All categories

4 years ago

PLEASE HURRY WILL MARK BRAINLIEST IF CORRECT

Physics

1 answer:

Readme [11.4K]4 years ago

6 0

The answer to the question is 1.48 x 10^-11 J

You might be interested in

Which of the following indicates a heat transfer?

lesya [120]

Answer:

TEMPERATURE CHANGES

Explanation:

WELL ITS BASIC. WHEN TEMPERATURE CHANGES. IT MEANS THAT HEAT IS BEING TRANSFERRED AS I HEARD THAT COLD CANNOT BE TRANSFERRED

3 0

3 years ago

Echnician a says that to prevent injuries in an auto accident, all steering columns have a break-off steering wheel. technician

Oliga [24]

<span>Technician a says that to prevent injuries in an auto accident, all steering columns have a break-off steering wheel. technician b says that to prevent injuries in an accident, all steering columns are now fitted with a flexible rubber tube. Both technicians are correct. The </span>vehicle manufacturers use break away steering column mounting brackets to protect the driver in an accident. The <span>vehicle manufacturers are required to use collapsible shafts in the steering column. </span>

4 0

3 years ago

The length and mass of the arm are Larm = X1 = 50 cm and Marm = 0.3 kg, X2 = 15 cm, and the mass of the object is MObject = 0.25

max2010maxim [7]

Answer: 0.5N

Explanation: if the system is at equilibrium, sum of the torque will be equal to zero.

But if they are not in equilibrium.

U will find the difference in the two torque

find the attached file for solution

3 0

3 years ago

An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu

choli [55]

Answer:

Wnet, in, = 133.33J

Explanation:

Given that

Pump heat QH = 1000J

Warm temperature TH= 300K

Cold temperature TL= 260K

Since the heat pump is completely reversible, the combination of coefficient of performance expression is given as,

From first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power required to drive the the heat pump is given as

Wnet, in= QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

So the 133.33J was the amount heat that was originally work consumed in the transfer.

Extra....

According to the first law, the rate at which heat is removed from the low temperature reservoir is given as

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

5 0

3 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$