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3 years ago

What is the mass of a crate if a net force of 12 N gives the crate an acceleration of 0.20 m/s2?

1 answer:

8 0

A :-) for this question , we should apply
F = ma
Given - F = 12 N
a = 0.20 m/s^2
Solution -
F = ma
12 = m x 0.20
m = 12 by 0.20
m = 60 kg

.:. The mass is 60 kg.

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Answer:

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3 years ago

What is the total energy equation?

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3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

A skydiver is falling at constant velocity. If a force of 600 N is pulling down on the skydiver, how much force must be acting u

skelet666 [1.2K]

Well, if the skydiver is at constant velocity, than there’s no acceleration, as stated by Newton’s first law. Thus the total net force would equate to 0. In order to make this statement true, the answer would have to be exactly 600 N.

7 0

3 years ago

Read 2 more answers

Choose the scenario in which the sound frequency of the waves is higher.

mrs_skeptik [129]

Answer:

B) the sound source moves towards you at 100 m/sec

Explanation:

The Dopper Effect is a phenomenon that occur when there is relative motion between an observer and a source of a wave. When this situation occurs, there is an apparent shift in frequency of the wave, as observed by the observer.

The apparent frequency observed by the observer is given by

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the original frequency of the wave

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if moving towards the source of the wave, negative otherwise)

$v_s$ is the velocity of the source (negative if moving towards from the observer, positive otherwise)

In this problem, we want to find the scenario in which the sound frequency is higher.

We see that in all 4 scenarios, the sound source is moving: this means we have to find the scenario in which the denominator of the equation is smaller.

First of all, we notice the sound source moves towards the observer, $v_s$ is negative, so the denominator is higher: this means that the correct option must be either A or B.

Also, we notice that since $v_s$ is negative, a value larger in magnitude will mean a smaller denominator: therefore, the correct answer will be

B) the sound source moves towards you at 100 m/sec

Since this situation will make the denominator of the formula the smallest possible.

5 0

3 years ago