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3 years ago

Does the tide that the moon raises on the earth different?

Physics

1 answer:

alukav5142 [94]3 years ago

7 0

Answer:

No the gravity of the moon pulls the water making high tide

Explanation:

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AKS 8a - Phenomena-Based Question: Use the data and the graph to make a claim as to which person represents each letter on the g

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Answer:

,mhom

Explanation:

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3 years ago

A wave has frequency of 50 Hz and a wavelength of 10 m. What is the speed?

Sunny_sXe [5.5K]

Answer:

v=500 m/s

<h3>Solution:</h3>

v=(10m)(50/s)

5 0

3 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$

$v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)$

$v_f^2 = 567.9$

$v_f = 23.8 m/s$

Part c)

Relative speed of two balls is given as

$v_{12} = v_1 - v_2$

$v_{12} = (13.9) - (-13.9) = 27.8 m/s$

now the distance between two balls in 0.8 s is given as

$d = v_{12} t$

$d = 27.8 \times 0.8$

$d = 22.24 m$

7 0

3 years ago

A learner sets up a circuit to determine the conductivity of certain solids. What piece of apparatus is not required for the nex

Lilit [14]

It is D as u dont need a stop watch aft that

4 0

3 years ago

a 100 g cart initially moving at 0.5 m/s collides elastically from a stationary 180 g cart. a) using the equation in the theory,

cupoosta [38]

A 100 g cart is moving at 0.5 m/s that collides elastically from a stationary 180 g cart. Final velocity is calculated to be 0.25m/s.

Collision in which there is no net loss in kinetic energy in the system as a result of the collision is known as elastic collision . Momentum and kinetic energy both are conserved quantities in elastic collisions.

Collision in which part of the kinetic energy is changed to some other form of energy is inelastic collision.

For an elastic collision, we use the formula,

m₁V₁i+ m₂V₂i = m₁V1f + m₂V₂f

For a perfectly elastic collision, the final velocity of the 100g cart will each be 1/2 the velocity of the initial velocity of the moving cart.

Final velocity = 0.5/2

=0.25 m/s.

To know more about elastic collision, refer

brainly.com/question/7694106

#SPJ4

7 0

1 year ago