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Alexandra [31]
3 years ago
11

Two skaters stand facing each other. One skater's mass is 60 kg, and the other's

Physics
1 answer:
jeyben [28]3 years ago
3 0

Answer:

v' = 2.4 m/s

Explanation:

Given that,

Mass of one skater, m = 60 kg

Mass of the other's skater, m' = 60 kg

The two skaters push off each other. After the push, the smaller  skater has a velocity of 3.0 m/s.

When there is no external force acting on a system, the momentum remains conserved. It means initial momentum is equal to the final momentum. Let v' is the velocity of the larger skater.

mv = m'v'

v'=\dfrac{mv}{m'}\\\\v'=\dfrac{60\times 3}{75}\\\\v'=2.4\ m/s

So, the velocity of the larger skater is 2.4 m/s.

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anastassius [24]

Answer:

Option (c)

Explanation:

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The bullet which is fired from the gun has some initial velocity but the bullet which is dropped has zero initial velocity.

the acceleration is acting on both the bullets which is equal to the acceleration due to gravity and they both in motion in the influence of gravity.

8 0
3 years ago
A square sheet of rubber has sides that are 20 cm long. What is the area of the square of rubber in cm squared?
Alinara [238K]

Answer:

400cm^2

Explanation:

sides are 20cm long Area for a square is a squared

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Hope this helps :)

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3 years ago
Why do nations should establish a set of rules and principles for responsible lunar/moon explortions? ASAP pleaseeee :(
Marianna [84]

So that we do not contaminate it with microorganisms or garbage or other human stuff.

5 0
3 years ago
A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

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k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

8 0
3 years ago
Which parts of the graph would increase if all of earth's polar ice melts? Which part would decrease?
Sedbober [7]

Answer:

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8 0
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