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Vesna [10]
3 years ago
8

Guys please help me on the rest of the numbers

Physics
1 answer:
wel3 years ago
8 0

Answer:

C. 100

D.3

E. 33.3

Explanation:

C. Mechanical Advantage=Load / Effort

= 200N

--------

100N

Therefore,. = 100

D. I. Velocity Ratio= distance moved by the effort / distance moved by load

= 30cm/10cm

= 3

II. Efficiency= M.A / V.R

= 100/3

= 33.33

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Which of the following changes will always increase the efficiency of a thermodynamic engine? Choose all correct statements.
Ilya [14]

Answer:B,C,D

Explanation:

Thermodynamic  efficiency is given by

\eta =1-\frac{T_C}{T-H}

\eta efficiency can be increased by Keeping _c constant and increasing T_H

Keeping T_H constant and decreasing T_c

by increasing \Delta T=T_H-T_c

by decreasing \frac{T_C}{T_H} ratio          

5 0
3 years ago
Read 2 more answers
Where on this diagram does the ball have the highest point of gravitational potential energy?
mixer [17]
It should be at the very top since it has more space to fall which gives it more potential energy
3 0
3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
A crane used 250,000 Joules of work to move a beam to the top of a building in 20 seconds. How much power did the crane use?
maria [59]

Answer:

12500W

Explanation:

Given parameters:

Work done  = 250000J

Time taken  = 20s

Unknown:

Power of the crane = ?

Solution:

Power is the defined as the rate at which work is being done;

  Mathematically;

        Power = \frac{work done}{time }

 insert the parameters and solve;

      Power  = \frac{250000}{20}   = 12500W

7 0
3 years ago
Sound is a type of mechanical wave and must have a medium through which to travel. Sound will move at different speeds when whic
Alex787 [66]
Hey there!

The answer would be B. The sound moves from air to water.

Sound travels through different mediums. It goes fastest in solids, a little slower in liquids, and slowest in air. Sound is a very fast wave, but remember that mediums can differ that. In a vacuum space, there is no sound at all. (ex. outer space)

Hope this helps !
5 0
3 years ago
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