A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor

Answers

Answer 1
Answer:

The magnitude of impulse will be "9.6 Ns".

According to the question,

Mass,

  • m = 2.4 kg

Final velocity,

  • v = 2.5 m/s

Initial velocity,

  • u = -1.5 m/s

By using Newton's 2nd law of motion, we get

Impulse, I = m(v-u)

By substituting the values, we get

                     = 2.4[2.5-(1.5)]

                     = 2.4(2.5+1.5)

                     = 2.4* 4

                     = 9.6 \ Ns

Thus the above answer is right.    

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Answer 2
Answer:

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns


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what happens when an electric current passes through a coil of wire instead of a single straight peice of wire

Answers

Answer:

An electric current passing through a coil of wire gives a strong form of magnetism called electromagnetism. When the electric current passes through a single straight piece of wire the electromagnetism is weak.

Explanation:

Final answer:

Passing an electric current through a coil of wire generates a magnetic field. The strength of this field can be modified by changing the amount of current or the number of turns in the coil.

Explanation:

When an electric current passes through a coil of wire, as opposed to a straight piece, it creates a magnetic field around the coil. This is the principle behind electromagnets and many electrical appliances we use on a daily basis. The strength of the magnetic field depends on the amount of current and the number of turns in the coil. For example, the more turns the wire has, the stronger the magnetic field.

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The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference.
b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference.
c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference.
d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.

Answers

Answer:

the answers the correct one is c

Explanation:

The diffraction pattern for a slit is

         a sin θ = m λ

Where a is the width of the slit, λ the wavelength, m the order of destructive interference and θ the angle where the interference occurs.

The expression for multi-slit diffraction (diffraction grating) is

          d sin θ = m λ

Where d is the distance between slits, λ the wavelength m the order of the diffraction maximums and θ the angle for these maximums.

When we compare the expressions of the answers the correct one is c

John, who has a mass of 65kg stands at rest on the ice. He catches a 10kg ball that is thrown to him at 5m/s.

Answers

The momentum of John after catching the ball is 50 kg.m/s.

"Your question is not complete, it seems to be missing the following information";

find John's momentum

The given parameters;

  • mass of John, m = 65 kg
  • mass of the ball caught by John, m' = 10 kg
  • initial velocity of John, u = 0
  • initial velocity of the ball, v = 5 m/s

Apply the principles of conservation of linear momentum to determine the momentum of John.

The momentum of John is calculated as follows;

P =  mu + mv

P = (65 x 0)   +  (10 x 5)

P = 0 + 50

P = 50 kg.m/s

Thus, the momentum of John after catching the ball is 50 kg.m/s.

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Suppose that 600 W of radiation in a microwave oven is absorbed by 250 g of water in a very lightweight cup. Approximately how long will it take to heat the water from 20 C to 80 C?(A) 50 s
(B) 100 s
(C) 150 s
(D) 200 s

Answers

Answer:

option (B)

Explanation:

Power, P = 600 W

mass of water, m = 250 g = 0.250 kg

T1 = 20° C

T2 = 80° C

ΔT = 80 - 20 = 60

specific heat of water, c = 4200 J/kg °C

Let the time taken is t.

Power x time = mass of water x specific heat of water x rise in temperature

600 x t = 0.250 x 4200 x 60

t = 105 second

option (B)

Final answer:

To heat 250g of water from 20°C to 80°C using a 600W microwave, it would approximately take 100 seconds.

Explanation:

In order to solve this problem, we first need to know the specific heat capacity of water, which is approximately 4.18 J/g°C. This value represents the amount of energy required to raise 1 gram of water by 1 degree Celsius. Given this value, we'll need to use the formula q = mcΔT, where q is the energy transferred (in joules), m is the mass of the water (in grams), c is the specific heat capacity (in J/g°C), and ΔT is the change in temperature (in °C).

We're given that the initial temperature of water is 20°C and we want to heat it to 80°C, so ΔT = 80°C - 20°C = 60°C. Substituting the known values into the formula, we get: q = 250g * 4.18 J/g°C * 60°C = 62700 J. Now, we know that power (P) = q/t. Given that the microwave oven operates at 600 W (or 600 J/s), we can solve for t: 62700 J ÷ 600 J/s ≈ 104.5 seconds. So, the closest answer would be (B) 100 seconds, considering the approximate value.

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Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)A. What is the magnitude of the change in the momentum, Δp1, of mass M1?
B. What is the change in the total momentum of the pair?
C. What is the magnitude of the change in the momentum Δp2, of mass M2?

Answers

Answer:

a). ΔP1=-2.4 x10^(3)  (D*m)/(s)

b). Pp=0 F=0

c). ΔP2=2.4 x10^(3)  (D*m)/(s)

Explanation:

Initial momentum

P_(1)=m_(1)*v_(i1)

Final momentum

P_(1f)=m_(1)*v_(f1)=-m_(1)*v_(i1)

The change of momentum m1 is:

a).

ΔP1=P_(1f)-P_(1)

ΔP1=-m_(1)*v_(i1)-m_(1)*v_(i1)

ΔP1=-2*m_(1)*v_(i1)

ΔP1=-2*6 D*200(m)/(s)

ΔP1=-2.4x10^(3)(D*m)/(s)

b).

The law of conservation of energy in this case there is not external forces so the momentum of the pair change is equal to zero

P=0

Fx=0

c).

ΔP1+ΔP2=0

ΔP2=-ΔP1

ΔP2=--2.4x10^(3)(D*m)/(s)

ΔP2=2.4x10^(3)(D*m)/(s)

Final answer:

The magnitude of the change in momentum of mass M1 is 2400 Daltons*m/s. The change in the total momentum of the pair is 2000 Daltons*m/s. The magnitude of the change in momentum of mass M2 is -400 Daltons*m/s.

Explanation:

A. To find the magnitude of the change in momentum of mass M1, we use the formula Δp1 = m1 * Δv1, where m1 is the mass of M1 and Δv1 is the change in velocity of M1. Since M1 simply changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, Δp1 = m1 * (2v1) = 6 * (2 * 200) = 2400 Daltons*m/s.

B. The change in the total momentum of the pair is equal to the sum of the changes in momentum of M1 and M2. Since M2 also changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, the change in the total momentum is Δp1 + Δp2 = 2400 Daltons*m/s + (-400 Daltons*m/s) = 2000 Daltons*m/s.

C. To find the magnitude of the change in momentum of mass M2, we use the same formula as in part A, but with the values for M2. Δp2 = m2 * Δv2 = 1 * (2 * (-200)) = -400 Daltons*m/s.

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Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.

Answers

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

or

(A_1)/(A_2) =(T_(2)^(4))/(T_(1)^(4))

substituting the values in the above question we get

(A_1)/(A_2) =(2100_(2)^(4))/(2700_(1)^(4))

or

(A_1)/(A_2) }=0.3659