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Naddik [55]
2 years ago
9

A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height of 3200 km fr

om the earth's surface? (Radius of the earth is 6400 km)​
Physics
1 answer:
natali 33 [55]2 years ago
4 0

Answer:

you are kind of better than I am I don't read much but with brainly I will be intelligent

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The wheels of a car have radius 12 in. and are rotating at 600 rpm. Find the speed of the car in mi/h.
guajiro [1.7K]

Answer:

21.4 mph

Explanation:

Circumference of tire in FEET   = pi * d =  pi * 1 ft = pi  feet

pi feet  x  600 rot/min  *  60 min /hr  *  1 mile / 5280 feet = 21.4 mph

7 0
1 year ago
How much of earths gravitational force acts on everyone and everything onboard of the international space station?
liraira [26]

B) Approx. 90%

The actual value is 0.89g

4 0
3 years ago
A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni
dimaraw [331]

Answer:

The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})

Put the value into the formula

E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})

E=5.75\ N/C

The direction is toward positive x- axis.

Hence, The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

7 0
2 years ago
Please help me and show work, will mark Brainliest!!
Licemer1 [7]

Speed is distance over time, learn that formula and look at the image

3 0
3 years ago
Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

        R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

         L = W

Area  

         A = W W = W²

        R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

    R₂ / R₁ = 2W²/L

6 0
3 years ago
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