2. Which is NOT a unit of speed?

A ball with a mass of 0.585 kg is initially at rest. It is struck by a second ball having a mass of 0.420 kg , initially moving

Y_Kistochka [10]

Answer:

a) $(v_1)=0.3989m/s$

b) $\theta_1=80.5 \textdegree$

c) $K.E=0.036J$

Explanation:

From the question we are told that:

Initial speed of 1st ball $u_{1}=0 m/s$

Mass of 1st ball $m_1=0.585kg$

Mass of 2nd ball $m_2=0.420kg$

Initial speed of 2nd ball $u_{2}=0.270 m/s$

Final speed of 2nd ball $v_{2}=0.220 m/s$

Angle of collision $\angle=36.9 \textdegree$

a)

Generally the equation for law of conservation is mathematically given by

$m_1u_1+m_2u_2=m_1v_1^2+m_2v_2^2$

The final velocity $v_1$ is given as

$0+(0.420)(0.270)=(0.585)(v_1)^2+(0.420)(0.220)^2$

$(v_1)^2=\frac{(0.420)(0.270)-(0.420)(0.220)^2}{0.585}$

$(v_1)^2=0.1591$

$(v_1)=0.3989m/s$

b)

Generally the equation for law of conservation is mathematically given by

$m_1u_1+m_2u_2=m_1v_1cos\theta_1+m_2\theta_2$

$0+(0.420)(0.270)=(0.585)(1.511)cos\theta_1+(0.420)(0.220)cos36 \textdegree$

$cos\theta_1= \frac{(0.420)(0.270)-(0.420)(0.220)cos36 \textdegree}{(0.585)(0.3989)}$

$cos\theta_1=0.1656$

$\theta_1=80.5 \textdegree$

c)

Generally the equation for kinetic energy is mathematically given by

$K.E=\frac{1}{2} mv^2$

1st Ball

$K.E=\frac{1}{2} (0.585)(0.3989)^2$

$K.E=0.0465J$

2nd ball

$K.E=\frac{1}{2} (0.420)(0.220)^2$

$K.E=0.101J$

Therefore the change in the total kinetic energy of the two balls as a result of the collision is

$0.101-0.0465$

$K.E=0.036J$

3 0

3 years ago

What is the pressure if a force of 200 N is applied over an area of

Gelneren [198K]

The pressure is $4 N / c m^{2}$ .

Answer: Option A

<u>Explanation:</u>

Pressure defines the force acting per single area. So depending upon the amount of force acting over a given area in the perpendicular direction of the object, the pressure acting on that area will vary.

So the pressure is directly proportionate to the applied force on a particular region and inversely proportionate into the areas of the region. Hence, the unit of pressure is considered as $N / c m^{2}$ or Pascal. Pressure can also be termed as the number of air molecules present in a specified closed region. Thus, for the present condition

$\text { Pressure }=\frac{\text { Force }}{\text { Area }}$

Thus, by appling the given values in the above equation, we get,

$\text { Pressure }=\frac{200}{50}$ = $4 N / c m^{2}$

So the pressure is $4 N / c m^{2}$ .

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goodluck my friend :-)

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