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3 years ago

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The roads are icy, and you observe a head-on collision on Summit, at the corner with Rhodes: a 1ton car swerves out of his lane

and slides through a stop sign at 41 mph straight into a 3 ton SUV traveling at 16mph in the other direction. The car and the SUV crumple from the collision, and stick together.
1. What is the final velocity, in MPH (you don't need to enter MPH in your answer) , of the SUV/car entanglement (the positive direction is the direction the car was initially going)

1 answer:

NNADVOKAT [17]3 years ago

8 0

Answer:

final velocity = - 1.75 mph

Explanation:

given data

mass m1 = 1 ton

mass m2 = 3 ton

velocity v = 41 mph

velocity u = 16 mph

to find out

what is final velocity V

solution

we will apply here Conservation of momentum that is express as

mv + Mu = (m + M) × V ...........................1

put here value we get

1 × 41 - 3 × 16 = (1 + 3 ) × V

solve it we get

41 - 48 = 4 V

V = $\frac{-7}{4}$

final velocity = - 1.75 mph

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Answer: Work done = 153.125Joules, Work done = 0.003Nm

Explanation:

Kinetic energy of a body is the energy possessed by a body by virtue of its motion.

Mathematically,

K.E = 1/2MV²

Where;

M = mass of the body = 2.5g = 0.0025kg

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Substituting this values in the formula, we have;

K.E = 1/2× 0.0025×350²

K.E = 153.125Joules

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Work = Force × distance

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Force = 0.025N

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To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circus performer wh

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Answer:

$v = 15.45 m/s$

Explanation:

As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

$W_f = (U_i + K_i) - (U_f + K_f)$

here we know that

$W_f = F_f . d$

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now from above equation

$W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)$

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A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

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$\Delta U=-W$

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$\Delta U=-q(E\cdot d)$

Put the value into the formula

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We need to calculate the change in electric potential energy

Using formula of electric potential energy

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$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

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