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3 years ago

7

If the 50.0 kg boy were in a spacecraft 5r from the center of the earth, what would his weight be? (use the 50.0 kg boy for sign

ificant figures)

2 answers:

jeka57 [31]3 years ago

5 0

The boy weighs 50.0 kg when he is 1R from the center of the Earth. To get his weight 5R away, we can use the formula:

$W_{boy}$ = G * m1 * m2 / $r^{2}$

Where G, m1, and m2 are constants. By ratio and proportion:

$W_{boy}$ * ( $(5R)^{2}$ = 50.0 kg * 9.81 m/ * ( $(1R)^{2}$

Canceling R on both sides and by computation:

$W_{boy}$ = 19.6 N

Oduvanchick [21]3 years ago

3 0

Answer:

19.6 N

Explanation:

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Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, $t=0.50\ s$

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, $v=0\ m/s$

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, $a=g=-9.8\ m/s^2$

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

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It decreses Decreases

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Akimi4 [234]

Answer:

E = 3.8 kJ

Explanation:

Given that,

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2 years ago

A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

aivan3 [116]

Answer:

Part a)

$F_n = 306 N$

Part b)

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

$mg - F_n = ma_c$

now we will have

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

now we have

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Part b)

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$v = 12.1 m/s$

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3 years ago