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Tcecarenko [31]
2 years ago
15

What units must the constant G have in order for the u it’s F to be newtons (N)?

Physics
1 answer:
EleoNora [17]2 years ago
5 0

The units for G must be [N][m^2][kg^{-2}]

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

F is the force

G is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between the objects

We know that:

  • The units of F are Newtons (N)
  • The units of m_1,m_2 are kilograms (kg)
  • The units of r are metres (m)

So, we can rewrite the equation in terms of G, to find its units:

G=\frac{Fr^2}{m_1 m_2}=\frac{[N][m]^2}{[kg][kg]}=[N][m^2][kg^{-2}]

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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Answer:

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) =?

Initial velocity (U) =?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V)/time(t)

a = V/t

Velocity (V) = distance (s) /time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √(s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) =..?

t = √(s/a)

t = √(256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

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Time (t) = 3.14 sec

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0 = U – 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s

7 0
3 years ago
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
IrinaK [193]

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

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