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3 years ago

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A car is traveling at a constant speed on the highway. Its tires have a diameter of 65.0 cm and are rolling without sliding or s

lipping. If the angular speed of the tires is 49.0 rad/s , what is the speed of the car, in SI units

1 answer:

lesya692 [45]3 years ago

4 0

Answer:

Explanation:

The car is rolling without slipping so Vcm= R×ω = 0.325×49 = 16

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A 100 kg cart is moving at 3 m/s. Calculate the cart’s kinetic energy.

STALIN [3.7K]

Answer:

450

Explanation:

Given,

Mass= 100kg

Velocity= 3 m/s

Kinetic Energy= ?

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= 1/2× 100× 3^2

= 1/2× 900

= 450.

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2 years ago

Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t

salantis [7]

Explanation:

Mass of bumper cars, $m_1=m_2=110\ kg$

Initial speed of car A, $u_1=8\ m/s$

Initial speed of car Z, $u_2=-10\ m/s$

Final speed of car A after the collision, $v_1=-10\ m/s$

We need to find the velocity of car Z after the collision. Let it is equal to $v_2$ . Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$110\times 8+110\times (-10)=110\times (-10)+110v_2$

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So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

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3 years ago

A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re

fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

$PV=nRT$

For a gas

$P_{1}V_{1}=nRT_{1}$

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

$10\times V=nR\times286$ ....(I)

When the temperature of the gas is increased

Then,

$P_{2}V_{2}=\dfrac{n}{2}RT_{2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}$

$\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}$

$P_{2}=\dfrac{10\times368}{2\times286}$

$P_{2}= 6.433\ atm$

$P_{2}=6.4\ atm$

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

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2 years ago

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balandron [24]

Answer:

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Explanation:

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