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olga nikolaevna [1]

2 years ago

7

Explain the changes in energy as a ball is held in the air , and then dropped

1 answer:

Nostrana [21]2 years ago

3 0

Answer:

Potential Energy to Kenetic Energy

Explanation:

When holding a ball in the air, the ball has potential energy. Once you drop the ball, the ball gains Kenetic Energy

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Need a little help here :(

Goshia [24]

Answer:

The output out be 200

Explanation:

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8 0

2 years ago

Read 2 more answers

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

2 years ago

Which of the following has the greatest kinetic energy? A. a mass of 4m at velocity v. B. a mass of 3m at velocity 2v. C. a mass

Artist 52 [7]

Answer:

Option C

Explanation:

Kinetic energy is the energy that the body possesses by virtue of its motion.

The formula for Kinetic energy is given by $\frac{1}{2} mv^2$

Using this formula let us find kinetic energy for the bodies given and find out which is the greatest

A) KE = $\frac{1}{2} (4m)(v^2) = 2mv^2$

B) KE = $\frac{1}{2} (3m)(2v)^2 = 6mv^2$

C) KE = $\frac{1}{2} (2m)(3v)^2 = 9mv^2$

D) KE = $\frac{1}{2} (3)(4v)^2 = 8mv^2$

Comparing these we find that 9mv^2 is the highest.

Hence option C is the answer.

4 0

2 years ago

Suppose A=BnCm, where A has dimensions LT, B has dimensions L2T-1, and C has dimensions LT2. Then the exponents n and m have the

julsineya [31]

Explanation:

The expression is :

$A=B^nC^m$

A =[LT], B=[L²T⁻¹], C=[LT²]

Using dimensional of A, B and C in above formula. So,

$A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}$

Comparing the powers both sides,

2n+m=1 ...(1)

2m-n=1 ...(2)

Now, solving equation (1) and (2) we get :

$n=\dfrac{1}{5}\\\\m=\dfrac{3}{5}$

Hence, the correct option is (E).

5 0

2 years ago

The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the

natta225 [31]

Answer:

50 N

Explanation:

Efficiency of a machine can't be more than 1, so I assume you mean 40%. (Remember, efficiency and mechanical advantage are not the same).

Efficiency is the ratio of work out of a system to the work in to the system.

e = Wout / Win

Work is force times distance, so:

e = (Fout × Dout) / (Fin × Din)

Rearranging:

Fin = (Fout × Dout) / (e × Din)

Fin = (Fout / e) × (Dout / Din)

Fin = (Fout / e) / (Din / Dout)

We know that e = 0.40, and Fout = 120 N. Since there are 6 pulleys, we also know that Din/Dout = 6.

F = (120 N / 0.4) / 6

F = 50 N

5 0

2 years ago