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3 years ago

8

An astronaut drops a rock from the top of a crater on the moon. When the rock is halfway down to the bottom of the crater, its s

peed is what fraction of its final impact speed?

2 answers:

Alexxx [7]3 years ago

8 0

Answer: vf1/vf2= 1/ sqrt(2)

Explanation :on the moon no drag force so we have only the force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics

if the rock travels H to the bottom we can calculate velocity:

vo=0m/s (drops the rock) , yo=0

vf*vf= vo*vo+2g(y-yo)

when the rock is halfway y = H/2 so:

vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)

when the rock reach the bottom y=H so:

vf2*vf2=2*g*H so vf2 = sqrt(2gH)

so vf1/vf2= 1/ sqrt(2)

good luck from colombia

Margarita [4]3 years ago

4 0

Answer:1/√2

Explanation:

the rock has lost half of its gravitational potential energy, its kinetic energy at the halfway point is half of its kinetic energy at impact. Since K.E is proportional to v^2, if K.E at halfway point is equal to 1/2 x K.E,

Then,

The K.E at impact, then the rock’s speed at the halfway point

its speed at impact will be 1/√2 its impact speed

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If a flea can jump straight up to a height of 0.410 m , what is its initial speed as it leaves the ground?

aivan3 [116]

Initial velocity = $v_0$

acceleration in the downward direction = -9.8 $\frac {m}{s^2}$

Final velocity at the highest point = 0

Maximum height reached = 0.410 m

Now, Using third equation of motion:

$\(v^2 = {v_0}^{2} + 2aH$

$\(0^2 = {v_0}^{2} - 2 \times 9.8 \times 0.410$

${v_0}^{2} = 2 \times 9.8 \times 0.410$

$v_0 = 2.834 \frac {m}{s}$

Speed with which the flea jumps = $2.834 \frac {m}{s}$

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3 years ago

the power rating of an electric lawn mower is 2200 watts. if the lawn mower was used for 30 mins, and 650 Newtons of force was u

Tasya [4]

Answer:

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Explanation:

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2 years ago

Which of the following CANNOT be

Nezavi [6.7K]

Answer:

Well, there is a kind of magnet to pick up a coin.. I think you can pick up a needle with one too.. I think safety pins. depending on what its made of though.

Explanation:

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2 years ago

Read 2 more answers

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

1 year ago

A baseball is thrown at an angle of 20° relative to the ground at a speed of 25 m/s if the ball was caught 50 m from the thrower

scoray [572]

Answer:

2.1 s

Explanation:

The motion of the ball is a projectile motion. We know that the horizontal range of the ball is

$d = 50 m$

And that the initial speed of the ball is

$u=25 m/s$

at an angle of

$\theta=20^{\circ}$

So, the horizontal speed of the ball (which is constant during the entire motion) is

$u_x = u cos \theta = 25 \cdot cos 20^{\circ} = 23.5 m/s$

And since the horizontal range is 50 m, the time taken for the ball to cover this distance was

$t=\frac{d}{u_x}=\frac{50}{23.5}=2.1 s$

which is the time the ball spent in air.

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2 years ago