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Arada [10]
3 years ago
13

Why is the earth sphered but we cant fall off

Physics
1 answer:
anygoal [31]3 years ago
3 0
Because of Gravity, Basically a force so strong it constantly pulls us to the earth with 1 G (Maybe 100 pounds of force constantly pulling us to the earth)
You might be interested in
A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.
Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

                                 M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mb*g*H = 0.5*mb*vb^2

                                  vb = √2*g*H

                                  vb = ( 2*9.81*15 ) ^0.5

                                  vb = 17.15517 m/s

- The angular momentum of system before collision is:

                                  M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

                                  M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

                                  M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

                                 M1 = M2

                                 vo = 247.03448 / (5.10*3)

                                 vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mo*g*h = 0.5*mo*vo^2

                                  h = vo^2 / 2*g

                                  h = 16.14604^2 / 2*(9.81)

                                  h = 13.3 m

3 0
3 years ago
An object of mass 100kg is raised 2m above the ground using an Inclined Plane of length 10m calculate the effort parallel to inc
MArishka [77]

Answer:

Slope = 2 m / 10 m = 1/5

For every  5 m of effort the object will be raised 1 m

W = work done on object = M g h      increase in PE of object

E S = W      where E is effort and S the distance thru which the effort acts

E S = M g H

E = 100 kg * 9.8 m/s^2 * 2 m / 10 m = 196 kg m / s^2 = 196 N

Check: total work = 2 * 9.8 * 100 = 1960 J

Force Needed = 1960 J / 2 m = 980 Newtons

Mechanical advantage = 980 / 196 = 5   as one would expect since the object is raised 1 m for every 5 m of force input

8 0
2 years ago
Which of the following describes the mass of an object precisely ? *
timurjin [86]

Answer:

2.345 would be the most precious because you have more numbers to work with and exact numbers

8 0
3 years ago
You have a source of energy containing 21 gj of energy at 600k how much this energy can be converted to work when rejecting heat
sweet [91]

Answer:

Available energy = 35 x 10⁶ J

Explanation:

Given:

Amount of energy (Q) = 21 gj = 21 x 10⁹ J

Temperature T1 = 600 k

Temperature T0 = 27 + 273 = 300k

Find:

Available energy

Computation:

Available energy = Q[1/T0 - 1/T1]

Available energy = 21 x 10⁹ J[1/300 - 1/600]

Available energy = 35 x 10⁶ J

4 0
2 years ago
a car headed north at 15 m/s accelerates for 4.25 s to reach a velocity of 28.3 m/s. What is the acceleration of the car?
Liono4ka [1.6K]

<u>Answer:</u>

The acceleration of the car is 3.13 m/s^2

<u>Explanation:</u>

In the question it is given that car initially heads north with a velocity 15 m/s. It then accelerates for 4.25 s and in the end its velocity is 28.3 m/s.

initial velocity u = 15 m/s

time t=4.25 s

final velocity v=28.3 m/s

The equation of acceleration is

a= \frac{(v-u)}{t}

= \frac{(28.3-15)}{4.25} =  \frac {13.3}{4.25} =3.13m/s^2

The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.

7 0
2 years ago
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