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mel-nik [20]
3 years ago
9

dam is used to block the passage of a river and to generate electricity. Approximately 58.4 x 103 kg of water falls each second

through a height of 20.1 m. If one half of the gravitational potential energy of the water were converted to electrical energy, how much power (in MW) would be generated
Physics
1 answer:
mrs_skeptik [129]3 years ago
3 0

Answer:

8.049 MW

Explanation:

The expression for gravitational potential energy is given as

Ep = mgh............. Equation 1

Ep = gravitational potential energy, m = mass of water, h = height, g = acceleration due to gravity.

Given: m = 58.4×10³ kg, h = 20.1 m, g = 9.81 m/s²

Substitute into equation 1

Ep =  58.4×10³(20.1)(9.81)

Ep = 1.6098×10⁷ J.

If one half the gravitational potential energy of the water were converted to electrical energy

Electrical energy = Ep/2

Electrical energy = (1.6098×10⁷)/2

Electrical energy = 8.049×10⁶ J

In one seconds,

The power generated = 8.049×10⁶ W

Power generated = 8.049 MW

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A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the
ICE Princess25 [194]

Answer:

energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is =25.72J; The impact loss at first impact is 6,316.64J

Explanation:

First find the velocity of the bullet after the first impact using

M1V1 + 0 = (M1 + M2)v'

Where M1 is the mass of the bullet

M2 is the mass of the block B

M3 is the mass of the carrier

v' is the velocity

v' = M1V1/(M1 + M2)

v'= (30 × 10^-3 kg)(460 m/s) / (30 × 10^-3 kg + 3 kg)

v' = 13.8/3.03

v'= 4.55m/s

Also calculate final velocity of the carrier v2'

v2' = M1V1/(M1 + M2 + M3)

v2'= (30 × 10 kg)(460 m/s) / (30 × 10 kg + 3 kg + 30kg)

v2' =0.42m/s

Now to calculate energy loss due to friction

Normal force

N= W1 + W2 = (m1 + m2)g

Where W1 and W2 is the weight of the bullet and block respectively

g is gravitational acceleration for taken as 9.81m/s

= (0.030 kg + 3 kg)(9.81 m/s) 29.724 N

Friction force = coefficient of kinetic× normal force

Where coefficient of kinetic = 0.2

Ff = (0.2)(29.724)= 5.945 N

Now

Energy loss due to friction = frictional force × distance

Assume distance is 0.5 m.

Energy loss due to friction = 5.945 N × 0.5 m

= 2.97J

Kinetic energy of block with embedded bullet immediately after first impact:

1/2 × (m1 + m2)(v')^2

1/2 × (30 × 10^-3 kg + 3 kg)(4.55m/s)^2

= 1/2 × (3.03kg) × (4.55m/s)^2

= 31.36 J

Final kinetic energy of bullet, Block, and Carrier together

1/2 × (m1 + m2 + m3)(v2')^2

1/2 × (30 × 10^-3 kg + 3 kg + 30kg) (0.42m/s)^2

1/2 × (33.03kg) × (0.42m/s)^2

= 2.91 J

Therefore

Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

= 31.36 J - 2.91 J

= 28.69 J

Impact loss due to AB impacting the carrier = loss due to friction- energy due to friction

28.69J - 2.97J

=25.72J

Initial kinetic energy of system ABC = 1/2(m1vo)

=1/2(0.030 kg)(460 m/s)^2 = 6,348J

Therefore

Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

= 6,348J - 31.36 J

= 6,316.64J

3 0
3 years ago
An electroplating solution is made up of nickel(II) sulfate. How much time would it take to deposit 0.500 g of metallic nickel o
ELEN [110]

Answer:

1.52 hour

Explanation:

M = 0.5 g, I = 3 A

Electrochemical equivalent of nickel

Z = 3.04 × 10^(-4) g/C

By use of Faraday's laws of electrolysis

M = Z I t

t = M / Z I

t = 0.5 / (3.04 × 10^-4 × 3)

t = 5482.45 second = 1.52 hour

6 0
3 years ago
A Venn diagram shows the similarities and the differences of two things. In the area where the two circles overlap, the similari
Mariana [72]

Answer:

found in the nucleus, has mass of one amu

Explanation:

4 0
3 years ago
If there are 3 resistors of .5 ohms, 1 ohms, and 1 ohms connected in parallel then what is the equivalent resistance
Ede4ka [16]

Answer:0.45ohms

Explanation:

Let R be there equivalent resistance

1/R=1/r+1/r+1/r

1/R=1/5+1/1+1/1

1/R=1/5+2

1/R=(1+10)/5

1/R=11/5

Cross multiplying we get

11R=5

Divide both sides by 11

11R ➗ 11=5 ➗ 11

R=0.45ohms

3 0
3 years ago
A TV is rated at 0.1 kW. This TV is used 3 hours per day. How much energy does the TV use per month (in kWh)?
exis [7]

Answer:

<em><u>A.9</u></em><em><u> </u></em><em><u>kWh</u></em>

Explanation:

Tv is rated 0.1 kW

in one day energy used will be 0.1kW ×3hr=0.3kwh

in 30 days energy used will be 0.3kwh×30=9kwh

6 0
3 years ago
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