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frutty [35]
10 months ago
10

a ball drops some distance and loses 30 j of gravitational potential energy. do not ignore air resistance. how much kinetic ener

gy did the ball gain? a ball drops some distance and loses 30 j of gravitational potential energy. do not ignore air resistance. how much kinetic energy did the ball gain? more than 30 j less than 30 j exactly 30 j
Physics
1 answer:
aniked [119]10 months ago
5 0

Thus, more than 30 J of potential energy can be loosed by the ball. Thus, the gravitational potential energy of the ball is more than 30 J.

If there is no air resistance, the ball's potential energy is entirely transformed into kinetic energy. When air resistance is taken into account, a portion of the potential energy is used to overcome it. Thus, AU > AKE. In the current scenario, a ball gains 30 J of kinetic energy while falling and is treated as encountering air resistance. The energy that an object retains due to its position in relation to other objects, internal stresses, electric charge, or other factors is known as potential energy in physics. The potential energy will be transformed into kinetic energy if the stones fall. High on the tree, branches have the potential to fall, which gives them energy. Chemical potential energy exists in the food we eat.

Learn more about potential energy here:

brainly.com/question/24284560

#SPJ4

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A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T
sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

\Delta U = Q-W

\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)

Where

\dot{m} = Mass flow

h_1 =Specific Enthalpy (IN)

h_2 = Specific Enthalpy (OUT)

g = Gravity

z_{1,2} = Heigth state (In, OUT)

V_{1,2} =Velocity (In, Out)

Our values are given by,

T_i = 10\°C

P_1 = 100kPa

\dot{m} = 5kg/s

z_2 = 20m

For this problem we know that as pressure, temperature as velocity remains constant, then

h_1 = h_2

V_1 = V_2

Then we have that our equation now is,

\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}

\dot{W} = \frac{(5)(9.81)(0-20)}{1000}

\dot{W} = -0.98kW

8 0
3 years ago
Wilma can mow a lawn in 80 minutes. Rocky can mow the same lawn in 120 minutes. Construct an equation that would allow you to de
xeze [42]

Answer:

t = 96 minutes

Explanation:

Time to mow 1 lawn by Wilma is 80 minutes

so work done in 1 minute by Wilma is given as

W_1 = \frac{1}{80}

Similarly Rocky mow same lawn in 120 minute

so work done in 1 minute by Rocky is given as

W_2 = \frac{1}{120}

now we know that they both worked by "t" time

so total work performed by them

W = \frac{t}{80} + \frac{t}{120}

they both mow 2 lawns then it is given as

2 = (\frac{1}{80} + \frac{1}{120})t

t = 96 minutes

5 0
3 years ago
Can someone explain to me #4.
nalin [4]
A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.
4 0
2 years ago
A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of
alekssr [168]

Answer:the

8/9 h

Explanation:

Height  =   1/2 a T^2     now change to T/3

 now height = 1/2 a (T/3)^2   =<u> 1/9</u>  1/2 a T^2     <===== it is 1/9 of the way down   or   8/9 h

 

3 0
2 years ago
A helicopter flies with an air speed of 175 km/h, heading south. The wind is blowing at 85 km/h to the east relative to the grou
spayn [35]

Answer:

154° at 195 km/h

Explanation:

The helicopter is moving south at 175 km/h, relative to the wind.

But the wind is moving east at 85 km/h, relative to the ground.

This means that the helicopter is moving south east relative to the ground.

Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.

This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.

Refer to the triangle b1.

The distance traveled by the helicopter in 1 hour is denoted by d.

d is the hypotenuse of the right triangle.

Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)

Hence the helicopter is traveling at 195 km/h relative to the ground.

To calculate the direction we use,

tan (x) = opposite/adjacent = 85/175

So the angle x is,

x = arctan (\frac{85}{175} ) = 25.9°

Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)

8 0
2 years ago
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