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marissa [1.9K]
3 years ago
7

a pebble is dropped down a well and hits the water 1.5 seconds later. using the equations for motion with constant acceleration,

determine the distance from the edge of the wall to the water's surface.
Physics
2 answers:
Effectus [21]3 years ago
7 0
Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.
Because the pebble hits the surface of the water after 1.5 s, therefore
h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

Answer:  11.025 m
yawa3891 [41]3 years ago
4 0

Answer: 11.025

Explanation:

Let h = the distance from the edge of the wall to the water surface (m).

Use g = 9.8 m/s² and neglect air resistance.

The initial vertical velocity of the pebble is zero.

Because the pebble hits the surface of the water after 1.5 s, therefore

h = (1/2)*(9.8 m/s²)*(1.5 s)² = 11.025 m

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Electrical energy is used to turn the blades of a fan. The amount of energy transformed is seen here: 750 j electrical energy is
MAVERICK [17]

Here is the energy that is left after the quantity of energy is transformed: 750 j of electrical energy is changed into 400 j of kinetic or mechanical energy, which is then turned into 0.32 j of efficient energy.

To run the fan, electrical energy is utilized.

Here, under the specified circumstances, 750 J of electrical energy is utilized to operate the fan, which is transformed into 400 J of kinetic energy. As a result, 350 J of energy is wasted due to various frictional and resistive losses.

Therefore, we may conclude that only 400 J of the 750 J available energy is used to power the fan, with the remaining energy being wasted as a result of friction.

Additionally, we can state that this fan's effectiveness will be

n = Useful ÷ Total

n = 400 ÷ 750

n = 8 ÷ 25

n = 0.32

Learn more about energy at

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4 0
1 year ago
An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.
Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

The \ weight, W  =G\dfrac{M \times m}{R^{2}} = 800 \ N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

W_1   =G\dfrac{\dfrac{M}{2}  \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2}  \times m \times 4}{ R ^{2}} = 2 \times G \times  \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0
3 years ago
2. If you want 0. 250 a (250 milliamps) to flow around a circuit with a resistance of 400 ohms, what voltage do you need?
grigory [225]

Answer:

0.000625 V

Explanation:

The formula linking current , resistance and voltage is :

V = I/R

Voltage = Current / Resistance

Now we substitute values given in question :

Voltage = 0.250 / 400

Voltage (V) = 0.000625

Our final answer is 0.000625 V

Hope this helped and have a good day

5 0
2 years ago
3. Describe the flow of one molecule of water through the water cycle, beginning in the ocean.
scZoUnD [109]

molecules of water are never destroyed - they go through various uses in a cycle of re-use. beginning in the ocean. a water molecue is attached to the wet suit of a deep sea diver. when the diver gets back on his boat, the water molecule leaves the ocean. Diver dry his suit under the sun. The water molecule is evaporated to the air. It meets up with more water molecules to form cloud. Cloud becomes rain over ground. Rain drains into stream which merges into river. River runs out to the ocean and the water cycle starts anew.

6 0
3 years ago
Read 2 more answers
a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

7 0
3 years ago
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