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zubka84 [21]
3 years ago
6

you and 3 friends apply a combined force of 489.5n to push a piano. The amount of work done is 1762.2j. What distance did the pi

ano move
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
5 0

Answer: 3.6m

Explanation:

Work done = force * distance

Distance = workdone / force

D= 1762.2J / 489.5n

D= 3.6m

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konstantin123 [22]

Answer:

option A is correct because air friction is greater than gravity

Explanation:

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3 years ago
Read 2 more answers
While riding in a hot air balloon, which is steadily descending at a speed of 1.01 m/s, you accidentally drop your cell phone?
mote1985 [20]

While riding in a hot air balloon, which is steadily at a speed of 1.01 m/s, and your phone accidentally falls.

<span>(a)    </span>The speed of your phone after 4 s is:

V= u + at

V= 1.01 + (9.8)(4)

V= 40.21 m/s

<span>(b)   </span>The balloon is ____ far:

V = u + at

V= 1.01 + (9.8)(1)

V=10.81 –distance at 1 one second

V= u + at

V= 1.01 + (9.8)(2)

V= 20.61-distance at 2 seconds

V= u+ at

V= 30.41- distance at 3 seconds

V= 40.21- distance at 4 seconds

D= 102.04 m

<span>(c)    </span>If the balloon is rising steadily at 1.01 m/s:

V= -1.1 m/s

<span> </span>

5 0
3 years ago
A baseball diamond is a square (don’t be fooled that it is usually shown rotated by 45°). Each side of the square is 90 feet lon
gayaneshka [121]

A line from second to home creates a right triangle with 45 degree angles.

This makes the diagonal line the side length of the square multiplied by the sqrt(2)

Second to home = 90sqrt(2) = 127.279 feet ( round answer as needed)

8 0
3 years ago
A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 13° above the horizontal. (a) if
Blababa [14]
<span>Answer: Therefore, x component: Tcos(24°) - f = 0 y component: N + Tsin(24°) - mg = 0 The two equations I get from this are: f = Tcos(24°) N = mg - Tsin(24°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24°) = 0.47 * (mg - Tsin(24°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24°) + sin(24°) * static coefficient) Then plugging in the values... T = 283.52. Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/</span>
7 0
3 years ago
A probe orbiting Venus exerts a gravitational force of 2.58 × 10^3 N on Venus. Venus has a mass of 4.87 × 10^24 kg. The mass of
Crazy boy [7]
So this is the info that's given to us:
m1 = 4.87 x 10^24 kg
m2 = 4.87 x 10^24kg
F = 2.58 x 10^3 N

Now we can use this equation to solve: f_g= \frac{Gm_1m_2}{r^2}

You may be asking what g is. Well, it is equal to 6.67x10^-11 m3 /kg s2. G is gravitational constant. So :

2.58 x 10^3 = \frac{6.67x10^{-11} * 4.87 x 10^{24} * 655}{r^{2}} 

As you can see we only have one variable, so we solve for r! 

r = 9.08 x 10^6 km

I hope this helps you out. I remember when I was first learning Newton's law<span> of universal </span><span>gravitation and I had some trouble. But with some practice, I got better. =) </span>

6 0
3 years ago
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