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2 years ago

two skateboarders of mass 50 kg and 60 kg push each other with force 70N.what is the acceleration of each skaters

Physics

1 answer:

Free_Kalibri [48]2 years ago

5 0

Answer:

0 m/s²

Explanation:

Since each skater pushes the other with a force of 70 N, according to Newton's third law, there is an equal reaction and thus the other pushes back with a force of 70 N in the other direction, so we have forces of +70N and -70 N respectively. So, the net force on each skateboarder is F = + 70 N + (-70 N) = + 70 N - 70 N = 0 N.

Since force, F = ma where a = acceleration and m = mass,

a = F/m.

So, since for each skater, F = 0N,

a = 0 N/m

= 0 m/s²

So, the acceleration of each skater is 0 m/s²

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77julia77 [94]

Answer:

See Explanation

Explanation:

m1(v1) + m2(v2)

Opposite turns the plus to subtraction.

80(8) - 120(4.0)

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7 0

2 years ago

A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit

elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

= 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

$T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }$ second

Squaring the terms and solving it for 'g'

g = 4 π² $\frac{(R+h)^{3} }{R^{2}T^{2} }$ m/s²

Substituting the values in the above equation

g = 4 π² $\frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}$

g = 4.84 m/s²

Therefore, the weight

w = m x g newton

= 5832 Kg x 4.84 m/s²

= 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N

8 0

3 years ago

In Figure 4.24, a current of 0.3 A flows through the conductor CD, and a charge of 4C passes through a cross-section AB of the

olga2289 [7]

Answer:

13.33 seconds

Explanation:

I = Q/t

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2 years ago

Read 2 more answers

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$