The word centripeta/means "<br> -seeking."<br> Answer here

State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitati

juin [17]

F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)

6 0

3 years ago

If a ball is 10m high with what velocity will it fall?

Semmy [17]

14m/s

Explanation:

Given parameters:

Height of the ball = 10m

Unknown:

Velocity of fall or final velocity = ?

Solution:

We are going to use the appropriate equation of motion to solve this problem.

The object is falling with respect to gravity.

V² = U² + 2gH

where V is the final velocity

U is the initial velocity

g is the acceleration due to gravity 9.8m/s²

H is the height of fall

The initial velocity here is zero and

V² = 2 x 9.8 x 10 = 196

V = 14m/s

learn more:

Motion problems brainly.com/question/5248528

#learnwithBrainly

6 0

3 years ago

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh

Svetradugi [14.3K]

Answer:

The time needed is $T = 16.8 s$

Explanation:

From the question we are told that

The magnitude of the stimulated acceleration due gravity is $a = 0.5 g$

The diameter of the spaceship is $d = 35m$

Generally the force acting on the spaceship is

$F = ma$

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

$F = \frac{mv^2}{r}$

Thus

$ma = \frac{mv^2}{r}$

=> $\frac{v^2}{r} = a$

Generally the speed of this spaceship is mathematically represented as

$v = \frac{2 \pi}{T}$

=> $v^2 = [\frac{2\pi}{T}] ^2$

=> $\frac{\frac{4\pi^2 r^2}{T^2} }{r} = 0.5g$

=> $\frac{4 \pi^2 r }{T^2} = 0.5 g$

=> $T = \sqrt{ \frac{4\pi^2 r}{0.5g}}$

substituting values

$T = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}$

$T = 16.8 s$

4 0

3 years ago

Read 2 more answers

Convert 15 litre into cubic metre

atroni [7]

Answer:

0.015m^3

Explanation:

1 m^3 = 1000 liters

x m^3 = 15 liters

Cross multiply

xm^3 x 1000 l = 15 l

Divide both sides by 1000

xm^3 x1000/1000 = 15/1000

xm^3 = 0.015m^3

Therefore 15 liter = 0.015m^3

5 0

3 years ago

An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal

Ugo [173]

Answer:

The value is $T_t = 2.5659 \ s$

Explanation:

From the we are told that

The initial speed of the object is $u = 8 \ m/s$

The greatest height it reached is $h = 15 \ m$

Generally from kinematic equation we have that

$v^2 = u^2 + 2gH$

At maximum height v = 0 m/s

So

$0^2 = 8^2 + 2 * - 9.8 * H$

=> $H = 3.27 \ m$

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as

$s = h - H$

=> $s = 15 - 3.27$

=> $s = 11.73 \ m$

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

$v = u + (-g) t$

At maximum height v = 0 m/s

$0 = 8 - 9.8t$

=> $t = 0.8163 \ s$

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

$h = ut_1 + \frac{1}{2} gt_1^2$

Here the initial velocity is 0 m/s given that its the velocity at maximum height

Also g is positive because we are moving in the direction of gravity

So

$15 = 0* t + 4.9 t^2$

=> $t_1 = 1.7496$

Generally the total time taken is mathematically represented as

$T_t = t + t_1$

=> $T_t = 0.8163 + 1.7496$

=> $T_t = 2.5659 \ s$

6 0

3 years ago

The word centripeta/means " -seeking." Answer here