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3 years ago

Which statement BEST explains why the specific heat of water is higher than the specific heat of most other substances?

2 answers:

5 0

Due to its polarity and hydrogen bonding water can absorb heat without a significant temperature change.. The high specific heat of water helps regulate the rate at which air changes temperature, which is why the temperature change between seasons is gradual instead of sudden, especially near the oceans.

jeka57 [31]3 years ago

5 0

<span>A) Due to its polarity and hydrogen bonding water can absorb heat without a significant temperature change.

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A railroad freight car of mass 3.34 × 104 kg collides with a stationary caboose car. They couple together, and 19.0% of the init

Tresset [83]

Answer:

4.123 * 10∧4 kg

Explanation:

mass of car M' = 3.34 * 10∧4 kg

energy loss E = 19/100 K

from law of conservation of momentum ,

M' V' = ( M' + M'' ) v

V = M' V' / ( M' + M'' )

Initial kinetic energy is K' = (M' V')² / 2

final kinetic energy K" = 81 K' /100

= (M' V')² 81 / (200) = ( M' + M'' ) v²/ 2

therefore , M' / ( M' + M" ) = 0.81

mass of caboose is , M" = 1.234 M' - M'

M" = .234 M'

= 0,234 ( 3.34 * 10∧4 kg)

= 4.123 * 10∧4 kg

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3 years ago

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3 years ago

Look at the quote: "Newton's laws of motion can be summarized in two simple ideas...." According to Newton's laws of motion, the

sertanlavr [38]

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3 years ago

A cell phone manufacturer uses a trendy celebrity to advertise its product to teenagers. The company is relying on the route to

alina1380 [7]

Answer:

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Explanation:

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3 years ago

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

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

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3 years ago