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1 year ago

A particle moving at speed 0. 32 c has momentum p0. the speed of the particle is increased to 0. 72 c. what is its momentum now?

1 answer:

6 0

A particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c then its momentum would be 2.25P₀.

<h3>What is momentum?</h3>

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

Mathematically the formula of the momentum is

P = mv

where P is the momentum of the particle

m is the mass of the particle

v is the velocity by which the particle is moving

As given in the question a particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c

by using the formula of the momentum and substituting the values of the velocity

P =mv

As mentioned in the question when the particle is moving with 0.32c velocity it has a momentum of P₀

P₀ = m*(0.32c)

If the speed of the particle is increased to 0.72 c the momentum would be

P=m*(0.72c)

by dividing the second equation from the first one

P₀/P = 0.32c/0.72c

P₀/P = 0.44

P =2.25P₀

Thus, the increased momentum of the particle would be 2.25P₀

Learn more about momentum from here

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3 years ago

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A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

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Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

3 years ago

A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of

ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

$v^2=u^2+2gh$

$h =\dfrac{v^2}{2g}$ ....(I)

For second half,

We need to calculate the time

Using equation of motion

$h =ut+\dfrac{1}{2}gt_{2}^2$

$h=0+\dfrac{1}{2}gt_{2}^2$

$t_{2}=\sqrt{\dfrac{2h}{g}}$

Put the value of h from equation (I)

$t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}$

$t_{2}=\dfrac{v}{g}$

According to question,

$t_{1}+t_{2}=10$

$t_{1}=t_{2}$

Put the value of t₁ and t₂

$\dfrac{v}{g}+\dfrac{v}{g}=10$

$\dfrac{2v}{g}=10$

$v=\dfrac{10\times g}{2}$

Here, g = 10

The initial velocity is

$v=\dfrac{10\times10}{2}$

$v=50\ m/s$

Hence, The initial velocity is 50 m/s.

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3 years ago

If the mass of an object stays the same and you apply more force what will happen to the objects motion

vovikov84 [41]

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3 years ago

The area of the effort and load of the Piston of a hydrolic are 0.5 and 5m respectively. If a force of 100 Newton is applied on

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Answer:

Explanation:

Using Pascal laws, which states that pressure are the input equals the pressure at the output.

Pressure is given as force/area

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Cross multiply

F1A2=F2A1

Given that

Ae=0.5m² area of effort

Al=5m² area of load

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Fl/Al=Fe/Ae

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3 years ago