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malfutka [58]
3 years ago
15

If the mass of a

Physics
1 answer:
mixer [17]3 years ago
5 0

Answer:

The final acceleration becomes (1/3) of the initial acceleration.

Explanation:

The second law of motion gives the relationship between the net force, mass and the acceleration of an object. It is given by :

F=ma

m = mass

a = acceleration

According to given condition, if the mass of a  sliding block is tripled while a constant net force is applied. We need to find how much does the acceleration decrease.

a=\dfrac{F}{m}

Let a' is the final acceleration,

a'=\dfrac{F}{m'}

m' = 3m

a'=\dfrac{F}{3m}

a'=\dfrac{1}{3}\times \dfrac{F}{m}

a'=\dfrac{1}{3}\times a

So, the final acceleration becomes (1/3) of the initial acceleration. Hence, this is the required solution.

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A 0.210-kg metal rod carrying a current of 11.0 A glides on two horizontal rails 0.490 m apart. If the coefficient of kinetic fr
sergij07 [2.7K]

Answer:

The  magnetic field is  B = 0.0764 \ T

Explanation:

From the question we are told that  

    The mass of the metal is  m  =  0.210 \ kg

     The current is  I  =  11.0 \ A

      The distance between the rail(length of the rod ) is  d  =  0.490 \ m

      The coefficient of kinetic friction is  \mu_k  =  0.200

Generally the magnetic force is mathematically represented as

      F_b  =  B *  I  *  d

Given that the rod is moving at a constant velocity, it

=>    F_b  =  F_k

Where F_k is the kinetic frictional force which is mathematically represented as

       F_k  =  \mu_k  *  m *  g

So

    B *  I  *  d =  \mu_k  * m * g

=>   B =  \frac{\mu_k  * m * g}{I  *  d }

substituting values

=>   B =  \frac{0.200  * 0.210  * 9.8 }{ 11  *  0.490 }

=>   B = 0.0764 \ T

3 0
3 years ago
a car, which has a mass of 2000kg traveled a distance of 200 meters in 5 seconds. After 20 seconds the car was raveling at a spe
iragen [17]

Answer:

Explanation:

vi = 200/5 = 40 m/s

a = (vf - vi)/t = (60 - 40)/20 = 1 m/s²

F = ma = 2000(1) = 2000 N

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2 years ago
Which describes how electromagnetic waves travel?
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Travels through solid materials, and air.... but it can also travel through the vacuum of space

Explanation:

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3 years ago
1.78 km = __m<br> Help plz
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Answer:

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Suppose we placed a positive charge Q on the Moon and an equal positive charge Q on the Earth. What value of Q would be needed t
Rudiy27

Answer: q=5.70 x 10^13 C

Explanation:

gravitational attraction = electrostatic repulsion GMm/d^2 = kQ^2/d^2 as you can see the d^2 cancel out. that is why lunar distance is irrelevant. G is the universal gravitational constant = 6.67 x 10^-11 m^3 / kgs^2 M is earth's mass = 5.972 × 10^24 kg m is moon's mass = 7.342×10^22 kg Q is charge on earth and moon. k is coulomb's constant = 9 x10^9 N m^2 /C^2 On solving equation for Q. Q = sqrt (GMm/k) = sqrt ( 6.67 x 10^-11 x 5.972 x 10^24 * 7.342×10^22 / 9 x10^9) = 5.70 x 10^13 C

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3 years ago
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