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gtnhenbr [62]
3 years ago
12

If your mass is 63.7kg and standing 7.5m away from a boulder with a mass of 9750.6kg what is the gravitational force?

Physics
1 answer:
Anna11 [10]3 years ago
3 0

Answer:

7.37 x 10^--7 N

Explanation:

Did it on A P E X

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What is a Pulley why is it used​
aleksley [76]

Answer:

A simple pulley is a wheel with a rope that allows you to pull one end and have it lift whatever is on the other end. A modern, common example of this is a crane, often used in construction.

Explanation:

4 0
3 years ago
Someone plzzz helpppppp with this last question
vredina [299]

Answer:

I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL

C IS CORRECT

7 0
3 years ago
So, RCF, or relative centrifugal force should be equal to RCF = <img src="https://tex.z-dn.net/?f=11%2C18%2Ar%2A%28RPM%2F1000%29
zysi [14]

After plugging all the data into the equation, the result of the relative centrifugal force (RCF)  is measured in terms of g.

<h3>What is relative centrifugal force?</h3>

The relative centrifugal force (RCF) or the g force is the radial force generated by the spinning rotor as expressed relative to the earth's gravitational force.

RCF = ac/g

where;

  • ac is centripetal acceleration
  • g is acceleration due to gravity

RCF = \frac{\omega ^2 r}{g} = 1.118\times 10^{-5} \ (RPM)^2 r = 11.18r\ (RPM/1000)^2

where;

  • r is radius in cm

<h3>For example, </h3>

Find the maximum RCF of the JS-4.2 rotor can be obtained from its maximum speed (4200 rpm) and its rmax (250 mm);

RCF = 11.18 \times 25\ cm \times (\frac{4200 \ RPM}{1000} )^2 = 4,930.3 \times g

Thus, after plugging all the data into the equation, the result is measured in terms of g.

Learn more about relative centrifugal force here: brainly.com/question/26887699

#SPJ1

6 0
2 years ago
Help me please this is for physics
Yuri [45]
<h2>Hello there! :)</h2>

It's a pleasure to be helping you today with your<u> physics question!</u>

Answer:

23.1m/s

Explanation:

We want to find the initial speed of the ball.

To do this, we have to apply the formula for the time of flight of a projectile:

T=\frac{2_{v0~sin 0} }{g}

where θ = angle of flight

g = acceleration due to gravity

v0 = initial speed

Therefore, substituting the given values into the formula, we have that:

\boxed{4.2=\frac{2~x~_{v0~sin63} }{9.8}}

⇒ 2 ×_{v0} ×0.8910= 9.8 × 4.2

⇒\boxed{{v0}=\frac{9.8~times~4.2}{2~times~0.8910}}

\boxed{{v0} =23.1m/s}

That is the initial speed of the ball.

<em />

<em>I hope this helps you!</em>

<em>Good Luck with your Assignment!</em>

3 0
3 years ago
A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m verticall
lorasvet [3.4K]

Explanation:

Given:

t = 20 seconds

x = 3000 m

y = 450 m

a) To find the vertical component of the initial velocity v_{0y}, we can use the equation

y = v_{0y}t - \frac{1}{2}gt^2

Solving for v_{0y},

v_{0y} = \dfrac{y + \frac{1}{2}gt^2}{t}

\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}

\:\:\:\:\:\:\:=120.5\:\text{m/s}

b) We can solve for the horizontal component of the velocity v_{0x} as

x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}

or

v_{0x} = 150\:\text{m/s}

4 0
3 years ago
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