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Sergeeva-Olga [200]
3 years ago
11

Question....... don't spam ​

Physics
1 answer:
kati45 [8]3 years ago
3 0
4.0 cm

Hope this helps (:
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In a hydroelectric dam, water falls 30.0m and then spins a turbine to generate electricity.
ozzi

Answer:

(a) 294 J

(b) 229591.84 kg

Explanation:

height, h = 30 m

g = 9.8 m/s^2

(a) U = m x g x h

U = 1 x 9.8 x 30

U = 294 J

(b) Let the mass of water is m

U = m x g xh

80% of this energy per second is generating power

0.8 x m x 9.8 x 30 = 54 x 10^6

m = 229591.84 kg

Thus, the mass of water required is 229591.84 kg.

4 0
3 years ago
A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost
densk [106]

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

<em>Substitute these values into equation (i) as follows;</em>

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = \frac{1}{2}m₁u₁² +  \frac{1}{2}m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = (\frac{1}{2} x 2.0 x 5.0²) +  (\frac{1}{2} x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = \frac{1}{2}(m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = \frac{1}{2} ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

3 0
4 years ago
If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if
umka21 [38]

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

P'=P+\rho gh

Put the value into the formula

P'=1.01\times10^{5}+1025\times9.8\times10

P'=201450\ Pa

We need to calculate the volume at the surface

Using equation of ideal gas

PV= RT

So, for both condition

PV=P'V'

Put the value into the formula

V=\dfrac{201450\times5.5}{1.01\times10^{5}}

V=10.97\ L

Hence, The volume at the surface is 10.97 L.

3 0
3 years ago
Read 2 more answers
Gather data: Select Mercury from the Solar system menu at left. Turn on Additional data. In the table below, record Mercury’s Ma
ki77a [65]

Answer:

volume of substance of weight of mercury is 13593 kilograms

3 0
3 years ago
1 A steel ball of mass 5 kg is released from rest at a height of 3 m above the ground. It rolls down the frictionless section AB
KonstantinChe [14]

The speed of the ball when it reached the point B is 7.7 m/s.  The angle of inclination of section BC is  11.54⁰.

<h3>What is frictional force?</h3>

The force between the two mating surfaces and having the relative motion is called the frictional force.

The speed of ball at point B is equal to the relation

v  = √(2gh)

Substitute the values of height h = 3m and acceleration due to gravity g =9.8 m/s², we get the speed at B

v =√(2 x 9.8 x 3)

v =7.7 m/s

Thus the speed of ball when it reaches the point B is 7.7 m/s.

Given is the mass of steel ball is 5 kg and the frictional force is 10N.

The frictional force is equal to the product of coefficient of friction and the normal force on the steel ball \.

f = μN

f =μmg

Substitute the values into the equation, we get the coefficient of friction.

10 = μ x 5 x 9.8

μ = 0.2041

Angle of inclination for the section BC is

θ = tan⁻¹(0.2041) = 11.54⁰

Thus, the angle of inclination of section BC is  11.54⁰.

Learn more about frictional force.

brainly.com/question/14662717

#SPJ1

7 0
2 years ago
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