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3 years ago

7

Bumper car A (331 kg) moving east at 3.87 m/s hit stationary car B (208 kg). After, the car A moves .88 m/s east. What is the ve

locity of car B?

1 answer:

Gre4nikov [31]3 years ago

4 0

Answer:

Ub = 221.99m/s

Explanation:

(Ma x Ua)+(Mb x Ub) = (Ma + Mb)V

(331 x 3.87)+(208 x Ub) = (331 + 208)88

(1257.8) + (208 x Ub) = 47432

(208 x Ub) = 47432 - 1257.8

(208 x Ub) = 46174.2

Ub = 46174.2/208 = 221.99m/s

Ub = 221.99m/s

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Answer:

v = 8.57 m/s

Explanation:

As we know that the wagon is pulled up by string system

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So as per work energy theorem we know that

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$F_t . d - (mgsin\theta)(d) = \frac{1}{2}mv^2 - 0$

so we have

$F_t = 129 N$

$\theta = 17.5^o$

m = 38.2 kg

d = 85.4 m

so now we have

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3 years ago

A charging RC circuit controls the intermittent windshield wipers in a car. The emf is 12.0 V. The wipers are triggered when the

lilavasa [31]

Answer:

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Explanation:

We have the following information,

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We apply the equation for capacitor charging the voltage across it,

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Replacing values,

$R=-\frac{1.8}{ln(10/12)*1.25*10^{-6}}$

$R=803k\Omega$

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3 years ago

the power rating of an electric lawn mower is 2200 watts. if the lawn mower was used for 30 mins, and 650 Newtons of force was u

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3 years ago

A company records authors reading their books aloud. These recordings are shared with libraries around the world for people to l

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Answer:

c

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2 years ago

Read 2 more answers

A guitar string with a linear mass density of 2.0 g/m is stretched between two vertical rods that are 65 cm apart. The string ho

Rudiy27

Answer:

717 Hz

Explanation:

<u>solution:</u>

The wave with three antinodes has m = 3. Thus, f_3 = 3f_1 and so the fundamental frequency of the string is

f_1 =f_3/3

=430 Hz/3

=143 Hz

Thus, the frequency of the fifth harmonic is

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= 717 Hz

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4 years ago