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4 years ago

A 100 W engine runs for 1 minute. How much work does it do?

Physics

2 answers:

ANEK [815]4 years ago

6 0

I think its B becausee 1min times 100 is 100.

Svet_ta [14]4 years ago

4 0

d 6000 j because every second multiplied by that would equal out to be 6000

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A square, single-turn coil 0.132 m on a side is placed with its plane perpendicular to a constant magnetic field. An emf of 27.1

marta [7]

Answer:

0.35 T

Explanation:

Side, a = 0.132 m, e = 27.1 mV = 0.0271 V, dA / dt = 0.0785 m^2 / s

Use the Faraday's law of electromagnetic induction

e = rate of change of magnetic flux

Let b be the strength of magnetic field.

e = dФ / dt

e = d ( B A) / dt

e = B x dA / dt

0.0271 = B x 0.0785

B = 0.35 T

6 0

3 years ago

A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent

Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A

At all times, angular velocity is $\omega = 0\,{\rm{rad/s}$

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 0\,{\rm{rad/s}}$

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero hence statement A is valid.

Statement B

Angular acceleration is the rate of change in angular velocity with respect to time.

Angular velocity and angular acceleration are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

Which when re-arranged becomes

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

There’s no change in angular velocity anytime when the angular velocity is $\omega = 10\,{\rm{rad/s}}$ .The final and initial velocities remain the same.

The equation can be modified as follows:

$\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement B is valid

Statement C

Angular velocity is defined as the change in the angular position with respect to time.

Angular velocity and angular displacement are related by

$\theta = \omega t$

Which can also be modified as:

${\theta _{\rm{f}}} - {\theta _{\rm{i}}}$

Note that the final position is ${\theta _{\rm{f}}}$ and initial position is ${\theta _{\rm{i}}}$

Modifying the equation to find the angular velocity we obtain

$\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}$

When the angular displacement has the same value at all times, the equation becomes

$\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}$

The angular velocity becomes zero.

Angular acceleration and angular velocity are related by

${\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t$

The expression above can be rearranged as follows:

$\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}$

At all times, the angular velocity is $\omega = 0\,{\rm{rad/s}}$ hence initial and final velocities remain the same

We obtain

$\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}$

Therefore, the angular acceleration becomes zero and statement C is valid.

Therefore, statements A,B and C are consistent .

4 0

4 years ago

The process of making alloys involves

sergeinik [125]

Answer:

The process of making alloys involves ( Heating ) pure metals to remove impurities. Then the pure metals are(mixed) with other components. An alloy is a mixture of metals or a mixture of a metal and another element. Alloys are defined by a metallic bonding character.

Explanation:

7 0

3 years ago

An LP turntable must spin at 3.500 rad/s to play a record. How much torque must the motor deliver if the turntable is to reach i

zmey [24]

Answer:

0.00321 Nm

Explanation:

$\omega_f$ = Final angular velocity = 3.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 2 rev

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{3.5^2-0^2}{2\times 2\pi \times 1.8}\\\Rightarrow \alpha=0.54156\ rad/s^2$

Moment of inertia is given by

$I=mr^2\\\Rightarrow I=0.25\times 0.154^2\\\Rightarrow I=0.005929\ kgm^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=0.005929\times 0.54156\\\Rightarrow \tau=0.00321\ Nm$

The torque required is 0.00321 Nm

8 0

4 years ago

The amount of heat needed

Simora [160]

Q = mcθ

Where Q = Amount of heat in J
   m = Mass of substance in kg
      θ = Temperature rise in °C or K
      C = Specific heat capacity in J/kgK

From given data:
Q = 33 kJ = 33 000 J, m= 2.0 kg, θ = 80 K, c = ?

33000 = 2 * c * 80
33000 = 160c
160c = 33000
   c = 33000/160 = 206.25

Therefore specific heat capacity is 206.25 J/kgK

4 0

4 years ago