All categories

3 years ago

The kinetic energy of the pendulum bob in figure 15-1 increases the most between locations

Physics

2 answers:

Ket [755]3 years ago

7 0

Answer:A and C

Explanation:

Fudgin [204]3 years ago

3 0

Answer:

A and C

Explanation:

Kinetic energy of a body is given by the expression, $KE =\frac{1}{2}mv^2$ , where v is the velocity and m is the mass of body.

In case of a simple pendulum the maximum velocity is at it's stationary position.

So maximum velocity i at position C, which means maximum kinetic energy is at position A.

The velocity of pendulum reduces as the angle between vertical stationary and current position increases.

So, velocity at E and A are the minimum, so at A and E the kinetic energy value is minimum.

Now examining the options given, we will understand the kinetic energy increase is maximum in case of A and C.

You might be interested in

What is kinematics ??<br>how do we spell it ?? <br>don't spam !! xD

34kurt

Answer:

Kinematics is the study of motion.

6 0

3 years ago

Read 2 more answers

What is the only bone of the skull that moves?

Darina [25.2K]

Answer:

Mandible.

Explanation:

The mandible in the bone that allows you to open and close your mouth, otherwise known ad the jawbone.

3 0

3 years ago

Read 2 more answers

A 46 g domino slides down a 30 degrees incline at a constant speed. What is the coefficient of friction?

blondinia [14]

Answer:

Explanation:

6 0

3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface: