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4 years ago

Given a force of 88 N and an acceleration of 4 m/s2, what is the mass?

1 answer:

7 0

Newton's second law of motion says:

Force = (mass) x (acceleration)

88 = (mass) x (4)

Divide each side by 4 :

Mass = 22 kg.

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A 30 kg child rides a 20 kg bicycle. Together, the child and the bicycle have a momentum of 110 kg-m/s. What is the velocity of

love history [14]

Answer:

The velocity of the boy and the bicycle is 2.2 m/s.

Explanation:

We have,

Mass of child is 30 kg and the mass of bicycle is 20 kg. The combined momentum of the child and the bicycle is 110 kg-m/s.

It is required to find the velocity of the boy and the bicycle. The momentum of an object is given in terms of mass and its velocity. So,

$p=Mv$

M is combined mass of child and bicycle

$v=\dfrac{p}{M}\\\\v=\dfrac{110}{30+20}\\\\v=2.2\ m/s$

So, the velocity of the boy and the bicycle is 2.2 m/s.

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3 years ago

What would you use to stop itches from a bite

sukhopar [10]

You can use apple cider vinegar. If you don't have it, you can go to shoppers drug mart and they will have an ointment to sooth itches.

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3 years ago

Read 2 more answers

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: