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alekssr [168]
3 years ago
13

What is gravitational force

Physics
2 answers:
Salsk061 [2.6K]3 years ago
5 0
Gravitational forces are forces of attraction. It's like the Earth pulling on you and keeping you on the ground.
sdas [7]3 years ago
4 0
What keeps us on the ground and doesn't let us float in the air
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How does a toaster oven use micro waves
vekshin1
It uses microwaves as little heatwaves and uses them to heat the food
5 0
3 years ago
At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to (a) 2 times that of the Celsius and (b) 1/4 ti
castortr0y [4]

Answer:

(a) F = 320

(b) = F = -5.1625

Explanation:

The formula that converts degree Celsius (C) to degree Fahrenheit (F) is:

F = 1.8C + 32

Solving (a): F = 2C

Substitute 2C for F in the above equation

F = 1.8C + 32

2C = 1.8C + 32

Collect like terms

2C - 1.8C = 32

0.2C = 32

Multiply both sides by 5

5 * 0.2C = 32 * 5

C = 160

Recall that F = 2C

F = 2 * 160

F = 320

Solving (b): F = ¼C

Substitute ¼C for F in the above formula

F = 1.8C + 32

¼C = 1.8C + 32

Convert fraction to decimal

0.25C = 1.8C + 32

Collect like terms

0.25C - 1.8C = 32

-1.55C = 32

Divide both sides by -1.55

C = 32/(-1.55)

C = -32/1.55

C = -20.65

Recall that: F = ¼C

F = -¼ * 20.65

F = -5.1625

5 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
A 2 kg ball is moving 3 m/s when it starts rolling up a hill.
AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]

Replacing the values on the equation we have:

Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\

Now we have:

h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0
3 years ago
What situation would give you a mechanical advantage? *
const2013 [10]
D. Using a fixed-pulley system
3 0
3 years ago
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