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3 years ago

What happens to the magnitude of the gravitational force as the distance between two bodies increase?

Physics

1 answer:

Anna [14]3 years ago

7 0

Answer:

The magnitude of the force will decrease

Explanation:

The gravitational force is one of the four fundamental forces of nature. It is an attractive force exerted between every object having mass.

Its magnitude is given by the equation:

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1 is the mass of the first object

m2 is the mass of the second object

r is the separation between the objects

As we see from the equation, the magnitude of the gravitational force is inversely proportional to the square of the distance between the objects:

$F\propto \frac{1}{r^2}$

Therefore, this means that as the distance between two bodies increases, the gravitational force will decrease.

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Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

5 0

3 years ago

When you do work on an object some of your energy is ____ to that object

HACTEHA [7]

Maybe the word could be converted?

8 0

3 years ago

Can the velocity of a body revese the direction when acceleration is constant?

TEA [102]

Answer:

Yes, the velocity of the object can reverse direction when its acceleration is constant. For example consider that the velocity of any object at any time t is given as: ... At At t = 0 sec, the magnitude of velocity is 2m/s and is moving in the forward direction i.e.v (t) = -2.

7 0

2 years ago

A roller coaster cart of mass m = 223 kg starts stationary at point A, where h1 = 26.8 m and a while later is at B, were h2 = 14

Tresset [83]

Answer:

vB = 15.4 m/s

Explanation:

Principle of conservation of energy:

Because there is no friction the mechanical energy is conserve

ΔE = 0

ΔE : mechanical energy change (J)

K : Kinetic energy (J)

U: Potential energy (J)

K = (1/2)mv²

U = m*g*h

Where :

m: mass (kg)

v : speed (m/s)

h : hight (m)

Ef - Ei = 0

(K+U)final - (K+U)initial =0

(K+U)final = (K+U)initial

((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:

((1/2)vB² + g*hB = (1/2 )vA²+ g*hA

(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )

(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)

(vB)² = (2)(9.8)(26.8 - 14.7)

(vB)² = 237.16

$v_{B} = \sqrt{237.16}$

vB = 15.4 m/s : speed of the cart at B

4 0

3 years ago

What is Diabetic Retinopathy? Describe it in at least 1 paragraph.

Irina18 [472]

Answer:

Diabetic Retinopathy is a form of diabetes that affects the eyes. It can be caused by damage to the retinas, and can cause permanent damage to the eyes, and even blindness. Initially the patient is asymptomatic and become more visibly affected in later stages. It can be treated if caught early, or in mild cases.

Explanation:

4 0

2 years ago