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2 years ago

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An electric heater,rated 15 volts 40 watts fitted into a metal block supplies heat to the block of mass 1.5kg and specific heat

capacity of 460. calculate the temperature rise of in the block if the current flows for 10 minutes

2 answers:

Ghella [55]2 years ago

7 0

The temperature rises of in the block if the current flows for 10 minutes is, 34.78K.

To find the answer, we need to know about the power of the electric heater.

<h3>How to find the temperature rise of in the block if the current flows for 10 minutes?</h3>

We have given with the variables,

$P=40 W\\m=1.5kg\\S=460\\t=10minutes=10*60s=600s$

We have an expression the heat given to the system as,

$P*t= mS$ ΔT

We have to find the temperature rise ΔT,

ΔT $=\frac{P*t}{mS} =\frac{40*600}{1.5*460} =34.78K$

Thus, we can conclude that, the temperature rises in the block if the current flows for 10 minute is 34.78K.

Learn more about the power of the electric heater here:

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notsponge [240]2 years ago

5 0

If the current flows for 10 minutes, the temperature in the block rises by 34.78 K.

We must understand the electric heater's power in order to determine the solution.

<h3>How can you determine the block's temperature increase if the current flows for ten minutes?</h3>

With the variables that we have provided,

$P=40Watts\\m=1.5kg\\s=460\\t=10 min=600s$

The heat supplied to the system is referred to as,

$Pt=ms$ ΔT

We must determine the increase in temperature, T.

ΔT $=\frac{Pt}{ms} =34.78 K$

Thus, we can infer that if the current flows for 10 minutes at 34.78K, the temperature in the block will increase.

Learn more about the effectiveness of an electric heater here:

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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

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

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4 years ago

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VashaNatasha [74]

Answer:

$a = 13.758\,\frac{m}{s^{2}}$

Explanation:

First, the instant associated to the angular displacement is:

$(1.10\,\frac{rad}{s} )\cdot t + (6.30\,\frac{rad}{s^{3}} )\cdot t^{2} - 0.628\,rad = 0$

Roots of the second-order polynomial are:

$t_{1} \approx 0.240\,s, t_{2} \approx -0.415\,s$

Only the first root is physically reasonable.

The angular velocity is obtained by deriving the angular displacement function:

$\omega (0.240\,s) = 1.10\,\frac{rad}{s}+ (12.6\,\frac{rad}{s^{2}})\cdot (0.240\,s)$

$\omega (0.240\,s) = 4.124\,\frac{rad}{s}$

The angular acceleration is obtained by deriving the previous function:

$\alpha (0.240\,s) = 12.6\,\frac{rad}{s^{2}}$

The resultant linear acceleration on the rim of the disk is:

$a_{t} = (0.650\,m)\cdot (12.6\,\frac{rad}{s^{2}} )$

$a_{t} = 8.190\,\frac{m}{s^{2}}$

$a_{n} = (0.650\,m)\cdot (4.124\,\frac{rad}{s} )^{2}$

$a_{n} = 11.055\,\frac{m}{s^{2}}$

$a = \sqrt{a_{t}^{2}+a_{n}^{2}}$

$a = 13.758\,\frac{m}{s^{2}}$

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Answer:

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ad-work [718]

Answer:

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3 years ago