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2 years ago

A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s.

2 answers:

8 0

Momentum = mass x velocity

Before collision
Momentum 1 = 2 kg x 20 m /s = 40 kg x m/s
Momentum 2 = 3 kg x -10m/s = -30 kg x m/s

After collision
Momentum 1 = 2 kg x -5 m/s = -10 m/s
Momentum 2 = 3 kg x V2 = 3V2

Total momentum before = total momentum after
40 + -30 = -10 + 3V2
V2 = 6.67 m/s

Total kinetic energy before
= (1/2) [ 2 kg * 20 m/s * 2 + 3 kg * ( -10 m/s) *2 ]
= 550 J

Total kinetic energy after
= (1/2) [ 2 kg * ( - 5 m/s) * 2 + 3 kg * 6.67 m/s *2 ]
= 91.73 J

Total kinetic energy lost during collision
=550 J - 91.73 J
= 458.27 J

Stells [14]2 years ago

4 0

Kinetic energy before:

E = 0.5mv^2 + 0.5mv^2

= 0.5*2*20^2 + 0.5*3*(-10)^2

= 400 + 150

= 550J

Kinetic energy after:

E = 0.5mv^2

= 0.5*2*(-5)^2

= 25J

Kinetic energy lost during the collision:

E = 550 - 25

= 525J

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6–23 an automobile engine consumes fuel at a rate of 22 l/h and delivers 55 kw of power to the wheels. if the fuel has a heating

Anna007 [38]

Explanation & answer:

Given:

Fuel consumption, C = 22 L/h

Specific gravity = 0.8

output power, P = 55 kW

heating value, H = 44,000 kJ/kg

Solution:

Calculate energy intake

E = C*P*H

= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)

= (22/3600)*1000*0.8*44000 j/s

= 215111.1 j/s

Calculate output power

P = 55 kW

= 55000 j/s

Efficiency

= output / input

= P/E

=55000 / 215111.1

= 0.2557

= 25.6% to 1 decimal place.

8 0

2 years ago

A gymnast dismounts the uneven parallel bars with some angular momentum about her transverse axis. Just after release, she is in

Luda [366]

Answer:

a. Her moment of inertia increases and she rotates slower.

Explanation:

As we know that initially when she starts her motion she is in piked position due to which her whole mass is concentrated near the axis of rotation

So here the rotational Inertia of her body will be smaller

Now when is comes closer to the position of landing she extends into layout position due to which her mass will move away from the axis of rotation

Due to this the rotational inertia of her body will increase

now we know that there is no external torque on the system

so here angular momentum must be conserved

So we will have

$I\omega = constant$

so if rotational inertia is increasing then angular speed must be slower

so correct answer will be

a. Her moment of inertia increases and she rotates slower.

7 0

2 years ago

A small rivet connecting two pieces of sheet metal is being clinched by hammering. Determine the impulse exerted on the rivet an

kykrilka [37]

Answer:

a) the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

b) the impulse exerted by the rivet when the anvil has a support weight of 9 lb = 0.799 lb.s

the energy absorbed by the rivet under each blow when the anvil has a support weight of 9 lb is = 7.99 ft.lb

Explanation:

The first picture shows a schematic view of a free body momentum diagram of the hammer head and the anvil.

Using the principle of conservation of momentum to determine the final velocity of anvil and hammer after the impact; we have:

$m_Hv_H + m_Av_A = m_Hv_2+m_Av_2$

From the question given, we can deduce that the anvil is at rest;

∴ $v_A = 0$ ; then, we have:

$m_Hv_H + 0 = (m_H+m_A) v_2$

Making $v_2$ the subject of the formula; we have:

$v_2 = \frac{m_Hv_H}{m_H + m_A}$ ------- Equation (1)

Also, from the second diagram; there is a representation of a free body momentum of the hammer head;

From the diagram;

F = impulsive force exerted on the rivet

Δt = the change in time of application of the impulsive force

Using the principle of impulse of momentum to the hammer in the quest to determine the impulse exerted (i.e FΔt ) on the rivet; we have:

$m_Hv_H - F \delta t = m_Hv_2$

$- F \delta t = - m_Hv_H + m_Hv_2$

$F \delta t = m_Hv_H - m_Hv_2$

$F \delta t = m_H(v_H - v_2)$ ------- Equation (2)

Using the function of the kinetic energy of the hammer before impact $T_1$ ; we have:

$T_1 = \frac{1}{2} m_Hv_H^2$ -------- Equation (3)

We determine the mass of the hammer $m_H$ by using the formula from:

$W_H = m_Hg$

where;

$W_H$ = weight of the hammer

$m_H$ = mass of the hammer

g = acceleration due to gravity

Making $m_H$ the subject of the formula; we have:

$m_H = \frac{W_H}{g}$

$m_H = \frac{1.5 \ lb}{32.2 \ ft/s^2}$

$m_H = 0.04658 \ lb.s^2/ft$

Now;

$T_1 = \frac{1}{2} m_Hv_H^2$

$T_1 = \frac{1}{2}*(0.04658 \ lb.s^2 /ft) *(20 \ ft/s)^2$

$T_1 = \frac{18.632 }{2}$

$T_1 = 9.316 \ ft.lb$

After the impact $T_2$ ; the final kinetic energy of the hammer and anvil can be written as:

$T_2 = \frac{1}{2}(m_H +m_A)v^2_2$

Recall from equation (1) ; where $v_2 = (\frac{m_Hv_H}{m_H+m_A})$ ; if we slot that into the above equation; we have:

$T_2 = \frac{1}{2}(m_H +m_A)( \frac{m_Hv_H}{m_H+m_A})^2$

$T_2 = \frac{1}{2} \frac{m^2_H +v^2}{m_H+m_A}$

$T_2 = \frac{1}{2} ({m^2_H +v^2})(\frac{m_H}{m_H+m_A})$

Also; from equation (3)

$T_1 = \frac{1}{2} m_Hv_H^2$ ; Therefore;

$T_2 = T_1 (\frac{m_H}{m_H+m_A})$ ----- Equation (4)

Now; To calculate the impulse exerted by the rivet FΔt and the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support; we have the following process

First , we need to find the mass of the anvil when we have an infinite mass support;

mass of the anvil $m_A = \frac{W_A}{g}$

where we replace; $W_A \ with \ \infty$ and g = 32.2 ft/s²

$m_A = \frac{\infty}{32.2 \ ft/s}$

However ; from equation (1)

$v_2 = \frac{m_H v_H}{m_H + m_A}$

$v_2 = \frac{0.04658*20}{0.04658+ \ \infty}$

$v_2 = 0$

From equation (2)

$F \delta t = m_H(v_H + v_2)$

$F \delta t = (0.04658 lb .s^2 /ft )(20ft/s - 0)$

$F \delta t = \ 0.932 \ lb.s$

Therefore the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

For the energy absorbed by the rivet ; we have:

$T_2 = T_1 (\frac{m_H}{m_H+m_A} )$

where;

$T_1= 9.316 \ ft.lb$

$m_H = 0.04658 \ lb.s^2/ft$

$m_A = \infty$

Then;

$T_2 = (9.316 \ ft.lb) (\frac{0.04658\ lb.s^2/ft)}{0.04658 \ lb.s^2/ft+ \infty} )$

$T_2 = (9.316 \ ft.lb)* 0$

$T_2 = 0$

Then the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support

ΔT = $T_1 - T_2$

ΔT = 9.316 ft.lb - 0

ΔT ≅ 9.32 ft.lb

Therefore; we conclude that the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

Due to the broadness of this question, the text is more than 5000 characters, so i was unable to submit it after typing it . In the bid to curb that ; i create a document for the answer for the part b of this question.

The attached file can be found below.

5 0

2 years ago

A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

6 0

2 years ago

If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens woul

Brilliant_brown [7]

Answer:

3×10^7 m/s or 0.10c (e)

Explanation: If the actual value of the speed of light were to be put into consideration.

Given that the speed of light is c = 3.0×10^8m/s

The alien spaceship is approaching at the rate of 10% of the speed of light.

10% of 3.0×10^8m/s

10/100 × 3.0×10^8m/s

0.1 ×3.0×10^8m/s

3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c

7 0

2 years ago