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Sidana [21]
2 years ago
10

A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later.

How fast was the pebble going when it hit the water?
Physics
1 answer:
Natali5045456 [20]2 years ago
3 0

Answer:

19.62 ms

Explanation:

t = Time taken = 2 s

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive)

Equation of motion

v=u+at\\\Rightarrow v=0+9.81\times 2\\\Rightarrow v=19.62\ m/s

The speed of the pebble when it hit the water is 19.62 ms

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An object travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the ac
Allisa [31]

Answer:

The formula to calculate velocity in this case:

v = v0 + at

=> a = (v - v0)/t

       = (50 - 0)/4

       = 50/4 = 12.5 (m/s2)

Hope this helps!

:)

3 0
2 years ago
7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has
NNADVOKAT [17]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, \dfrac{m}{t}=30\ kg/s

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

F=\dfrac{\Delta P}{\Delta t}

F=\dfrac{m(v-u)}{\Delta t}

F=30\ kg/s\times (-16-16)\ m/s

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

6 0
2 years ago
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
B) Explain the method of preparing electromagnet. How do you test the
kap26 [50]

Answer:

An electromagnet is made by forming a coil around a soft iron bar (known here as the metal) such as a nail or  screw and connect with an insulated copper wire (known here as the electric current conductor) the ends of the wound copper is then connected separately to the positive and negative terminals of a battery (known here as the source of electric current)

The north seeking needle of the magnetic compass will move away when brought close to the north pole of the formed electromagnet which can then be labelled N

The magnetic compass needle will be attracted to the south pole of the electromagnet which can then be labelled S

Explanation:

An electromagnet is an electric powered magnet that is formed (temporarily) by the perpendicular movement of electric current with respect to a metal core

The magnitude and the poles of an electromagnet can be changed by changing the magnitude and the direction of flow of the electric current respectively.

5 0
2 years ago
A 6.0-ohm resistor that obeys Ohm’s Law is connected to a source of variable potential difference. When the applied voltage is d
leva [86]

Is halved. A 6Ω resistor connected to a voltage source which voltage is decreased from 12V to 6V the current passing through the resistor is halved.

The key to solve this problem is applying Ohm's Law V = R I, clearing I from the equation, we obtain I = V/R. Then, the current is directly proportional to the voltage and inversely proportional to the resistance.

V = 12V and R = 6Ω

I = 12V/6Ω = 2A

V = 6V and R = 6Ω

V = 6V/6Ω = 1A

As we can see the current is halved if the voltage descreased from 12V to 6V

5 0
2 years ago
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