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3 years ago

Keaton is asked to solve the following physics problem:

Physics

1 answer:

RideAnS [48]3 years ago

5 0

Answer:

The answer is C "think about the problem first, systematically consider all factors, and form a hypothesis"

Explanation:

In physics there is some basic fomula that sir Isacc Newton proposed under the topic of motion. The three formulas are below;

1) v=u+at

2)v^2=u^2+2as

3)s=ut+(1/2)(at^2)

the variables are explained below;

u= initial velocity of the body

a=acceleration/Speed of the body

t= time taken by the body while travelling

s= displacement of the body.

Therefore to solve keatons problem, the factors(variables) in the formulas above need to be systematically considered. Since the ball was dropped from the top of the building, the initial velocity is 0 because the body was at rest. Also the acceleration will be acceleration due to gravity (9.8m/s^2)

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A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will

alukav5142 [94]

Answer:

The change in momentum is $\Delta p = kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A =$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = * (22.9 - 22.6)$

$\Delta p = * (0.3)$

$\Delta p = kg \cdot m/s$

6 0

3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

What is Force!????!?

spin [16.1K]

A force is a push or pull upon an object resulting from the object's interaction with another object.

4 0

3 years ago

Read 2 more answers

Pavlova-9 [17]

Explanation:

momentum = mass x velocity

initial momentum = 100 x 15 = 1500kgm/s

after momentum = 100 x 20 = 2000kgm/s

a =(v-u)/t

a = (20-15)/10

a = 5/10

a = 0.5m/s²

f = ma

f = 100 x 0.5

f = 50N

7 0

3 years ago

A constant friction force of 25 N acts on a 65-kg skier for 15 s on level snow. What is the skier’s change in velocity

nikdorinn [45]

Answer:

$\Delta v=5.77m/s$

Explanation:

Newton's 2nd Law relates the net force F on an object of mass m with the acceleration a it experiments by F=ma. In our case the net force is the friction force, since it's the only one the skier is experimenting horizontally and the vertical ones cancel out since he's not moving in that direction. Our acceleration then will be:

$a=\frac{F}{m}$

Also, acceleration is defined by the change of velocity $\Delta v$ in a given time t, so we have:

$a=\frac{\Delta v}{t}$

Since we want the change in velocity, mixing both equations we conclude that:

$\Delta v=at=\frac{Ft}{m}$

Which for our values means:

$\Delta v=\frac{Ft}{m}=\frac{(25N)(15s)}{(65Kg)}=5.77m/s$

4 0

3 years ago