epidemic. In summary, Comprehensive User Engagement Sites play a significant role in addressing the drug epidemic in Philadelphia. They provide harm reduction services, treatment programs, and community support, all while promoting a safer and healthier environment for the city's residents.
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Which of these symbols is for an element?
A. Au
B. Naci
C. КСІ
D. H20
Please answer this ASAP
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27. Julie and Ashwin were doing an experiment in the classroom. They
each had to move a 15 kg box. Julie was able to move her box, but
Ashwin was only able to move his after removing half the items in the
box. Which is true regarding the force he applies after removing the
items from the box?
a
He has to apply more force to lift the box
because the mass is greater.
b
He has to apply less force to lift the box
because the mass has decreased.
С
He has to apply a different type of force after
changing the mass.
d
He has to apply the same amount of force to
the box to overcome gravity.
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Answer:
b. He has to apply less force to lift the box because the mass has decreased.
Explanation: i just took a ck12 test and it was right lol
What is the conversion of how many kilograms are in 48 ounces
Answers
Answer:1.36078
Explanation: