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elena-14-01-66 [18.8K]

2 years ago

12

Consider two slides, both of the same height. One is long and the other is short. From which slide will a child have a greater f

inal speed when sliding off? Assume that there is no friction acting.

1 answer:

Lunna [17]2 years ago

8 0

Answer:

The final speed will be the same for the children on the shorter side and on the longer side.

Explanation:

This is because since the they are the same distance above the ground, their potential energy which is a function of mass, acceleration due to gravity and vertical height are the same.

PE = Mass × gravity × vertical height

At this point, we can deduce that the horizontal length of the slide has no effect on the potential energy. Only the vertical height does.

All this potential energy is converted to kinetic energy at the end of the slide. Since the potential energy is the same, then the kinetic energy will be the same and thus their velocity is the same.

Mathematically, consider that PE = mgh and KE = $\frac{1}{2}$ m $v^{2}$

at the bottom of the slide, since energy has to be conserved, PE must be equal to KE.

mgh = $\frac{1}{2}$ m $v^{2}$

final velocity of the child , v = $\sqrt{2gh}$

It shows the final velocity is only a function f acceleration due to gravity and height.

Thus, making their velocities equal.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is $\theta = 24.2^o$

Explanation:

From the question we are told that

The length of eire let out is $L = 250 \ m$

The emf generated is $\epsilon = 40 V$

The earth magnetic field is $B = 5.0 *10^{-5} T$

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The emf generated is mathematically represented as

$\epsilon = L\ v\ B\ sin \ \theta$

making $\theta$ the subject of the formula

$\theta = sin ^{-1}[ \frac{\epsilon}{L * B *v} ]$

substituting values

$\theta = sin ^{-1}[ \frac{40}{250 * (5*10^{-5}) *(7.80 *10^{3})} ]$

$\theta = 24.2^o$

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2 years ago

You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that

oksano4ka [1.4K]

As we know that reaction time will be

$t = 0.50 s$

so the distance moved by car in reaction time

$d = vt$

$d = 20 \times 0.50$

$d = 10 m$

now the distance remain after that from intersection point is given by

$d = 110 - 10 = 100 m$

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

$v_f = 0$

$v_i = 20 m/s$

now from kinematics equation we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 20^2 = 2(a)100$

$a = \frac{-400}{200} = -2 m/s^2$

so the acceleration required by brakes is -2 m/s/s

Now total time taken to stop the car after applying brakes will be given as

$v_f - v_i = at$

$0 - 20 = -2 (t)$

$t = 10 s$

total time to stop the car is given as

$T = 10 s + 0.5 s = 10.5 s$

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2 years ago

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2 years ago

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An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

kolbaska11 [484]

Answer:

$mg=200.4 N$ .

Explanation:

This problem can be solved using Newton's law of universal gravitation: $F=G\frac{m_{1}m_{2}}{r^{2}}$ ,

where F is the gravitational force between two masses $m_{1}$ and $m_{2}$ , $r$ is the distance between the masses (their center of mass), and $G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2})$ is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting $w_{s}$ be the weight on the surface:

$w_{s}=mg$ ,

$mg=8*10^{2}$ ,

$m=(8*10^{2})/g$ ,

since we now that $g=9.8m/s^{2}$ we get that the mass is

$m=81.6kg$ .

Now we can use Newton's law of universal gravitation

$F=G\frac{Mm}{r^{2}}$ ,

where $m$ is the mass of the astronaut and $M$ is the mass of the earth. From Newton's second law we know that

$F=ma$ ,

in this case the acceleration is the gravity so

$F=mg$ , (<u>becarefull, gravity at this point is no longer</u> $9.8m/s^{2}$ <u>because we are not in the surface anymore</u>)

and this get us to

$mg=G\frac{Mm}{r^{2}}$ , where $mg$ is his new weight.

We need to remember that the mass of the earth is $M=5.972*10^{24}kg$ and its radius is $6.37*10^{6}m$ .

The total distance between the astronaut and the earth is

$r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6}$ meters.

Now we can compute his weigh:

$mg=G\frac{Mm}{r^{2}}$ ,

$mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}}$ ,

$mg=200.4 N$ .

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Answer:

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Explanation:

After the string breaks, the ball now disconnected from the centripetal force that was exerted via the string, continues its travel in a straight line in the direction of the tangential velocity it had at the moment the string broke.

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