A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 meters per second. What is the magnitude of the total momentum of the two balls after collision?

The energy released when 10 g of steam is spilled on the hand is approximately 25 kJ

The heat required to raise the temperature of 10 g of water from 33 °C to 100 °C is

Q₁ = m x c x ΔT

= 10 g * 4.2 * (100 °C - 33 °C)

= 2772 J

The heat required to vaporize 10 g of water at 100 °C is given by:

Q₂ = m x L

= 10 * 2.2

= 22,000 J

where L is the specific latent heat of vaporization of water

The total energy released

Q = Q1 + Q2

= 2772 J + 22,000 J

= 24,772 J

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Answer:

stress,depression and anxiety by improving self esteem.

Segment a, which is drawn from the center of the polygon perpendicular to one of its sides, is called the apothem.

The typical hexagon has a surface area of around 874.6 square meters.

To find the area of a regular polygon, use the formula:

Area = (1/2) x Perimeter x Apothem

Find the perimeter of the polygon. Since the polygon has n sides, use the formula:

Perimeter = n x s

where s = length of one side.

Since s = 18m, find n by using the formula:

n = 360° / (180° - (360° / n))

where n = number of sides.

Plugging in the values:

n = 360° / (180° - (360° / n))

n = 360° / (180° - (360° / 6))

n = 6

So the polygon has 6 sides, which makes it a hexagon.

Now find the perimeter:

Perimeter = n x s

Perimeter = 6 x 18

Perimeter = 108m

Next, find the apothem, use the formula:

Apothem = s / (2 x tan(π/n))

Plugging in the values:

Apothem = 18 / (2 x tan(π/6))

Apothem = 9√3 m

Now use the formula for the area:

Area = (1/2) x Perimeter x Apothem

Area = (1/2) x 108 x 9√3

Area ≈ 874.6 m²

Therefore, the area of the regular hexagon is approximately 874.6 square meters.

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The correct statement is Molecular collisions transfer thermal energy between the system and its surroundings. Thus, option D is correct.

The energy moves between a system of reacting substances and the surroundings by the collision of molecules. The transfer of heat or thermal energy between the system and its surroundings by the process of Conduction. Conduction is the process of transmitting the heat to the neighboring atoms or collisions by the process of collisions.

The conduction takes place more steadily in solids and liquids where the molecules are closer together. When the molecules are collided with the nearby molecules, the potential energy is converted into kinetic energy and hence the heat energy is transferred between the system and its surroundings.

Hence, Molecular collisions transfer thermal energy between the system and its surroundings. Thus, the correct option is D.

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The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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The atmospheric blurring can negatively affect the view of objects and requires a larger separation to distinguish forms. It is caused by turbulent air in motion.

The atmospheric blurring is a natural phenomenon caused by turbulent air in motion, which negatively affects the view of astronomical objects.

The atmospheric blurring may lead to larger percent errors in objects with smaller angular diameters because in these cases it is required a larger separation to distinguish forms.

Smaller angular diameters produce larger squared visibility, thereby increasing the effects of distortion caused by atmospheric blurring.

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Answer:

A.) 4.52s

B.) 44.3 m/s

Explanation:

Given that a cargo of mass 50kg fall freely from rest at a height of 100m and came to rest having penetrated 5.0cm of the ground

The total height = 100 + 5/100

The total height = 100 + 0.05

The total height = 100.05 m

Since the cargo fall from rest, the initial velocity U will be equal to zero. That is,

U = 0

Use second equation of motion

h = Ut + 1/2gt^2

Substitutes g = 9.8 and others parameters into the formula

100.05 = 0 + 1/2 × 9.8 × t^2

100.05 = 4.9t^2

t^2 = 100.05/4.9

t^2 = 20.42

t = sqrt( 20.42)

t = 4.52 s

B.) To calculate the velocity of the body when it hits the ground,

You can achieve it by first calculating its kinetic energy. Or by using the first equation of motion.

So, let use first equation of motion

V = U + at

But U = 0

Substitute the time and g into the formula

V = 9.8 × 4.52

V = 44.29 m/s

Therefore, the velocity of the body when it hits the ground is 44.3 m/s approximately.

Answer:

hlo buddy

Explanation:

please Msg me nk175285

my G m a i l id

It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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Answer: To find the total resistance of the wires, we need to use a formula that takes into account the length and cross-sectional area of the wires and the resistivity of the material they are made of. By plugging in the appropriate values for the length, cross-sectional area, and resistivity of the wires in the formula, we can find the individual resistances of the wires. Then, by using another formula for the total resistance of a parallel combination, we can find the effective resistance of the wires in parallel.

Explanation:R = ρ * (L / A)

where R is the resistance of the wire, ρ is the material's resistivity (resistance per unit length), L is the length of the wire, and A is the cross-sectional area of the wire.
according to the given data
AP= 40 cm
px=60cm

Next, we need to calculate the wire PX's length and cross-sectional area. The length is given as 60 cm, and we can assume the same diameter of 1 mm, giving a cross-sectional area of 0.785 mm^2.

Now we can calculate the resistance of each wire using the formula above:

R_AX = ρ * (80 cm / 0.785 mm^2) = 101.91 * ρ

R_PX = ρ * (60 cm / 0.785 mm^2) = 76.43 * ρ

The total resistance of the parallel combination is given by the formula:

1/R_total = 1/R_AX + 1/R_PX

Substituting the values of R_AX and R_PX, we get:

1/R_total = 1/(101.91 * ρ) + 1/(76.43 * ρ)

1/R_total = (1.55 / ρ)

R_total = ρ / 1.55

therefore, the effective resistance of the parallel combination is (ρ / 1.55) ohms

Answer:

U calculate from CREST to REST POSITION

Option C. The heat absorbed or lost by a substance while it's changing state  correctly describes latent heat

Latent heat is the heat absorbed or lost by a substance while it is changing state.

The latent heat is a type of heat that is transferred during phase change, i.e., while a substance undergoes a change of state.

For example, when ice melts into liquid water, or when liquid water evaporates into water vapor, heat is absorbed from the surroundings.

Latent heat is not associated with a temperature change; rather, it's associated with a change of state.

For instance, the temperature of water remains at 100°C while boiling.

When water is boiling, the latent heat of vaporization is absorbed and utilized to break the hydrogen bonds holding water molecules together to change water from the liquid phase to the gaseous phase.

When the water is boiling, adding more heat won't increase the water's temperature, instead, the extra heat will be absorbed to change the phase of water molecules.

Therefore, the correct answer to the given question is option C: The heat absorbed or lost by a substance while it is changing state.

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The skateboarder will travel a distance of 9.19 m

We know that,

a = ( v - u ) / t

Δv = ( v + u ) / 2 ( If acceleration is constant )

Δv = d / t

where,

a = Acceleration

v= Final velocity

u = Initial velocity

t = Time taken

d = Displacement

Δv = Average velocity

Given that,

u = 1.8 m / s

a = 1.5 m / s²

t = 2.5 s

a = ( v - u ) / t

1.5 = ( v - 1.8 ) / 2.5

v = 3.75 + 1.8

v = 5.55 m / s

Δv = ( 5.55 + 1.8 ) / 2

Δv  = 3.675 m / s

d = Δv * t = 3.675 * 2.5

d = 9.19 m

If the initial and final velocity is known, average velocity can be calculated by taking the average value of initial and final velocity, if the acceleration is constant.

Therefore, the skateboarder will travel a distance of 9.19 m

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(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

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Answer:

Explanation:

The Law of Energy Conservation states that K1 + U1 = K2 + U2

m= 72.0 kg

h= 3.90 m

a)

K1 + U1 = K2 + U2

0 + mgh = 1/2mvf^2 + 0

mass cancels out so gh=1/2vf^2

(9.8 m/s^2)(3.9 m)=(.5)(vf^2)

vf= 8.74 m/s

b)

1/2mv^2 + mgh = 1/2mv^2 + 0

mass cancels again

(.5)(2.9^2 m/s) + (9.8 m/s^2)(3.9 m) = (.5)(vf^2)

vf= 9.21 m/s

c)

This would be the same as the past problem as the velocity gets squared so direction along the axis doesn't matter. Thus, vf= 9.21 m/s

The direction of the net force is up.

The net force on the charges can be determined using Coulomb's law as follows;

\(F_{AC} = + \frac{KQ^2}{r^2}\)

\(F_{BC} = - \frac{KQ^2}{r^2}\)

F(net) = F(AC) + F(BC)

F(net) = 0

The, direction of the net force on charge C will be parallel to charge C to achieve a zero net force. Thus, the direction of the net force is up.

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Answer:

density = mass/volume

so . . .  

density = (36 g)/(15 cm³) = 2.4 g/cm³

Explanation:

The gravitational force that the moon has on a person with a mass of 50 kg is approximately 1.15 N.

The gravitational force between two objects depends on their masses and the distance between them. This force is given by the formula:

F = (G × m₁ × m₂) / r² where F is the gravitational force, m₁ and m₂ are the masses of the two objects, r is the distance between them, and G is the gravitational constant, which has a value of 6.7 × 10⁻¹¹ N m²/kg².

Using this formula, we can find the gravitational force that the moon has on a person with a mass of 50 kg.

The mass of the moon is 7 × 10²² kg, and the distance between the moon and the person is 380,000,000 m.

Therefore, we have:

m₁ = 50 kg

m₂ = 7 × 10²² kg

r = 380,000,000 m

G = 6.7 × 10⁻¹¹ N m²/kg²

Substituting these values into the formula, we get:

F = (G × m₁ × m₂) / r²

F = (6.7 × 10⁻¹¹ × 50 kg × 7 × 10²² kg) / (380,000,000 m)²

F = 1.15 N

Therefore, the gravitational force that the moon has on a person with a mass of 50 kg is approximately 1.15 N.

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Answer:

The mass of the feather would be 0.0075 kg.

Explanation:

If the feather were to be on earth, it would have experienced the gravitational pull of the earth downwards. So that from Newton's second law of motion,

F = mg

where f is the force on the object, m is the mass of the object and g is the acceleration due to gravity of the earth.

But, F = 0.075 N, g = 10 m/\(s^{2}\).

So that,

0.075 = m x 10

m = \(\frac{0.075}{10}\)

   = 0.0075 kg

The mass of the feather would be 0.0075 kg.

Cause: According to this law, the skateboard requires an external source of force to move, which the skateboarder indirectly provides.

Effect: On the pavement, the skateboarder reverses forward (that is he applies a force on the road in a direction opposite the direction of intended motion).

What connection exists between a cart's mass and its acceleration after the spring is let go?

The acceleration of the cart is inversely proportional to its mass. As a result, the acceleration of the cart will decrease as its mass increases.

The mass and acceleration are inversely correlated: the acceleration decreases as mass increases.

Newton's second law states that force and acceleration are directly proportional to one another, so if a cart's mass is held constant while the force acting on it increases, the cart will accelerate. When an object moves, it has kinetic energy, which it then expends.

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Answer:

Muscle Tissue

Explanation:

Sorry if I'm wrong

Explanation:

Sorry I don't know the answer

The difference in length between the two rods = 0.0025m

Linear expansivity of a material is given by the formula:

For the Aluminium rod:

For the Nickel rod:

The difference in length between the two rods will be given as:

The difference in length between the two rods = 0.0025m

Answer:

a cold front i think

Explanation:

Answer:

5.51 m/s^2

Explanation:

Initial scale reading = 50 kg  

assume the greatest scale reading = 78.09 kg

Determine the maximum acceleration for these elevators

At rest the weight is = 50 kg

Weight ( F ) = mg = 50 * 9.81 = 490.5 N

At the 10th floor weight = 78.09 kg

Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N

F = change in weight

Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)

50 * a = 275.61

Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2

Answer:

We cannot tell from the information given

Explanation:

Given;

mass of the box, m = 5 kg

first force, F₁ = 10 N

second force, F₂ = 5 N

(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;

∑Fx = 10 N - 5 N

      = 5 N

Apply Newton's second law of motion;

∑Fx = ma

a = ∑Fx/m

a = 5 / 5

a = 1 m/s² in the direction of the 10 N force.

(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;

∑Fx = 10 N + 5 N

∑Fx = 15 N

a = 15 / 5

a = 3 m/s²

Therefore, the information given is not enough to determine the acceleration of the box.

Answer:

1) c.

2) a.)

Explanation:

1)

2)

       \(v = \lambda * f (1)\)

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

Refraction

Explanation:

because the pencil looks magnified or distorted in water

Answer:

(B) Refraction

Explanation:

the fact or phenomenon of light being deflected in passing obliquely through the interface between one medium and another or through a medium of varying density.

In physics, refraction is the redirection of a wave as it passes from one medium to another. The redirection can be caused by the wave's change in speed or by a change in the medium.

Hope this helps!! :) (Also there is an example attached :D)