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3 years ago

An engine converts 95% of its energy to mechanical. What happens to the other 5% of its energy?

Physics

1 answer:

Elza [17]3 years ago

6 0

Answer:

Here, 5% of energy wasted as a heat lost or work done against frictional forces.

Explanation:

No engine has yet made that can convert its all input energy into mechanical energy. Some of the input energy goes in vain in the form of heat and friction.

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A block is at rest on the incline shown in the gure. The coefficients of static and kinetic friction are 0.6 and 0.51, respec- t

Alekssandra [29.7K]

Normal reaction force on the block while it is at rest on the inclined plane is given as

$F_n = mgcos\theta$

here we know that

m = 46 kg

$\theta = 29^o$

now we will have

$F_n = 46*9.8*cos29 = 394.3 N$

now the limiting friction or maximum value of static friction on the block will be given as

$F_s = \mu_s * F_n$

$F_s = 0.6 * 394.3 = 236.56 N$

Above value is the maximum value of force at which block will not slide

Now the weight of the block which is parallel to inclined plane is given as

$F_{||} = mg sin\theta$

here we know that

$F_{||} = 46*9.8 sin29 = 218.55 N$

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.

So here <u>friction force on the given block will be same as its component on weight which is 218.55 N</u>

5 0

3 years ago

5. A stream of air at 12 bar and 900 K is mixed with another stream of air at 2 bar and 400 K with 2 times the mass flowrate. If

Softa [21]

Answer:

The anserrs to the question are

(a) The temperature would be 566.67 K and

(b) The pressure of the resulting air stream is 14 bar

Explanation:

Here the two streams of gases meet ad form a single stream

The steady-flow energy equation can be implemented at the mixing point of the to streams as follows

mₐhₐ₁ + mₙ hₙ₁ + Q° + W° = mₐhₐ₂ + mₙhₙ₂

Where the flow is adiabatic, we have

Q = 0 and W = 0 hence

mₐhₐ₁ + mₙ hₙ₁ = mₐhₐ₂ + mₙhₙ₂ where h = cp×T we have

mₐcpₐ×Tₐ + mₙcpₙ×Tₙ = mₐcpₐ×T + mₐcpₙ×T

Therefore the final temperature T is given by

$T = \frac{m_ac_{pa}T_a + m_nc_{pn}T_n}{m_ac_{pa} + m_nc_{pn}}$ for the same kind of gas cpₐ =cpₙ therefore

$T = \frac{m_aT_a + m_nT_n}{m_a + m_n}$ where mₐ and mₙ are mass flow rate

Therefore we have where Tₐ = = 900 K and Tₙ = 400 K and mₙ = 2mₐ gives

$T = \frac{m_a*900 + 2*m_a*400}{m_a + 2*m_a}$ = $T = \frac{900 + 800}{1 + 2}$ = 1700/3 = 566.67 K

From Dalton's law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the components of the mixture

Therefore the total pressure of the combined stream = pₐ + pₙ = p

= 12 bar + 2 bar = 14 bar

Stream pressure = 14 bar

4 0

3 years ago

Humans can see several thousand shades of color but have cone photoreceptors that are sensitive to only three (perhaps four) wav

kipiarov [429]

Answer:

B. Color perception is achieved by activation of various combinations between the tree cone types.

Explanation:

Human eye have only three photo receptor ( three wavelength) but they can see several thousands of shades of color because color perception is gained by combination of these three wave length cone types ( RED, BLUE AND GREEN) into various combinations. Our retina have two types of cell:- cone cells and cylindrical cells. Cone cells are responsible for the identification of color.

4 0

3 years ago

Brainly i can you please help me

Art [367]

Answer:

bet

Explanation:

4 0

3 years ago

A 2.4-kg cart is rolling along a frictionless, horizontal track towards a 1.7-kg cart that is held initially at rest. The carts

inessss [21]

Answer:

Explanation:

Mass of first cart M1=2.4kg

Velocity of first cart U1=4.1m/s

Mass of second cart M2=1.7kg

Second cart is initially at rest U2=0

After an instant, the velocity of the second cart is U2=-2.8m/s

Now after collision the two cart move together with the same velocity I.e inelastic collision

Using conservation of momentum

Momentum before collision, = momentum after collision

M1U1 + M2U2 = (M1+M2)V

2.4×4.1 + 1.7× -2.8 =(2.4+1.7)V

9.84 - 4.76 = 4.1V

5.08=4.1V

V=5.08/4.1

V=1.24m/s

The momentum of the two cart at that instant is

M1U1+M2U2

2.4×4.1 + 1.7× -2.8

9.84 - 4.76

5.08kgm/s

So the momentum at the instant the velocity is 4.1m/s for cart 1 and -2.8m/s for cart 2 is 5.08kgm/s

3 0

3 years ago