All categories

3 years ago

What is the potential energy of a 150 kg diver standing on a diving board that is 10 m high?

Physics

2 answers:

igomit [66]3 years ago

7 0

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 150 × 10 × 10

We have the final answer as

Hope this helps you

EastWind [94]3 years ago

4 0

Answer:

14,700

Explanation:

You might be interested in

1. ________________electricity is the type of electricity commonly used in homes and businesses throughout the world.

amid [387]

Number 1 is letter A

3 0

3 years ago

A charged particle of mass m = 5.00g and charge q = -70.0μC moves horizontally to the right at a constant velocity of v = 30.0 k

pychu [463]

Answer:

The magnitude of the force, B = 5 Tesla, Up (North) direction

Explanation:

Magnetic force F= Eq where Electric field, E = 750 NC

and charge, q = -70 μC = -7 × $10^{-5}$ C

F = 750 × -7 × $10^{-5}$

F = 0.0525

But F = qvB; B = $\frac{F}{qv}$

where B is the magnetic field

= 0.0525 ÷ ( -7 × $10^{-5}$ × 30)

B = 5.0 Teslas

The force on a negative charge is in exactly the opposite direction to that on a positive charge.

Hence the direction of the charge is up (North).

8 0

3 years ago

Read 2 more answers

What is the function of the bronchioles?<br> please help

beks73 [17]

Answer:

The bronchioles function is to deliver air to tiny sacs called alveoli where oxygen and carbon dioxide are exchanged.

Explanation:

Bronchioles are air passages inside the lungs that branch off like tree limbs from the bronchi—the two main air passages into which air flows from the trachea (windpipe) after being inhaled through the nose or mouth. The bronchioles deliver air to tiny sacs called alveoli where oxygen and carbon dioxide are exchanged.

4 0

3 years ago

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0

3 years ago

Can someone help pleaseeee

Eduardwww [97]

Answer:

Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .

Explanation:

The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.

4 0

3 years ago