I need help with these two not sure on them

A 1250 kg car is stopped at a traffic light. A 3550 kg truck moving at 8.33 m/s hits the car from behind. If bumpers lock, how f

Radda [10]

Answer:

the two vehicles will be moving at a speed of 6.16 m/s

Explanation:

This is a case of completely inelastic collision, therefore, the conservation of momentum can be written as:

$m_1\,v_1+m_2\,v_2=(m_1+m_2)\,v_f$

which given the information provided results into:

$m_1\,v_1+m_2\,v_2=(m_1+m_2)\,v_f\\(1250)\,(0)+(3550)\,(8.33)=(1250+3550)\,v_f\\29571.5=4800\,v_f\\v_f=6.16\,\,m/s$

7 0

3 years ago

A father racing his son has 1/3 the kinetic energy of the son, who has 1/4 the mass of the father. The father speeds up by 1.5 m

Feliz [49]

Explanation:

Let the speeds of father and son are $v_f\ and\ v_s$ . The kinetic energies of father and son are $K_f\ and\ K_s$ . The mass of father and son are $m_f\ and\ m_s$

(a) According to given conditions, $K_f=\dfrac{1}{3}K_s$

And $m_s=\dfrac{1}{4}m_f$

Kinetic energy of father is given by :

$K_f=\dfrac{1}{2}m_fv_f^2$ .............(1)

Kinetic energy of son is given by :

$K_s=\dfrac{1}{2}m_sv_s^2$ ...........(2)

From equation (1), (2) we get :

$\dfrac{v_f^2}{v_s^2}=\dfrac{1}{12}$ ..............(3)

If the speed of father is speed up by 1.5 m/s, so the ratio of kinetic energies is given by :

$\dfrac{K_f}{K_s}=\dfrac{1/2m_f(v_f+1.5)^2}{1/2m_sv_s^2}$

$v_s^2=4(v_f+1.5)^2$

Using equation (3) in above equation, we get :

$v_f=\dfrac{1.5}{\sqrt3-1}=2.04\ m/s$

(b) Put the value of $v_f$ in equation (3) as :

$v_s=7.09\ m/s$

Hence, this is the required solution.

8 0

4 years ago

Which statement is true?

miv72 [106K]

B might be the correct answer

8 0

3 years ago

A quantity found by multiplying the force by the distance moved

Andrej [43]

The quantity that is calculated from the product of the force and the distance traveled due to the force is called work. It has SI units of Joules (J) which is equivalent to Newton-meter (N-m). It is the energy that happens when an object is being moved by an external force.

7 0

3 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$