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gladu [14]
2 years ago
8

Two objects are moving downward at a speed of 3 cm/s. Object A has a mass of 2kg and Object B has a mass of 1kg. You want both o

bjects to move to the left at a speed of 3cm/s.
In what direction or (directions) should you exert the forces that will make the desired change?
Physics
1 answer:
kompoz [17]2 years ago
4 0
Yeah that’s how i feels better now tho haha it’s all okay cool tho haha it’s cool cool bro bro yeah bro lol lol bro bro lol lol yeah yeah that’s okay good lol
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A current of 1.75 A flows through a 55.4 Ω resistor for 9.5 min. How much heat was generated by the resistor? Answer in units of
kirill115 [55]

Answer:

9.67\times 10^4 J

Explanation:

We are given that

Current=I=1.75 A

Resistance =R=55.4 ohm

Time=t=9.5 min=9.5\times 60=570 s

1 min=60 s

We have to find the heat generated by the resistor.

We know that

Heat=E=I^2Rt

Using the formula

Heat,E=(1.75)^2\times 55.4\times 570=9.67\times 10^{4} J

Hence, the heat generated by the resistor =9.67\times 10^4 J

6 0
3 years ago
Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then
Andreas93 [3]

Answer:

The displacement is  \Delta H =    -  1 \ m

The distance  is  D =  4 \  m

Explanation:

From  the question we are told that

    The height from which the ball is dropped is  h  =  1 \ m

    The height attained at  the first bounce is  h_1  = 0.8  \  m

    The height attained at  the second bounce is   h_2 = 0.5 \  m

    The height attained at  the third bounce is h_3 = 0.2 \  m

Note  : When calculating displacement we consider the direction of motion

Generally given that upward is positive  the total displacement of the ball is mathematically represented as

            \Delta H =  (0  -  h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)

Here the 0 show that there was no bounce back to the point where Billy released the ball  

           \Delta H =  (0  -  1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)

=>          \Delta H =    -  1 \ m

Generally the distance covered by the ball is mathematically represented as  

                D =  h +  2h_2 + 2h_3 + 2h_3

The 2 shows that the ball traveled the height two times

              D =  1 +  2* 0.8  + 2* 0.5 + 2* 0.2

=>           D =  4 \  m

     

5 0
3 years ago
How do ocean currents on the water's surface transfer energy?
Anit [1.1K]
Exactly what the first person said. Hope this helps!
6 0
3 years ago
Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pi
Serga [27]

Answer:

Flow Rate = 80 m^3 /hours  (Rounded to the nearest whole number)

Explanation:

Given

  • Hf = head loss
  • f = friction factor
  • L = Length of the pipe = 360 m
  • V = Flow velocity, m/s
  • D = Pipe diameter = 0.12 m
  • g = Gravitational acceleration, m/s^2
  • Re = Reynolds's Number
  • rho = Density =998 kg/m^3
  • μ = Viscosity = 0.001 kg/m-s
  • Z = Elevation Difference = 60 m

Calculations

Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)

The energy equation for this system will be,

Hp = Z + Hf

The other three equations to solve the above equations are:

Re = (rho*V*D)/ μ

Flow Rate, Q = V*(pi/4)*D^2

Power = 15000 W = rho*g*Q*Hp

1/f^0.5 = 2*log ((Re*f^0.5)/2.51)

We can iterate the 5 equations to find f and solve them to find the values of:

Re = 235000

f = 0.015

V = 1.97 m/s

And use them to find the flow rate,

Q = V*(pi/4)*D^2

Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours

7 0
3 years ago
In a hydroelectric dam, water falls 35.0 m and then spins a turbine to generate electricity. Suppose the dam is 80% efficient at
anyanavicka [17]

Answer:

Height through the water falls h = 33 m

Efficiency of the of the unit

η

=

80

%

Power of the production unit

P

=

45

×

10

6

W

Acceleration due to gravity

g

=

9.8

m

/

s

Potential energy of one kg of water

Δ

U

=

m

∗

g

∗

h

=

1

∗

9.8

∗

33

=

323.4

J

Eighty percentage of the above energy is converted into electrical energy.

So eighty percentage of the potential energy of one kg of water

Δ

U

1

=

258.72

J

So mass of water required to flow per second to produce 45 MW of electricity in kilograms

M

=

P

Δ

U

1

=

45

∗

10

6

258.72

=

173933.2096

k

g

/

s

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Hydroelectric Energy: Definition, Uses, Advantages & Disadvantages

from Earth Science 101: Earth Science

Chapter 23 / Lesson 9

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