All categories

3 years ago

What is true about an object in free fall?

1 answer:

5 0

The only force being acted upon it is gravity. A ball that was thrown downward off a building isn't in freefall, because it had initial velocity. A piece of paper can't necessarily be in free fall because it is affected by air resistance.

You might be interested in

Temperature flows from __ to __ temperatures

timama [110]

Answer:hot to cold

Explanation:

Temperature flow from hot to cold temperatures

8 0

3 years ago

A racetrack has the shape of an inverted cone, as the drawing shows. On this surface the cars race in circles that are parallel

sergey [27]

Answer:

The value of d is 183.51 m.

Explanation:

Given that,

Speed of car = 34.0 m/s

Suppose The car race in the circle parallel to the ground surface is at an angle 40°

The radius of circular path $r = d\cos\theta$

Normal force acting on the car = N

We need to calculate the value of d

Using component of normal force

The horizontal component of normal force is equal to the gravitational force.

$N\cos\theta=mg$ ....(I)

The vertical component of normal force is equal to the centripetal force

$N\sin\theta=\dfrac{mv^2}{r}$ .....(II)

Divided equation (I) by equation (II)

$\tan\theta=\dfrac{v^2}{gr}$

Put the value of g

$\tan\theta=\dfrac{v^2}{g\times d\cos\theta}$

$v^2=\tan\theta\times g\times d\cos\theta$

$v^2=g\times d\sin\theta$

$d=\dfrac{v^2}{g\sin\theta}$

Put the value into the formula

$d=\dfrac{(34.0)^2}{9.8\times\sin40}$

$d=183.51\ m$

Hence, The value of d is 183.51 m.

3 0

3 years ago

PLZ HELP Lab: Circuit Design

Masteriza [31]

Answer:

Explanation:

6 0

2 years ago

A 2.0-kg pistol fires a 1.0-g bullet with a muzzle speed of 1000 m/s. The bullet then strikes a 10-kg wooden block resting on a

Andreyy89

Answer:

1000 N

Explanation:

An impulse results in a change of momentum

FΔt = mΔv

F = 0.001 kg(1000 - 0) m/s / 0.001 s = 1000 N

4 0

3 years ago

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was

True [87]

This problem is looking for the minimum value of μs that is necessary to achieve the record time. To solve this problem:

Assuming the front wheels are off the ground for the entire ¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².

For a constant acceleration, distance = 402.3

m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2

µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18

5 0

3 years ago