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Lemur [1.5K]
4 years ago
9

Find the magnitude of the gravitational force a 68.4 kg person would experience while standing on the surface of earth with a ma

ss of 5.98 à 1024 kg and a radius of 6.37 à 106 m. the universal gravitational constant is 6.673 à 10â11 n · m2 /kg2 . answer in units of n.
Physics
1 answer:
luda_lava [24]4 years ago
6 0
Given: Mass of a person M = 68.4 Kg.

           Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

           G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: F = ?

Formula: F = GMeM/r²

               F = (6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴)(68.4 Kg)/(6.37 x 10⁶ m)²

               F = 672.41 N




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Where V represents electric potential, K is coulomb constant, q  is Charge and r is distance between any  two around charge to the point charge.

Electric potential at O due to four charges is given by,

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a bends each wavelength of white light slightly differently so that each wavelength color comes out separated forming a rainbow
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A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

Horizontal component of force(Normal reaction) :

mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07

Therefore, the coefficient of static friction between the box and the plane is 1.07.

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3 years ago
A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w
Scilla [17]

Answer:

v_f = 20 m/s

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2

now for pure rolling condition we will have

v = R\omega

also we have

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now by work energy theorem we can say

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842 J = mv_f^2 - mv_i^2

842 = 3(v_f^2) - 3\times 11^2

now solve for final speed

v_f = 20 m/s

3 0
3 years ago
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