Ask question

Ask question

All categories

3 years ago

9

A ball is thrown straight up from the ground with a speed of 12 meters per second. What is the maximum height reached by the bal

l?

1 answer:

Kazeer [188]3 years ago

3 0

Answer:

the maximum height of object is 7.34 meters

You might be interested in

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:

A)

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:

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

B)

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:

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

C)

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

2 years ago

Two balls, one twice as massive as the other, are dropped from the roof of a building (freefall). Just before hitting the ground

Katena32 [7]

KE= 1/2 mv^2

Kinetic Energy is equal to 1/2 x mass x velocity squared

The mass of the larger ball has TWICE
the kinetic energy. KE is directly proportional to the mass.

6 0

2 years ago

How can heat be transferred across empty space?

yawa3891 [41]

The answer is: Heat can<span> be </span>transmitted<span> though </span>empty space<span> by thermal radiation. Thermal radiation (often called infrared radiation) is a type electromagnetic radiation (or light). Radiation is a form of energy transport consisting of electromagnetic waves traveling at the speed of light.</span>

4 0

3 years ago

Read 2 more answers

A cube made Of an unknown material has a height of 9 symeters the mass of this cube is 3645 g. calculate the density of this cub

kherson [118]

Answer: $5 \frac{g}{cm^{3}}$

Explanation:

The density $\rho$ of a material is given by:

$\rho=\frac{m}{V}$ (1)

Where:

$m=3645 g$ is the mass of the cube

$V$ is the volume of the cube

Now, the volume of a cube is equal to the length $L$ of its edge to the power of 3:

$V=L^{3}$ (2)

If we know $L=9 cm$ , the volume of this cube is:

$V=(9 cm)^{3}=729 cm^{3}$ (3)

Substituting (3) in (1):

$\rho=\frac{3645 g}{729 cm^{3}}$ (4)

$\rho=5 \frac{g}{cm^{3}}$ This is the density of the cube

8 0

3 years ago

Would the solar panel work under a fluorescent or halogen light? explain your response being sure to relate your observations to

eduard

I think that the solar panel would work under a fluorescent or halogen light if the photons are being produced. These types of lights mimic sunlight so it would not work as good as the real thing but it could work. Just not be as powerful.

4 0

2 years ago