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3 years ago

The moon’s relative motion causes

Physics

1 answer:

Lerok [7]3 years ago

7 0

Answer: Different cycles of the moon and ocean currents are all I can think of.

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If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0

3 years ago

How often do the earth's magnetic poles switch?

RSB [31]

Reversals are the rule, not the exception. Earth has settled in the last 20 million years into a pattern of a pole reversal about every 200,000 to 300,000 years, although it has been more than twice that long since the last reversal.

3 0

3 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

Now, gravitational force on particle 3 due to particle 1:

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

Now the net force

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

For angle in counterclockwise direction from the +x-axis

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

4 years ago

Question 7 of 25

Crank

Answer: 350 ms

Explanation:

Just took the quiz:)

5 0

3 years ago

Read 2 more answers

A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that time?

Helen [10]

Answer:

the car have travelled 0.31 mile during that time

Explanation:

Applying the Equation of motion;

s = 0.5(u+v)t

Where;

s = distance travelled

u = initial speed = 0 mph

v = Final speed = 50 mph

t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour

Substituting the given values into the equation;

s = 0.5(0+50)×(1/80)

s = 0.3125 miles

s ~= 0.31 mile

the car have travelled 0.31 mile during that time

8 0

3 years ago