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3 years ago

A 1.5-kg ball is throw at 10 m/s. What is the balls momentum?

Physics

1 answer:

alexdok [17]3 years ago

3 0

Answer : The momentum of ball is, 15 kg.m/s

Explanation :

Momentum : It is defined as the motion of a moving body. Or it is defined as the product of mass of velocity of an object.

Formula of momentum is:

where,

p = momentum = ?

m = mass = 1.5 kg

v = velocity = 10 m/s

Now put all the given values in the above formula, we get:

Therefore, the momentum of ball is 15 kg.m/s

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An electric dipole consists of a particle with a charge of 6 x 10–6 c at the origin and a particle with a charge of –6 x 10–6 c

mote1985 [20]

An electric dipole consists of a particle with a charge of 6 x 10⁻⁶ c at the origin and a particle with a charge of –6 x 10⁻⁶ c on the x axis at x = 3 x 10⁻³ m. Its dipole moment is 18 x 10⁻⁹ Cm

Dipole moment of a dipole is dependent on the charge of the dipole and the distance between the two charges.

Electric Dipole consists of two charges which are equal and opposite in charge i.e. positive and negative charges.

Given,

Dipole moment, p = ?

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Distance between charges, d = 3 x 10⁻³ m

Dipole moment (p) is given by:

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p = 6 x 10⁻⁶ x 3 x 10⁻³ Cm

p = 18 x 10⁻⁹ Cm

The dipole moment for the given charge configuration is 18 x 10⁻⁹ Cm

Learn more about Dipole moment here, brainly.com/question/14058533

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2 years ago

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An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

poizon [28]

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$

=> $b = 0.08 - 0.05 9$

=> $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is

$k * b - mg = 0$

=> $k * 0.021 - 2 * 9.8 = 0$

=> $k = 933 \ N/m$

$A = 0.30 * \sqrt{\frac{2}{933} }$

=> $A = 0.014 \ m$

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