Ask question

Ask question

All categories

3 years ago

11

if you throw a 0.25-kg ball with a force of 20 n, what's the acceleration of the ball? a. 80 m/s2 b. 12.5 m/s2 c. 5 m/s2 d. 40 m

/s2

1 answer:

nata0808 [166]3 years ago

4 0

To answer this question, we can use Newton's Second Law:

F = ma

We know that:
F = 20N
m = 0.25kg

So, we can plug what we know into the equation and isolate for the acceleration.

20 = 0.25a

a = 80m/s^2 which corresponds to answer A

You might be interested in

Which term describes the object that a satellite revolves around?

m_a_m_a [10]

The object that a satellite revolves around is the <em>central body</em> of the system. <em>(C)</em>

For example:

-- The central body of the solar system is the Sun.

-- The central body for TV satellites, GPS satellites, weather satellites, and the International Space Station is the Earth.

-- The central body for Phobos and Deimos is Mars.

This should be a pretty easy question to answer by elimination, when you notice that "Orbit", "Period", and "Rotation" are not "Bodies".

6 0

3 years ago

Read 2 more answers

Let's start with an example from history. Listed below are a series of claims regarding United States President John F. Kennedy

ruslelena [56]

Answer:

Note that both of the falsifiable claims in this example happen to be true. The claim about Kennedy being the 35th President is falsifiable because it can be checked against historical records. The claim that Kennedy died from a bullet in his brain is falsifiable because it could have been shown false by the medical examiner. The remaining claims are not falsifiable: Statements that call on any type of supernatural being are by definition out of the realm of science. Similarly, a claim of something being "undetectable" could not be falsified, and a claim about what Kennedy would have done if he had lived is a conjecture that cannot be disproven.

Explanation:

7 0

4 years ago

A hockey puck is struck so that it slides at a constant speed and strikes the far side of the rink, 58.2 m away. The shooter hea

DENIUS [597]

Answer:

$v = 33.66 m/s$

Explanation:

Let hockey puck is moving at constant speed v

so here we have

$d = vt$

so time taken by the puck to strike the wall is given as

$t = \frac{58.2}{v}$

now time taken by sound to come back at the position of shooter is given as

$t_2 = \frac{58.2}{340}$

$t_2 = 0.17s$

so we know that total time is 1.9 s

$1.9 = t + t_2$

$1.9 = t + 0.17$

$1.9 - 0.17 = t$

$t = 1.73 s$

now we have

$1.73 = \frac{58.2}{v}$

$v = 33.66 m/s$

7 0

3 years ago

Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face

Sunny_sXe [5.5K]

Answer:

$\mu = 1.645$

Explanation:

By Snell's law we know at the left surface

$\theta_i = 19^o$

$\theta_r = ?$

$\mu_1 = 1$

$\mu_2 = \mu$

now we have

$1 sin19 = \mu sin\theta_r$

$0.33 = \mu sin\theta_r$

now on the other surface we know that

angle of incidence = $\theta_r'$

$\theta_e = 90$

so again we have

$\mu sin\theta_r' = 1 sin90$

so we have

$\theta_r = sin^{-1}\frac{0.33}{\mu}$

$\theta_r' = sin^{-1}\frac{1}{\mu}$

also we know that

$\theta_r + \theta_r' = 49$

$sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49$

By solving above equation we have

$\mu = 1.645$

3 0

3 years ago

Read 2 more answers

A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can

belka [17]

Answer: 21.91 s

Explanation:

Given that,

Maximum height of the car, h = 48 ft

Acceleration of the elevator, a = 0.6 ft/s²

Deceleration of the elevator, -a = 0.3 ft/s²

Maximum speed of the elevator, v = 8 ft/s

Initial speed of the elevator, u = 0

If when the elevator accelerate from 0 to maximum velocity, v.

Let s be the vertical distance traveled during acceleration.

v² = u² - 2as

s = (v² - u²) / 2a

s = (8² - 0) / 2*0.6

s = 64 / 1.2

s = 53.33 ft

If when the elevator decelerates from maximum velocity, v to zero.

Let S be the vertical distance traveled during deceleration

u² = v² + 2aS

S = (u² - v²) / 2a

S = (0 - 8²) / 2 * 0.3

S = -64 / 0.6

S = 106.67 ft

Since he sum of s and S (i.e s + S) is greater than 48 ft, then the elevator will switch from acceleration to deceleration

without reaching the maximum velocity. Below, the switching point is labeled y.

v² = u² + 2ay

y = v²/2a

Inserting this into the earlier deceleration equation, we have

-v²/2 = d * [48 - (v²/2a)], where

d = deceleration

a = acceleration

Therefore, v = [4.√6. a √-(a.b/a)] / b

Where b = acceleration - deceleration

v = 4.382 ft/s

Using this newly found v, we proceed to find our s

s = (u² + v²)/2a

s = 19.2 / 1.2

s = 16 ft

The transport times for each segment are found from

v = u + a*t, thus upward t1

4.382 = 0 + 0.6 * t

t = 4.382/0.6

t = 7.303 s

Also,

4.382 = 0 + 0.3 * T

T = 4.382/0.3

T = 14.607 s

The total travel time is then t + T =

7.303 + 14.607

Total time of travel is 21.91 s

5 0

3 years ago