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3 years ago

Hansel pushes a 2 lb. box 5 feet in 20 seconds. How much work has he done?

1 answer:

8 0

Work = force x distance

You can see time doesn’t matter (if we were talking about power, which is the RATE at which work is performed, that would be a different story).

W = 2 x 5 = 10 foot-pounds of work

Foot-pounds are gross units. Better to work in SI units when you can!

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Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

4 0

4 years ago

An airplane is flying north. What is the direction of the air friction force acting or

Nimfa-mama [501]

Answer: south

Explanation:

8 0

3 years ago

________ are more likely to be found near rural communities due to the large requirement for space.

nirvana33 [79]

Explanation/Answer:

Wind turbines are more likely to be found near rural communities due to the large requirement for space.

5 0

4 years ago

Read 2 more answers

A venturi meter is a device for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at a speed v2 thro

yarga [219]

Answer:

$V_2=6.17\ m/s$

$Q=0.49\ m^3/s$

Explanation:

Given that

$A_1=0.03m^2$

$A_2=0.08m^2$

We know that from continuity equation

$Q=A_1V_1=A_2V_2$

$A_1V_1=A_2V_2$

$0.3V_1=0.08V_2$

$V_1=2.66V_2$

Now from energy equation

$P_1+\dfrac{1}{2}\rho V_1^2=P_2+\dfrac{1}{2}\rho V_2^2$

$P_2-P_1=\dfrac{1}{2}\rho V_1^2-\dfrac{1}{2}\rho V_2^2$

$140=\dfrac{1}{2}\times 1.2\times 7.11\times V_2^2-\dfrac{1}{2}\times 1.2\times V_2^2$

$V_2=6.17\ m/s$

$Q=A_2V_2$

Q=6.17 x 0.08

$Q=0.49\ m^3/s$

7 0

3 years ago

A person drives north 8 blocks, then turns west and drives 4 blocks. The driver then turns south and drives 8 blocks. How could

stealth61 [152]

A person would be driving 4 blocks west from the starting point to make the shorter distance while maintaining same displacement.

<u>Explanation:</u>

The measurement of an object’s position change from its point is called displacement. It is usually calculated from starting to the end points and represented by ‘delta s’. In the given scenarios, the person drove in the way that he finals the driving by 4 blocks away from the west.

Means, the persons drive to 8 blocks north and then to 8 blocks south get cancelled. Hence, to make the shorter distance with maintaining same displacement he would be driving 4 blocks west from the starting point.

7 0

3 years ago