What is not a type of muscle? A) Cardiac B) Visceral C) Skeletal C) Epidural

Calculate the power required of a 60-kg person who climbs a tree 5 meters high in 10 seconds.

Alina [70]

Answer:

Explanation:

Power = Energy/time

-Don't have energy so I'm gonna solve for it

Gravitational Potential Energy = mass x gravity x height

= 60 kg x 9.8 m/s2 x 5m

= 2940 J

Power = Energy/time

=2940 J/10 s

= 294 W

3 0

2 years ago

Sound waves are a longitudinal wave that have a speed of about 340 m/s when traveling through room temperature air. What is the

AlexFokin [52]

The wavelength of the wave is 0.055 m

Explanation:

The relationship between speed, frequency and wavelength of a wave is given by the wave equation:

$v=f\lambda$

where

v is the speed

f is the frequency

$\lambda$ is the wavelength

For the sound wave in this problem we have

v = 340 m/s is the speed

f = 6,191 Hz is the frequency

Solving for $\lambda$ , we find the wavelength:

$\lambda=\frac{v}{f}=\frac{340}{6191}=0.055 m$

Learn more about waves and wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

8 0

4 years ago

A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp

raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination $\theta =11.1^{\circ}$

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction $\mu _k=0.39$

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using $v^2-u^2=2as$

Final velocity v=0

$0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s$

$s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}$

s=0.68 m

5 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: