In which regions can the gravitational field strength due to the two planets be zero? Explain.

Which of the following is a major way in which oceans contribute to weather systems?

mylen [45]

Since, the options are not given the question is incomplete the complete question is as follows.:

Which of the following is a major way in which oceans contribute to weather systems?

provide a diverse habitat for many organisms

experience changes in amounts of dissolved salts

store and transport the Sun's heat energy

reach depths that can be as much as 12000 meters

Answer: Store and transport the Sun's heat energy.

Explanation:

Oceanic currents are just like a conveyor belt. It helps in transportation of the warm water and the precipitation from the equator to the poles and the cold water in the poles towards the tropics. This way the oceans counteract the uneven distribution of the radiation of sun that reaches upto the surface earth. This will regulate the global climate.

4 0

3 years ago

An alpha particle (a helium nucleus, consisting of two protons and two neutrons) has a radius of approximately 1.6 × 10-15 m. A

Snezhnost [94]

Answer:

9.96x10^-20 kg-m/s

Explanation:

Momentum p is the product of mass and velocity, i.e

P = mv

Alpha particles, like helium nuclei, have a net spin of zero. Due to the mechanism of their production in standard alpha radioactive decay, alpha particles generally have a kinetic energy of about 5 MeV, and a velocity in the vicinity of 5% the speed of light.

From this we calculate the speed as

v = 5% 0f 3x10^8 m/s (speed of light)

v = 1.5x10^7 m/s

The mass of an alpha particle is approximately 6.64×10−27 kg

Therefore,

P = 1.5x10^7 x 6.64×10^−27

P = 9.96x10^-20 kg-m/s

8 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:

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

3 years ago

Meg walks with a velocity of 0.9 m/s west. She does so while riding on a train that is traveling with a velocity of 2.7 m/s east

Shkiper50 [21]

<span>Velocities are vectors so we can add them!

Let's let +x be East and -x be West.

-0.9 + 2.7 = 1.8

Since our answer is positive that means East so the answer is C.</span>

6 0

3 years ago

If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite d

Yuliya22 [10]

Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:

A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

ΔP=20 kg.m/s

Explanation:

Given data

Mass m=0.2 kg

Initial speed Vi=-44.5m/s

Final speed Vf=55.5 m/s

Required

Change in momentum ΔP

Solution

First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

$v_{i}=-44.5m/s\\v_{f}=55.5m/s$