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3 years ago

If an object undergoes a change in momentum of 10 kg m/s in 3 s ,then the force acting on it is

Physics

1 answer:

Paha777 [63]3 years ago

8 0

Answer:

Force = 3.333 Newton

Explanation:

Given the following data;

Change in momentum = 10 Kgm/s

Time = 3 seconds

To find the force acting on it;

In Physics, the change in momentum of a physical object is equal to the impulse experienced by the physical object.

Mathematically, it is given by the formula;

Force * time = mass * change in velocity

Impulse = force * time

Substituting into the formula, we have;

10 = force * 3

Force = 10/3

Force = 3.333 Newton

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3 years ago

What sport does not require a high level of fitness

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2 years ago

Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 350 newtons stretches a sprin

Bingel [31]

Answer:

Explanation:

350 N force stretches the spring by 30 cm

spring constant K = 350 / 0.30 = (350 / 0.3) N / m

To calculate work done by a spring force we proceed as follows

spring force when the spring is stretched by x = Kx

This force is variable so work done by it can be calculated by integration

Work done by it in stretching from x₁ to x₂

W = ∫ F dx

= ∫ Kx dx with limit from x₁ to x ₂

= 1/2 K ( x₂² - x₁² )

Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m

Work done

= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )

= 227.50 J

5 0

3 years ago

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$

B) the speed of the sphere with mass 107.0 kg is

$v_2=\frac{m_1v_1}{m_2}$

$=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s$

C) the magnitude of the relative velocity with which one sphere is

$v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s$

D) the distance of the centre is proportional to the acceleration

$\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962$

Thus,

$x_1=3.962x_2$

and

$x_2=0.252x_1$

When the sphere make contact with eachother

Therefore,

$x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r$

And

$x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r$

The point of contact of the sphere is

$32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m$

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3 years ago

Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp

maxonik [38]

Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = $\dfrac{N}{C}m$

c = Speed of light = m/s

Momentum is given by

$\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s$

The unit of MeV/c in SI fundamental units is kg m/s

5 0

3 years ago