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3 years ago

What causes the second high tide in the day?

Physics

1 answer:

steposvetlana [31]3 years ago

5 0

Answer:

This rotation generates a centrifugal force, which on the Earth is strongest at locations facing away from the Moon. This in turn causes the sea level in these locations to rise up, forming the second high tide during the course of a day.

Explanation:

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If a sound waves frequency is 100Hz. what is its period

agasfer [191]

Period = (1/frequency) .

If frequency is 100 per second, then

Period = (1) / (100 per second) = 0.01 second .

4 0

3 years ago

Which of the following is not considered a major work flow structure?

mixas84 [53]

Answer:

The correct option is;

D. Fabrication

Explanation:

A workflow flow is a detailed business process consisting of a series of required interconnected tasks in directed graph format that is executable by workflow management system.

Considering each of the options, we have

A. Work center

This consists of part of the transformation input to output. The location

B. Project

This is the unique identifier of the task to be processed

C. Assembly line

Forms part of the required input where transformation takes place and items are being processed within the assembly line

D. Fabrication

Here the item is fixed, without motion, therefore this is not considered a major work flow structure

E. Continuous flow

Here again, the items are being processed and are in motion, which constitutes a workflow structure.

8 0

3 years ago

Hans Eysenck makes a connection between __________ and __________in explaining personality development. A. genetic factors . . .

tekilochka [14]

The answer is A, genetic factors...conditioning. I just took the test. Good luck!!

6 0

4 years ago

Read 2 more answers

Find the total displacement of a mouse that travels 1.0 m north and then 0.8 m south.

andreev551 [17]

North: 1m
South: 0,8m

Direction:

1m>0,8m

so mouse moved north

Distance:

1m-0,8m=0,2m

so mouse traveled 0,2m

Answer: The mouse moved 0,2m north.

"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."

Regards M.Y.

3 0

3 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

melisa1 [442]

Answer:

$v_{2.6b}=11.18\ m.s^{-1}$

$v_{7.2b}=14.19\ m.s^{-1}$

$s_{bb}=226.3305\ m$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

$t_b=21.3956\ s$

$a_y=1.1065\ m.s^{-2}$

Explanation:

Given:

initial speed of blue car, $u_b=0\ m.s^{-1}$

initial speed of yellow car, $u_y=0\ m.s^{-1}$

acceleration rate of blue car, $a_b=4.3\ m.s^{-2}$

time for which the blue car accelerates, $t_{ab}=3.3\ s$

time for which the blue car moves with uniform speed before decelerating, $t_{ub}=14.3\ s$

total distance covered by the blue car before coming to rest, $s_b=253.26 \ m$

distance at which the the yellow car intercepts the blue car just as the blue car come to rest, $s_y=253.26 \ m$

Speed of blue car after 2.6 seconds of starting the motion:

Applying the equation of motion:

$v_{2.6b}=u_b+a_b.t$

$v_{2.6b}=0+4.3\times 2.6$

$v_{2.6b}=11.18\ m.s^{-1}$

Speed of blue car after 7.2 seconds of starting the motion:

∵The car accelerates uniformly for 3.3 seconds after which its speed becomes uniform for the next 14.3 second before it applies the brake.

so,

$v_{7.2b}=u+a_b\times t_{ab}$

$v_{7.2b}=0+4.3\times 3.3$

$v_{7.2b}=14.19\ m.s^{-1}$

Distance travelled by the blue car before application of brakes:

This distance will be $s_{bb}=$ (distance travelled during the accelerated motion) + (distance travelled at uniform motion)

Now the distance travelled during the accelerated motion:

$s_{ab}=u_b.t_{ab}+\frac{1}{2} a_{b}.t_{ab}^2$

$s_{ab}=0\times 3.3+0.5\times 4.3\times 3.3^2$

$s_{ab}=23.4135\ m$

Now the distance travelled at uniform motion:

$s_{ub}=14.19\times 14.3$

$s_{ub}=202.917\ m$

Finally:

$s_{bb}=s_{ab}+s_{ub}$

$s_{bb}=23.4135+202.917$

$s_{bb}=226.3305\ m$

Acceleration of the blue car once the brakes are applied

Here we have:

initial velocity, $u=14.19\ m.s^{-1}$

final velocity, $v=0\ m.s^{-1}$

distance covered while deceleration, $s_{db}=s_b-s_{bb}$

$\Rightarrow s_{db}=253.26 -226.3305=26.9295\ m$

Using the equation of motion:

$v^2=u^2+2a_{db}.s_{db}$

$0^2=14.19^2+2\times a_{db}\times 26.9295$

$a_{db}=-3.7386\ m.s^{-2}$ negative sign denotes deceleration.

The total time for which the blue car moves:

$t_b=t_a+t_{ub}+t_{db}$ ........................(1)

Now the time taken to stop the blue car after application of brakes:

Using the eq. of motion:

$v=u+a_{db}.t_{db}$

$0=14.19-3.7386\times t_{db}$

$t_{db}=3.7956\ s$

Putting respective values in eq. (1)

$t_b=3.3+14.3+3.7956$

$t_b=21.3956\ s$

For the acceleration of the yellow car:

We apply the law of motion:

$s_y=u_y.t_y+\frac{1}{2} a_y.t_y^2$

Here the time taken by the yellow car is same for the same distance as it intercepts just before the stopping of blue car.

Now,

$253.26=0\times 21.3956+0.5\times a_y\times 21.3956^2$

$a_y=1.1065\ m.s^{-2}$

7 0

3 years ago