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kkurt [141]
3 years ago
9

A racquet ball with mass m = 0.221 kg is moving toward the wall at v = 13.9 m/s and at an angle of θ = 25° with respect to the h

orizontal. The ball makes a perfectly elastic collision with the solid, frictionless wall and rebounds at the same angle with respect to the horizontal. The ball is in contact with the wall for t = 0.073 s.1) What is the magnitude of the initial momentum of the racquet ball?2) What is the magnitude of the change in momentum of the racquet ball?3) What is the magnitude of the average force the wall exerts on the racquet ball?4) Now the racquet ball is moving straight toward the wall at a velocity of vi = 11.8 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -7.8 m/s. The ball exerts the same average force on the ball as before.
What is the magnitude of the change in momentum of the racquet ball?5) What is the time the ball is in contact with the wall?6) What is the change in kinetic energy of the racquet ball?
Physics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:

1) 3.07kgm/s

2) 5.56kgm/s

3) 76.16N

4) 4.33kgm/s

5) 0.57s

6) -8.66J

Explanation:

Given

m = 0.221kg

v = 13.9m/s

θ = 25°

t = 0.073s

1) to get the magnitude of the initial momentum is the racquet ball. We use the formula,

P(i) = mv(i)

P(i) = 0.221 * 13.9

P(i) = 3.07kgm/s

2) Magnitude of the change in momentum of the ball,

P(i,x) = P(i) cos θ

P(i,x) = 3.07 * cos25

P(i,x) = 3.07 * 0.9063

P(i,x) = 2.78

ΔP = 2P(i,x)

ΔP = 2 * 2.78 = 5.56kgm/s

3) magnitude of the average force exerted by the wall,

F(ave) = ΔP/Δt

F(ave) = 5.56/0.073

F(ave) = 76.16N

4) ΔP(z) = mv(f) - mv(i)

ΔP(z) = 0.221*-7.8 - 0.221*11.8

ΔP(z) = -1.72 - 2.61

ΔP(z) = 4.33kgm/s

5) F(ave) = ΔP/Δt

Δt = ΔP/F(ave)

Δt = 4.33 / 76.16

Δt = 0.57s

6) KE(i) = 0.5mv(i)²

KE(f) = 0.5mv(f)²

ΔKE = 0.5m[v(f)² - v(i)²]

ΔKE = 0.5 * 0.221 [(-7.8)² - 11.8²]

ΔKE = 0.1105 ( 60.84 - 139.24 )

ΔKE = 0.1105 * -78.4

ΔKE = -8.66J

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<h2>Answer:</h2>

(a) 3.96 x 10⁵C

(b) 4.752 x 10⁶ J

<h2>Explanation:</h2>

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h          [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

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(b) The energy (E) involved in the process is given by;

E = Q x V           -----------------(i)

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Q = magnitude of the charge = 3.96 x 10⁵C

V = electric potential = 12V

Substitute these values into equation (i) as follows;

E = 3.96 x 10⁵ x 12

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8 0
3 years ago
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To solve this problem it is necessary to apply the definition given in Faraday's law in a solenoid for which it is noted that

\epsilon = - d\frac{\phi_B}{dt}

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Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

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We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

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6 0
2 years ago
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creativ13 [48]

Answer:

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in the water film    λ = 319.55 10⁻⁹ m

Explanation:

When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,

when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship

             v = λ f

in the void we have

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the refractive index is

             

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3 0
2 years ago
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Answer:

1.F = 256 N

2.a = 8 m/s/s

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F = ma

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m = 256 N/8 m/s/s

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6 0
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