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salantis [7]
3 years ago
10

Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3400 N on the car for 0.400 s. (Assume t

he initial velocity is in the positive direction.) (a) What impulse (in kg · m/s) is imparted by this force? (Indicate the direction with the sign of your answer.) kg · m/s (b) Find the final velocity (in m/s) of the bumper car if its initial velocity was 3.50 m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor. (Indicate the direction with the sign of your answer.) m/s
Physics
1 answer:
babunello [35]3 years ago
3 0

Answer:

a)   I = -1360 N s   b)  v_{f} = -3.3 m / s

Explanation:

a) The expression of the impulse is

      I = F t = Δp

The force applied by the rail is in the opposite direction to the movement, so it is negative

     I = -3400 0.400

     I = -1360 N s

b) We use the momentum relationship with the moment

    I = ΔP

    I = m v_{f}- mv₀

   v_{f} = (I + m v₀) / m

   v_{f} = (-1360 + 200 3.5) / 200

   v_{f} = -3.3 m / s

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Tatiana [17]

Answer: The correct answer is True.

Explanation:

Loudness of sound is referred to how soft or loud a sound is for the listener.

This term is measured in a unit known as decibels referred to as dB.

This unit is used to measure the relative intensity of sounds on a scale from zero to 100 dB.

More the value of decibels, it will be uncomfortable for a person to hear that sound.

So Yes, the loudness of sound is measured in decibels.

7 0
3 years ago
Give a real-world example of how energy is transformed from electrical energy to thermal energy. Describe how the heat can be tr
timofeeve [1]

Answer:

A microwave

Explanation:

Car

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3 0
1 year ago
A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)
bezimeni [28]

Answer:

205 V

V_{R} = 2.05 V

Explanation:

L = Inductance in Henries, (H)  = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

V_{L} = - IwLsin(wt)

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

 = 11.0 V / 500 rad/s (0.500 H)

 = 11.0 / 250

I = 0.044 A

Now

V_{R} = IR

    = (0.044 A) (93 Ω)

V_{R} = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V_{R} = V_{R} cos (wt)

Putting V_{R} = 4.092 V and w = 500 rad/s

V_{R} = V_{R} cos (wt)

    = (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V_{R} = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

    =  (4.092 V) (cos (500 rads/s)(0.00209))

    = (4.092 V) (cos(1.045))

    = (4.092 V)(0.501902)

    = 2.053783

V_{R} = 2.05 V

8 0
3 years ago
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jenyasd209 [6]
Σf = m a
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