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katrin2010 [14]

3 years ago

13

The gravitational attraction between a large object and a small object is...

1 answer:

Marina86 [1]3 years ago

5 0

Answer:

A) Greater than the attraction between two small objects the same distance apart.

Explanation:

The gravitational force between two objects is:

F = GMm / r²

where G is the gravitational constant,

M is the mass of one object,

m is the mass of the other object,

and r is the distance between the objects.

If the distance is the same, then two large objects will have a larger gravitational force between them than two small objects.

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Gauss’ law: a. Relates the surface charge density to the electric field.b. Relates the electric field at points on a closed surf

Mamont248 [21]

Answer:

b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface

Explanation:

Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.

8 0

3 years ago

kati45 [8]

Answer:

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8 0

3 years ago

Assuming there are no accidents or delays, the distance that a car travels down the interstate

Ganezh [65]

Explanation:

The distance that a car travels down the interstate can be calculated with the following formula:

Distance = Speed x Time

(A) Speed of the car, v = 70 miles per hour = 31.29 m/s

Time, d = 6 hours = 21600 s

Distance = Speed x Time

D = 31.29 m/s × 21600 s

D = 675864 meters

or

$D=6.75\times 10^5\ m$

(b) Time, d = 10 hours = 36000 s

Distance = Speed x Time

D = 31.29 m/s × 36000 s

D = 1126440 meters

or

$D=1.12\times 10^6\ m$

(c) Time, d = 15 hours = 54000 s

Distance = Speed x Time

D = 31.29 m/s × 54000 s

D = 1689660 meters

or

$D=1.68\times 10^6\ m$

Hence, this is the required solution.

7 0

3 years ago

Pls help me... taking a quiz.

saw5 [17]

Guess I recommend doing that

7 0

3 years ago

A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel

trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2$

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s$

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

$1\ mile=1609.34\ m$

$1\ h=3600\ seconds$

$18.03\times \frac{3600}{1609.34}=40.33\ mph$

Speed of the boat by when it will hit the dock is 40.33 mph

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m$

The distance at which the boat will have to start decelerating is 312.5 m

5 0

3 years ago