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Lelu [443]
3 years ago
8

Two masses are connected by a string which passes over a pulley with negligible mass and friction. One mass hangs vertically and

one mass slides on a horizontal surface. The horizontal surface has a coefficient of kinetic friction of 0.200. The vertically hanging mass is 3.00 kg and the mass on the horizontal surface is 3.00 kg. The magnitude of the acceleration of the vertically hanging mass is (the initial velocity of the horizontal mass is to the right)
Physics
1 answer:
monitta3 years ago
3 0

Answer:

a=2,5m/s^2

Explanation:

From the question we are told that:

Coefficient of kinetic friction \mu= 0.200

Vertical Mass M_v=3kg

Horizontal mass M_h=3.00kg  

Generally the equation for kinetic force F_k is mathematically given by

F_k=\mu N\\F_k=0.2*3\\F_k=0.6

Generally the equation for T is mathematically given by

For M_v=3kg3g-T=3a

For M_h=3kg

T=M_v V+F_k\\T=3.0a+0.6

Therefore substituting

3-3a-0.6=3a\\2.4g=6a

a=2,5m/s^2

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Answer:

60

Explanation:

Translation -

A book weighing 12 N is balanced on a table. Knowing that the static friction coefficient is 0.5, how much is the friction force worth?

Friction force is

f = u * n

f = 0.5 * 12N

f = 60

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What is the characteristic that allows an object to have kinetic energy?
Shtirlitz [24]
Kinetic energy is movement, thus movement would be a characteristic that allows an object to have kinetic energy
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Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it
olga2289 [7]
<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>
6 0
3 years ago
What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?
viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> V_{E} due to a point charge is expressed as:

V_{E}=k\frac{q}{r}

Where:

k=9(10)^{9}\frac{Nm^{2}}{C^{2}} is the <u>electric constant</u>

q=1.6(10)^{-19}C is the <u>electric charge of the hydrogen nucleus</u>, which is positive

r=5.292(10)^{-11}m is the <u>distance</u>

Rewritting the equation with the known values:

V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}

Finally:

V_{E}=27.21 V

5 0
3 years ago
Two identical satellites are in orbit about the earth. One orbit has a radius r and the other 2r. The centripetal force on the s
velikii [3]

Answer:

the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

Explanation:

Mass of satellite, m

orbit radius of first, r1 = r

orbit radius of second, r2  = 2r

Centripetal force is given by

F= \frac{mv^{2}}{r}

Where v be the orbital velocity, which is given by

v=\sqrt{gr}

So, the centripetal force is given by

F= \frac{mgr}}{r}}=mg

where, g bet the acceleration due to gravity

g=\frac{GM}{r^{2}}

So, the centripetal force

F= \frac{GMm}}{r^{2}}}

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F= \frac{GMm}{4r^{2}} .... (1)

Gravitational force on the satellite having smaller orbit

F'= \frac{GMm}{r^{2}} .... (2)

Comparing (1) and (2),

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So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

8 0
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