Given:
The mass of the satellite is,
\(\begin{gathered} m=900000\text{ kg} \\ \end{gathered}\)The mass of the earth is,
\(\begin{gathered} M=5.98\times10^{24}\text{ kg} \\ \end{gathered}\)The radius of the earth is,
\(\begin{gathered} R=6.38\times10^6\text{ m} \\ \end{gathered}\)The altitude of the satellite is,
\(\begin{gathered} h=4000\text{ km} \\ =4000\times10^3\text{ km} \end{gathered}\)To find:
The time period of the satellite
Explanation:
The orbital speed of the satellite is,
\(\begin{gathered} v=\sqrt{\frac{GM}{R+h}} \\ Here,\text{ G=6.67}\times10^{-11}\text{ N.m}^2.kg^{-2} \end{gathered}\)The orbital speed is,
\(\begin{gathered} v=\sqrt{\frac{6.67\times10^{-11}\times5.98\times10^{24}}{6.38\times10^6+4000\times10^3}} \\ =6.199\times10^3\text{ m/s} \end{gathered}\)The period is,
\(\begin{gathered} T=\frac{2\pi(R+h)}{v} \\ =\frac{2\pi(6.38\times10^6+4000\times10^3)}{6.199\times10^3} \\ \approx10521\text{ s} \end{gathered}\)The time period is 10521 s.
7. A particle of mass 3 kg is held in equilibrium by two light unextensible strings. One string is horizontal, as shown in Figure 7.30. The tension in the horizontal string is PN and the tension in the other string is N. Find a) the value of 0 b) the value of P.
The tension in the strings are 31.47 and 19.25 N respectively.
Mass of the block, m = 3 kg
From the figure, consider the vertical components,
T₁ sin45° + T₂ sin30° = mg
(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4
Also, consider the horizontal components,
T₁ cos45° = T₂ cos30°
T₁/√2 = T₂ x√3/2
T₁ = T₂ x √3/2 x √2
So,
T₁ = 0.612T₂
Applying in the first equation,
(T₁/√2) + (T₂/2) = 29.4
(0.612T₂/1.414) + 0.5T₂ = 29.4
0.434 T₂ + 0.5 T₂ = 29.4
0.934 T₂ = 29.4
Therefore, the tension,
T₂ = 29.4/0.934
T₂ = 31.47 N
So, the tension,
T₁ = 0.612 T₂
T₁ = 0.612 x 31.47
T₁ = 19.25 N
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the radiation that is least damaging to humans is
The radiation that is least damaging to humans is non-ionizing radiation.
What is non ionizing radiationNon-ionizing radiation refers to the type of radiation that does not have enough energy to remove tightly bound electrons from atoms or molecules, thus not causing significant damage to biological tissues.
Examples of non ionizing radiation include radio waves, microwaves, visible light and low energy ultraviolet (UV) radiation. while excessive exposure to any form of radiation can have adverse effects, non-ionizing radiation is generally considered to be less harmful compared to ionizing radiation, which includes X-rays and gamma rays.
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A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total energy.
Answer:
a = ω^2 A formula for max acceleration (ignoring sign)
V = ω A formula for max velocity
V^2 = ω^2 A^2 = a A from first equation
E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J
(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule
A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diameter of 1.00 mm. The wire length is
2.00 m and copper has a number density of 8.50 × 1028 m–3.
Find the change in mean drift velocity for electrons as they pass from one end of the wire to the other and therefore calculate the average acceleration of the electrons.
The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
The given parameters;
Current flowing in the wire, I = 4.00 mAInitial diameter of the wire, d₁ = 4 mm = 0.004 mFinal diameter of the wire, d₂ = 1 mm = 0.001 mLength of wire, L = 2.00 mDensity of electron in the copper, n = 8.5 x 10²⁸ /m³The initial area of the copper wire;
\(A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2\)
The final area of the copper wire;
\(A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2\)
The initial drift velocity of the electrons is calculated as;
\(v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s\)
The final drift velocity of the electrons is calculated as;
\(v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s\)
The change in the mean drift velocity is calculated as;
\(\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s\)
The time of motion of electrons for the initial wire diameter is calculated as;
\(t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s\)
The time of motion of electrons for the final wire diameter is calculated as;
\(t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s\)
The average acceleration of the electrons is calculated as;
\(a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2\)
Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
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Answer:
The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
The given parameters;
Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³
The initial area of the copper wire;
The final area of the copper wire;
The initial drift velocity of the electrons is calculated as;
The final drift velocity of the electrons is calculated as;
The change in the mean drift velocity is calculated as;
The time of motion of electrons for the initial wire diameter is calculated as;
The time of motion of electrons for the final wire diameter is calculated as;
The average acceleration of the electrons is calculated as;
Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².
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Explanation:
Two blocks, 1 and 2, are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105
. The figure illustrates the configuration.
A system of two blocks connected by a rope passing over a pulley. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle. Box M subscript 1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M sub 1 and the surface is mu subscript 1. Box M subscript 2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle. The coefficient of friction between box M sub 2 and the surface is mu subscript 2.
The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.
The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.
The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.
The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.
The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).
The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).
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Some help me with this pls
37. Segment B of the graph corresponds to the only solid state. ,38. Segment E of the graph corresponds to the melting point. and all answer given below:
What is Segment?Segment is a customer data platform (CDP) that enables businesses to collect, store, and analyze customer data from multiple sources. It allows companies to unify their customer data into one place, which can then be used to create more personalized customer experiences. With Segment, businesses can create custom segments of their customers and create targeted campaigns to meet their specific needs.
39. The melting point (temperature) is 300°C.
40. Segment F of the graph corresponds to the only liquid state.
41. Segment A of the graph corresponds to the only gas state.
42. Segment D of the graph corresponds to the boiling point.
43. No, the temperature decreases during freezing.
44. Segment C of the graph corresponds to the freezing point.
45. The boiling point (temperature) is 450°C.
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what is a likely consequence of continued human population growth
Answer:
Rapid human population growth has a number of consequences. Population grows fastest in the world's poorest countries. High fertility rates have historically had a strong relation with poverty, and high childhood mortality rates
Explanation:
i got it from goo.gle but i hope that helps:)
have a good day bro
what is the similaries between parallel and series circuits
Answer:
same power source
can have more than 1 light bulb in a circuit
requires a source of energy
Explanation:
Match the following activities.
1.foot-pound
2.work
3.steam
4.newton meter
5.muscular force
6.force of gravity
7.force of wind
1.amount of force times the distance it moves an object
2. kind of force used to operate engines
3.English unit of measuring work equal to moving one pound one foot
4.metric measurement of work equal to moving one newton
one meter
5. kind of force that pulls things toward the center of the earth
6. force used when walking
7. force used to turn a windmill
Answer:
amount of force times the distance it moves an object: work
kind of force used to operate engines: steam
English unit of measuring work equal to moving one pound one foot: foot-pound
metric measurement of work equal to moving one Newton one meter: Newton-meter
kind of force that pulls things toward the center of the earth: force of gravity
force used when walking: muscular force
force used to turn a windmill: force of wind
Explanation:
I found the answers in quizlet
Answer:
im in the same class so here's your answer
Explanation:
5. [6.67/10 Points] DETAILS
MY NOTES
ASK YOUR TEACHER
N/C
(a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the
magnitude of the electric field only.)
6.00 μC
1.50 μC -2.00 μC
3.00 cm
Submit Answer
PREVIOUS ANSWERS
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Need Help? Read It
(b) If a charge of -6.13 µC is placed at this point, what are the magnitude and direction of the force on it?
magnitude
N
direction
SERCP11 15.3.P.018. 2/5 Submissions Used
2.00 cm →
As the charge is negative, the force, which has a magnitude of 33.05 N, is directed to the left, against the electric field.
What is the electric field's intensity at a specific distance from the point charge E?E is a symbol for the magnitude of the electric field at a specific distance from a point charge. At twice the distance from the point charge, what is the electric field's strength? The field's strength is E/2 at twice the distance. The field's strength is still equal to E at a distance that is twice as great.
E = k*q/r²
r1 = 2.00 cm
r2 = 1.00 cm + 3.00 cm = 4.00 cm
r3 = 1.00 cm
Using these distances, we can calculate the electric field due to each charge:
E1 = kq1/r1² = (9.0 x 10⁹ Nm²/C²) * (1.50 x 10⁻⁶ C) / (0.02 m)² = 168.75 N/C (to the right)
E2 = kq2/r2² = (9.0 x 10⁹ Nm²/C²) * (-2.00 x 10^⁻⁶ C) / (0.04 m)² = -112.50 N/C (to the left)
E3 = kq3/r3² = (9.0 x 10⁹ Nm²/C²) * (6.00 x 10⁻⁶ C) / (0.01 m)² = 5.40 x 10⁶ N/C (to the right)
E = E1 + E2 + E3 = 168.75 N/C - 112.50 N/C + 5.40 x 10⁶ N/C = 5.39 x 10⁶ N/C (to the right)
F = q*E
F = (-6.13 x 10 C) * (5.39 x 10⁶ N/C) = -33.05 N
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14. How are an object's speed and velocity related to an object's kinetic energy and total mechanical energy
Speed and velocity are related to an object's kinetic energy, which is the energy associated with its motion.
What is kinetic energy?Kinetic energy is the energy that an object has due to its motion. It is a form of energy that is associated with the motion of an object, and is calculated by multiplying the mass of the object by the square of its velocity. Kinetic energy is one of the most fundamental forms of energy, and is found in everything from the movement of atoms and molecules to the motion of planets and stars. Kinetic energy is also the energy that is associated with everyday objects, such as a ball that is thrown or a car that is moving. Kinetic energy can be converted into other forms of energy, such as heat or electricity, and can also be used to do work, such as powering a car or a machine. Kinetic energy is a crucial part of the natural world, and is necessary for many of the processes that occur in nature.
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Distinguish between physical and chemical changes. Include examples in your explanations.
Answer:
Chemical changes are recognized when a substance changes its properties permanently and it cannot be the same substance as before.
Instead the physical changes implies that if you can return to the same substance through a reverse process.
Explanation:
A chemical change is, by example, a combustion, if a sheet of paper burns, its result is ashes, the ashes cannot go back to being a sheet of paper because its properties changed, heat energy was generated that changed matter permanently.
A physical change, by example, is that of freezing water, the water becomes ice, but this can easily become water again if the temperature is increased, its properties do not change and the chemistry of the substance does not change.
A chemical change takes place when a chemical reaction takes place, while when a matter changes forms but not the chemical identity then a physical change takes place.
• A product or a new compound formation takes place from a chemical change as the rearrangement of atoms takes place to produce novel chemical bonds.
• In a chemical change always a chemical reaction takes place.
• Some of the chemical changes examples are souring milk, burning wood, digesting food, mixing acid and base, cooking food, etc.
• In a physical change no new chemical species form.
• The changing of the state of a pure substance between liquid, gas, or solid is a physical change as there is no change in the identity of the matter.
• Some of the physical changes are melting of ice, tempering of steel, breaking a bottle, crumpling a sheet of aluminum foil, boiling water, and shredding paper.
Thus, a new substance is formed during a chemical change, while a physical change does not give rise to a new substance.
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Which of these is NOT an inherited trait of the plant?
Explanation:
I think you forgot to add the other part!
Answer:
uhm what plant...
Explanation:
WRITE PLEASE: What distinguishes the 6 kingdoms from each other, according to taxonomic system described in this unit? Be sure to be specific and name each of the kingdoms in your description of their traits. ill mark brainliest
Answer:
:P
Explanation:
the six kingdoms of life, as described in the taxonomic system, are distinct from one another in a variety of ways. each kingdom has its own unique characteristics that set it apart from the others.
the first kingdom is the kingdom animalia, which is composed of multicellular organisms that are heterotrophic and motile. animals are capable of movement and have specialized organs and tissues that allow them to interact with their environment.
the second kingdom is the kingdom plantae, which is composed of multicellular organisms that are autotrophic and sessile. plants are capable of photosynthesis and have specialized organs and tissues that allow them to interact with their environment.
the third kingdom is the kingdom fungi, which is composed of multicellular organisms that are heterotrophic and sessile. fungi are capable of absorbing nutrients from their environment and have specialized organs and tissues that allow them to interact with their environment.
the fourth kingdom is the kingdom protista, which is composed of unicellular organisms that are either autotrophic or heterotrophic and motile. protists are capable of movement and have specialized organelles that allow them to interact with their environment.
the fifth kingdom is the kingdom monera, which is composed of unicellular organisms that are autotrophic and motile. monerans are capable of movement and have specialized organelles that allow them to interact with their environment.
the sixth kingdom is the kingdom archaea, which is composed of unicellular organisms that are autotrophic and motile. archaeans are capable of movement and have specialized organelles that allow them to interact with their environment.
in summary, the six kingdoms of life are distinct from one another in a variety of ways. each kingdom has its own unique characteristics that set it apart from the others. animals are multicellular and heterotrophic, plants are multicellular and autotrophic, fungi are multicellular and heterotrophic, protists are unicellular and either autotrophic or heterotrophic, monerans are unicellular and autotrophic, and archaeans are unicellular and autotrophic.
you better mark me brainliest
What formulas can you use to solve for velocity, given displacement and time?
Answer:
velocity=displacement/time
Explanation:
A frictionless toy car is placed on a ramp, which is inclined at an unknown angle with respect to the horizontal. Starting from rest, the car is allowed to roll freely down the ramp. After a time interval of Δ=0.67 s, the car has traveled a distance of =1.2 m down the ramp and is moving with speed f. Find the speed of the car at the end of the 0.67 s time interval.
The final speed of the toy car at the end of the given time period is 3.58 m/s.
The given parameters;
distance traveled by the car, s = 1.2 mtime of motion of the car, t = 0.67 sinitial velocity of the car, u = 0The acceleration of the car is calculated as;
\(s = ut + \frac{1}{2} at^2\\\\1.2 = 0 + 0.5\times a\times (0.67)^2\\\\1.2 = 0.225a\\\\a = \frac{1.2}{0.225} \\\\a = 5.33 \ m/s^2\)
The final velocity of the toy car is calculated as;
\(v_f^2 = u^2 + 2as\\\\v_f^2 = 0 + 2\times 5.33 \times 1.2\\\\v_f^2 = 12.792\\\\v_f = \sqrt{12.792} \\\\v_f = 3.58 \ m/s\)
Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.
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3.) Brady walks 5 meters forward, turns right, walks 25 meters, turns right again, walks 5
meters, then turns left and walks 10 meters. What are Brady's distance and
displacement?
Answer:
Distance covered is: 45 meters
Displacement is 15 meters to the right of where he started
Explanation:
Notice that Brady has walk a path that looks like an incomplete rectangle of height 5 meters and length 25meters, although he actually didn't cover the full length (25 meters) when getting back to the point where he started (he made just 10 meters instead of 25 after the third turn right) See attached image.
Therefore, Brady's displacement is 15 meters to the right of where he started, and the total distance he covered is :
Distance = 5m + 25m + 5m + 10m = 45m
Number 1a and b and number 2(everything)
Answer:
Explanation:
The density of a substance is the ratio of its mass to its volume. Its unit of measurement is kg\(m^{-3}\).
i.e density, ρ = \(\frac{mass}{volume}\)
1a. To determine the density of the acetic acid, Rachael needs to know the mass and volume of the acid.
i. Measure the mass of the given beaker using the mass balance.
ii. Transfer the acetic acid into the beaker, and measure the new mass using the mass balance.
iii. Subtract the mass of the beaker from the new mass to determine the mass of the acetic acid.
iv. Measure the volume of the acid on the scale of the beaker.
v. Divide the value of the mass by its volume to determine its density of the acetic acid.
b. Given that the density is 1.05 g/\(cm^{3}\), and volume is 200 \(cm^{3}\).
Then,
density = \(\frac{mass}{volume}\)
1.05 = \(\frac{mass}{200}\)
mass = 1.05 x 200
= 210 g
mass = 210 g
2. Length of titanium = 0.40 m
Area of titanium = 0.05 \(m^{2}\)
mass = 90.0 kg
density = \(\frac{mass}{volume}\)
But,
volume = area x length
= 0.05 x 0.4
= 0.02 \(m^{3}\)
density of titanium = \(\frac{90}{0.02}\)
= 4500 kg\(m^{-3}\)
An unstretched ideal spring hangs vertically from a fixed support. A 0.4 kg object is then attached to the lower end of the spring. The object is pulled down to a distance of 0.35 m below the unstretched position and released from rest at time t= 0. A graph of the subsequent vertical position y of the lower end of the spring as a function of t is given above, where y= 0 when the spring was initially unstretched. At which time is the upward velocity of the object the greatest?
Answer:
The correct answer will be "0.25 sec".
Explanation:
The graph of the given question is attached below.
According to the graph of the question,
Time,
T = 1 sec
For the upward velocity,
⇒ \(t = \frac{T}{4}\)
By putting the value, we get
⇒ \(=\frac{1}{4}\)
⇒ \(=0.25 \ sec\)
hii Happy New year can you all please answer this
Answer:
truuuuuuuuuuuuueeeeeeeee
Three cars are driving at 29.0 m/s along the road shown in the figure (Figure 1). Car B is at the bottom of the hill and car C is at the top. Suppose each car suddenly brakes hard and starts to skid. Assume μk=1.00. What is the x-component of the tangential acceleration (i.e., the acceleration parallel to the road) of car A?
Since the car masses are unknown, we are unable to calculate the numerical value of the x-component of Car A's tangential acceleration.
How do kinetic and potential energy compare?The energy that is held in any object or system as a function of its position or component arrangement is known as potential energy. The object or system is unaffected by external factors like air pressure or altitude. Kinetic energy, on the other hand, describes the power of moving particles within a system or an object.
They are being affected by the kinetic frictional force, which is caused by:
f = μk * N
Therefore,
fB = μk * N = μk * mB * g
Car C is at its highest point at the top of the hill, where the normal force acting on it is equal to the force of gravity. Therefore,
fC = μk * N = μk * mC * g
where mC is the mass of Car C.
For Car A, the x-component of the tangential acceleration is given by:
aA = (fB - fC) / mA
where mA is the mass of Car A.
We can substitute the following values and simplify by assuming that the mass of each of the three automobiles is the same:
aA = (μk * mB * g - μk * mC * g) / mA
aA = μk * g * (mB - mC) / mA
Since μk = 1.00 and g = 9.81 m/s², we can plug in the values and get:
aA = (mB - mC) * 9.81 / mA
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An object is traveling with a constant speed of 35 m/s. How long does it take to cover 770 m?
Answer:
22 s
Explanation:
time = distance / velocity
We know that distance = 770 m and velocity = 35 m/s.
t = d / v
t = 770 m / 35 m/s
t = 22 s
It takes 22 seconds to cover 770 m. Hope this helps, thank you !!
Two bodies separated from
each other at a certain distance
started moving simultaneously
to meet each other - one with ar
acceleration of 2.4 m/s, and the
other with an acceleration of 4.8
m/s2. Determine the ratio of the
displacement module of the first
body to the displacement
module of the second body at
the moment of their meeting.
The result of the ratio of the displacement module of the second body at the point of meeting is 0.5.
How to find displacement ratio?To determine the ratio of the displacement of the first body to the displacement of the second body at the moment of their meeting, use the equation of motion:
d = vt + 1/2at²
where d is the displacement, v is the initial velocity, t is the time, and a is the acceleration.
Since the bodies are moving simultaneously towards each other, then assume that their initial velocities are zero. Also, at the moment of their meeting, their displacement will be the same, d₁ = d₂.
Assume that the time at which they meet is t, then:
d₁ = 1/2 * 2.4t²
And the equation for the displacement of the second body:
d₂ = 1/2 * 4.8t²
If d₁ = d₂
then, 1/2 * 2.4t² = 1/2 * 4.8t²
Solving this equation for t and substituting it into the equation for d₁ or d₂, the ratio of the displacement of the first body to the displacement of the second body: d₁/d₂ = 2.4/4.8 = 0.5 or 1/2
So, at the moment of their meeting, the displacement of the first body is half of the displacement of the second body.
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A body weighs 63N on the surface of the earth . What is the gravitational force on it due to the earth at height of 3200km from the earth surface ?Radius of the earth is6400
Answer:
Weight of the body,W=63 N
Acceleration due to gravity at height h from the Earths surface is given by the relation:
g ′ = (1+ R e h ) 2 g
Substituting h=R e /2, g ′ =4g/9
Weight of body of mass m at height h is given as: W ′ =mg ′
=4/9 mg
=4/9×63=28 N
Which would most likely form a homogenous mixture?
10 marbles mixed with 10 small balls
a pinch of sugar mixed with a cup of water
a pile of big rocks mixed with a bag of sand
1 cup of peas mixed with 1 cup of carrots
Answer:
A pinch of sugar mixed with a cup of water
Answer:
it is AAAAAAAA!!!!!!!!!!Explanation:
Two equal magnitude electric charges are separated by a distance d. The electric potential at the midpoint between these two charges is zero. A student considering this situation says: "The electric field at the midpoint between the two charges will be zero also, since the two charges are opposite in sign, so the fields will equal but opposite, and add to zero." There is something wrong with the student's statement. Identify any problem(s) and explain how to correct it/them.
Answer:
The charges under study are of the same sign
The calculation of the electric field for each charge separately, there is no relationship between the charges
Explanation:
Let's start by writing the equation for the electric field
E = k q / r²
where q is the charge under analysis and r the distance from this charge to a positive test charge.
When analyzing the statement the student has some problems.
* The charges under study are of the same sign, it does not matter if positive or negative.
* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.
* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero
A person standing at the edge of a seaside cliff kicks a rock horizontally of the cliff from a
height of 52 m and it lands a distance of 35 m from the base of the cliff. What is the speed at
which the rock was initially kicked?
The time will be the same for both horizontal and vertical component. The initial speed is 10.7 m/s
What is Speed ?Speed is a distance travel per time taken. It is a scalar quantity and it is measured in m/s
Given that a person standing at the edge of a seaside cliff kicks a rock horizontally of the cliff from a height of 52 m and it lands a distance of 35 m from the base of the cliff.
The rock will move vertically downward with initial velocity = 0. The time taken will be constant. That is, same horizontally.
Let us first calculate the time by using the formula
h = ut + 1/2gt²
Where
h = 52 mu = 0Range R = 35 mg = 9.8 m/s²Substitute all the necessary parameters into the formula
52 = 0 + 1/2 × 9.8 × t²
52 = 4.9t²
t² = 52/4.9
t² = 10.6
t = √10.6
t = 3.26 s
The speed at which the rock was initially kicked can be found by
R = Ut
35 = U × 3.26
U = 35/3.26
U = 10.7 m/s
Therefore, rock was initially kicked at a speed of 10.7 m/s
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Mr. Bateman creates a standing wave in the front of the classroom with the spring. S nodes form. The distance from Mr. Bateman to the cabinet is 6m. If a student times the spring moving back and forth and gets 0.2s for one cycle of the spring, how fast are the
waves moving?
Answer:
The speed of a wave is equal to the wavelength divided by the period. The wavelength is the distance between two consecutive nodes, and the period is the time it takes for one complete cycle of the wave.
In this case, the wavelength is 6 m and the period is 0.2 s. Therefore, the speed of the wave is 30 m/s.
The answer is 30 m/s.
Explanation:
Which of the following creates the night-and-day cycle experienced on Earth?
A) Orbit of Earth around the Sun
B) Rotation of the Sun
C) Rotation of the Earth
D) Tilt of the axis of Earth
What are five different types of objects that orbit the sun in the solar system?