What distance is required for a train

What is the momentum of a 100-kilogram fullback carrying a football on a play at a velocity of 3.5. m/sec

Olenka [21]

Answer:

350 kg m/s

Explanation:

Momentum = mass × velocity

p = mv

p = (100 kg) (3.5 m/s)

p = 350 kg m/s

6 0

2 years ago

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Which of the following is a unit of volume of solids?

Alla [95]

Answer:

Cubic meters is the unit of volume.

Explanation:

Volume :

Volume is the multiplication of height, width and length of the solids.

In mathematically term,

$V = w\times l\times d$

Where, w = width

h = height

l = length

Put the unit into the formula

$V = m\times m\times m$

$V =m^3$

This is the unit of volume.

We know that,

Square meters is the unit of area.

Liters per cubic gram is the unit of density.

Gram per cubic centimeter is the unit of density in CGS.

Hence, Cubic meters is the unit of volume.

4 0

3 years ago

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Which three types of electromagnetic waves carry most of the Sun's energy that strikes Earth?

pshichka [43]

Answer:

A.

Explanation:

3 0

3 years ago

Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass

Alja [10]

Answer:

3947.6 N

Explanation:

Centripetal Force: This is the force that tend to moves a body towards the center of a circle during circular motion.

The formula for centripetal force is

F = mω²r ........................ Equation 1

Where F = Centripetal force, ω = angular velocity, r = radius.

Where π = 3.1415

Given: m = 4 kg, ω = 0.5 rev/s = (0.5×2π) rad/s = 3.1415 rad/s, r = 100 m.

Substitute into equation 1

F = 4(3.1415)²(100)

F = 3947.6 N

Hence the centripetal force on the turbine blade = 3947.6 N

6 0

2 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

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

3 0

2 years ago