Extreme tides are

1. The diagram shows a satellite traveling in uniform circular motion around the Earth.

FromTheMoon [43]

Answer:

M V R = constant angular momentum is constant because no forces act in the direction of V

Since M (mass) = constant

V R = constant

The force is directed along the gravitational force vector (towards the center of rotation)

6 0

2 years ago

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

4 years ago

Combustion reactions are also ________ reactions. A. nuclear B. exothermic C. endothermic D. spontaneous

sveta [45]

I think is a) i hope this hlp

5 0

3 years ago

Read 2 more answers

A 60 Watt light bulb runs for 120 seconds. How much energy does it use?

sattari [20]

ENERGY = POWER X TIME
=60 X 120=7200KWh

6 0

3 years ago

Suppose the wavelength of the light is 550 nm . How much farther is it from the dot on the screen in the center of fringe E to t

Bogdan [553]

Complete question:

Suppose the wavelength of the light is 550 nm . How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit? Fringe C is the central maximum.

Check the image uploaded.

Answer:

The difference is 1100 nm

Explanation:

A bright fringe creates constructive interference, the wavelength is always in a multiple form.

At center fringe, the difference is (550 nm)(0) = 0

For the first maximum, the difference is (550 nm)(1) = 550 nm

For the second maximum, the difference = (550 nm)(2) = 1100 nm

Thus, for nth maxima, the difference is (550 nm)(n)

From the image uploaded, C is located on the second maximum, therefore the difference is given as (550 nm)(2) = 1100 nm

Therefore, the dot on the screen in the center of fringe E to the left slit is 1100 nm

3 0

4 years ago