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CaHeK987 [17]
3 years ago
12

For high and low tides differences would they be caused by the moon and how?

Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
8 0
The moon has a small amount of gravity. Low tides mean the moon is not pulling on the water. High tides mean that the moon is pulling on the water.
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PLZ HELP!
MrMuchimi

Answer:

The answer is ball D

Explanation:

because I said it was

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4 years ago
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A 4. 0 nc positive point charge is located at point a in the figure. (figure 1) what is the electric potential at point b?.
algol [13]

The electric potential at point b experienced by the charge cab be determined using the formulas given.

<h3>Electric potential</h3>

The electric potential of a point charge is the work done in moving the charge from infinity to certain point against the electric field.

V = Ed

V = (F/q)d

V = (Fd)/q

where;

  • V is the electric potential
  • F is electric force
  • E is the electric field
  • q is the charge

Thus, the electric potential at point b experienced by the charge cab be determined using the formulas given.

Learn more about electric potential here: brainly.com/question/14306881

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2 years ago
The zinc plate is coated with mercury ​
raketka [301]
Amalgamating is the coating of zinc plate with mercury.
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3 years ago
Problem 1. Cylinders and a pendulum A uniform solid cylinder of radius r and mass m can roll inside a hollow cylinder of radius
Nataliya [291]

Answer:

The two answers are in the explanation

Explanation:

Please find the attached files for the solution

7 0
4 years ago
Plz i need help for the 5 problems. plz show the work!!!
Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

8 0
3 years ago
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