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3 years ago

a car initially at 39.40 m/s accelerates at -3.00 m/s^2 how far has a car travel when it comes to a stop 9.35 seconds later

Physics

1 answer:

Kipish [7]3 years ago

6 0

Answer:

499.523. meter

I hope it's helps you

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One terminal of a car battery is said to be connected to “ground.” since it is not really connected to the ground, what is meant

GalinKa [24]

This expression means that the negative terminal (-) is connected to the metal chassis or engine, which means that all voltages used for the electrical devices in the car are measured with respect to the car's chassis or engine.
Today's vehicles have a negative ground system, which means that the vehicle's steel frame or chassis is directly connected to the negative side of the battery via the negative battery cable.

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3 years ago

An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on i

Sergeu [11.5K]

Answer:

$a = 2.77~{\rm m/s^2}$

Explanation:

Since the pulley has a mass concentrated on its rim, the pulley can be considered as a ring.

The moment of inertia of a ring is

$I = mr^2 = (2.3)(23.5\times 10^{-2})^2 = 0.127$

The mass on the left is heavier, that is the pulley is rotating counterclockwise.

By Newton's Second Law, the net torque is equal to moment of inertia times angular acceleration.

$\tau = I \alpha$

Here, the net torque is the sum of the weight on the left and the weight on the right.

$\tau = m_1gR - m_2gR = (1.65)(9.8)(23.5\times 10^{-2}) - (1)(9.8)(23.5\times 10^{-2}) = 1.497~{\rm Nm}$

Applying Newton's Second Law gives the angular acceleration

$\tau = I\alpha\\1.497 = 0.127\alpha\\\alpha = 11.78~{\rm rad/s^2}$

The relation between angular acceleration and linear acceleration is

$a = \alpha R$

Then, the linear acceleration of the masses is

$a = 11.78 \times 23.5\times 10^{-2} = 2.77~{\rm m/s^2}$

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3 years ago

. A student times a car traveling a distance of 2 m. She finds that it takes the car 5 s to

AveGali [126]

No, the car travels 1 metre in 5s at the start which is 0.2m/s, while the second meter it travels one metre in 8 seconds which is 0.125 m/s, the speed changes therefore it is not constant during the two metres the car travels

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3 years ago

Starting from rest, a crate of mass m is pushed up a frictionless slope of angle theta by a horizontal force of magnitude F. Use

ludmilkaskok [199]

Answer:

v = sqrt[2*(F*h*cot(theta)-mgh)/m]

Explanation:

Work = KE + Ug

F*r=1/2mv^2+mgh

1/2mv^2=F*r-mgh

v=sqrt[2(F*r-mgh)/m]

r=h/tan(theta)=h*cot(theta)

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