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Verdich [7]
3 years ago
14

If the mass of the Moon were suddenly to double, and the distance to the moon cut in half, how would the gravitational force bet

ween the moon and the Earth Change?
Physics
2 answers:
Andrei [34K]3 years ago
6 0

Isaac Newton compared the acceleration of the moon to the acceleration of objects on earth. Believing that gravitational forces were responsible for each, Newton was able to draw an important conclusion about the dependence of gravity upon distance. This comparison led him to conclude that the force of gravitational attraction between the Earth and other objects is inversely proportional to the distance separating the earth's center from the object's center. But distance is not the only variable affecting the magnitude of a gravitational force. Consider Newton's famous equation

Fnet = m • a

Newton knew that the force that caused the apple's acceleration (gravity) must be dependent upon the mass of the apple. And since the force acting to cause the apple's downward acceleration also causes the earth's upward acceleration (Newton's third law), that force must also depend upon the mass of the earth. So for Newton, the force of gravity acting between the earth and any other object is directly proportional to the mass of the earth, directly proportional to the mass of the object, and inversely proportional to the square of the distance that separates the centers of the earth and the object.

Alenkinab [10]3 years ago
6 0

Isaac Newton compared the acceleration of the moon to the acceleration of objects on earth. Believing that gravitational forces were responsible for each, Newton was able to draw an important conclusion about the dependence of gravity upon distance. This comparison led him to conclude that the force of gravitational attraction between the Earth and other objects is inversely proportional to the distance separating the earth's center from the object's center. But distance is not the only variable affecting the magnitude of a gravitational force. Consider Newton's famous equation

\

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Define density<br><br> Please help!!
Romashka-Z-Leto [24]

, density is how compact an object is. Put another way, density is the mass of an object divided by its volume.

5 0
4 years ago
A speed skater moving across frictionless ice at 9.2 m/s hits a 5.0 m wide patch of rough ice. She slows steadily, then continue
enot [183]

Answer:

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

Explanation:

The distance travelled on the rough ice is equal to the width of the rough ice.

distance d = 5.0 m

Initial speed u = 9.2 m/s

Final speed v = 5.8 m/s

The time taken to move through the rough ice can be calculated using the equation of motion;

d = 0.5(u+v)t

time t = 2d/(u+v)

Substituting the given values;

t = 2(5)/(9.2+5.8)

t = 2/3 = 0.66667 second

The acceleration is the change in velocity per unit time;

acceleration a = ∆v/t

a = (v-u)/t

Substituting the values;

a = (5.8-9.2)/0.66667

a = -5.099974500127

a = -5.10 m/s^2

her acceleration on the rough ice is -5.10 m/s^2

7 0
3 years ago
The plane of a rectangular coil, 3.7 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil
gizmo_the_mogwai [7]

Answer:3.77 T/s

Explanation:

Given

Area of cross-section=3.7\times 3.7 cm^2=13.69 cm^2

N=no of turns=61

Resistance=R=8 \Omega

current (i)=0.04 A

emf induced=i\times R=0.04\times 8=0.32 V

emf induced e=NA\frac{\mathrm{d} B}{\mathrm{d} t}

0.32=62\times 13.69\times 10^{-4}\times \frac{\mathrm{d} B}{\mathrm{d} t}

\frac{\mathrm{d} B}{\mathrm{d} t}=\frac{0.32\times 10^4}{848.78}=3.77 T/s

8 0
3 years ago
In 1939 or 1940, Emanuel Zacchini took his human-cannonball act to an extreme. After being shot from a cannon, he soared over th
Arlecino [84]
 We first determine the vertex by using the formula,<span>-b/2a = vertex, in order to get the values for the t-coordinate. That is why we got 
</span>
v_y=26.5 sin(53)=21.163v_x=26.5 cos(53)=15.948 
then

let x=0since you are going to land on a 3m tally=-.5(9.8)t^2+ 21.163*t
y=0=-4.9t+21.163t=4.31
vx*4.31= total distance travelled=68.88m
Then for the first wheel, you have 15.948m=vxdetermine the time when he reaches 23 meters, that is

23/15.948=1.44218 sec

substitute t with1.44218 sec, then determine the height.
h(1.44218)=20.329

determine vertex by using a graphing calculatort=2.1594s h=22.85m
using the time value of the vertex, determine horizontal distance travelled
34.438m away from cannon
8 0
3 years ago
A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t
julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

F_k = F_{W,E}

kx_1 = mg

The extension of the spring due to the weight of the object on Moon is a value of x_2, then

kx_2 = mg_m

Recall that gravity on the moon is a sixth of Earth's gravity.

kx_2 = m\frac{g}{6}

kx_2 = \frac{1}{6} mg

kx_2 = \frac{1}{6} kx_1

x_2 = \frac{1}{6} x_1

We have that the displacement at the earth was x_1 = 0.3m, then

x_2 = \frac{1}{6} 0.3

x_2 = 0.05m

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0
3 years ago
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