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3 years ago

A moving car skids to a stop with the wheels locked across a level roadway. Of the forces listed, identify which act on the car.

Physics

1 answer:

Vesnalui [34]3 years ago

4 0

Answer:

Normal, Gravity, Friction, and Air Resistance.

Explanation:

When a moving car skid to stop and its wheels are locked across, then the following forces will be applied on the car:

Normal force: It will act counter to gravity that pushes an object against a surface and acts perpendicular to the contact surface.

Gravity: Gravity force acts in each and every object having mass and it can not be avoidable. So, the gravity force will also apply to the car and attract it to the earth's surface.

Friction: Friction is a force that acts opposite to the motion and stops or slows motion. Friction will be applied to the car that will oppose the motion of the car and stop it.

Air resistance: air resistance is defined as the forces exerted by air that acts opposite to the relative motion of an object. Air resistance will also be applied to the car when it will skid to stop as we are always surrounded by the air.

Hence, the correct answers are "Normal, Gravity, Friction, and Air Resistance."

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For a relative frequency distribution, relative frequency is computed is computed as ____________.

Yuliya22 [10]

Answer:

For a relative frequency distribution, relative frequency is computed as the class frequency divided by the number of observations.

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2 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

$T=2\pi\sqrt{\dfrac{m}{k}}$

m is the mass of object

$m=\dfrac{W}{g}$

$m=\dfrac{2.45}{9.8}$

m = 0.25 kg

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}$

k = 24.09 N/m

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

6 0

3 years ago

A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

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3 years ago

Two point charges are separated by 6.4 cm . The attractive force between them is 10 N . Suppose that the charges attracting each

LenaWriter [7]

Answer:

Two point charges are separated by 6.4 cm . The attractive force between them is 10 N .

units.

Explanation:

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2 years ago

A. At what point in it's motion is the kinetic energy of the end of a pendulum greatest? b. At what point is the potential energ

Darya [45]

The kinetic energy will be greatest at the bottom of the swing motion.

The potential energy will be greatest at the highest position of the swing.

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1 year ago