The hereditary material is known as

Describe, in as much detail as you can, how the energy

GalinKa [24]

Answer:

As a type of thermal power station, a coal-fired power station converts chemical energy stored in coal successively into thermal energy, mechanical energy and, finally, electrical energy. The coal is usually pulverized and then burned in a pulverized coal-fired boiler.Coal-fired plants produce electricity by burning coal in a boiler to produce steam. The steam produced, under tremendous pressure, flows into a turbine, which spins a generator to create electricity.

8 0

3 years ago

A 25 n object requires a 5.0 n to start moving over a horizontal surface. what is the coefficient of static friction?

olganol [36]

Static friction is the friction that exists between two or more solids that are not moving with a relative speed. To calculate the static friction coefficient we use the formula Fs=us × n where Fs is the static friction , us is the coefficient of static friction and the n is the normal force.
thus the coefficient of static friction will be 5 N÷ 25 N = 0.2
Hence 0.2 is the coefficient of static friction

3 0

3 years ago

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Sometimes balance point may not be obtained on the potentiometer wire why

scoundrel [369]

Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage

.
The balance point is not on the potentiometer wire - this statement means that

. In that case ,
l > L
V > E

8 0

2 years ago

A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.

Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

= 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = $0.5mv^{2}$

= $0.5 x 25 x 5.6^{2}$ = 392 J

(C) average frictional force = $\frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}$

change in KE (ΔKE) = initial KE - final KE
ΔKE = $0.5mv^{2}$ - $0.5mvf^{2}$
when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE = $0.5x25x5.6^{2}$ - 0 = 392 J

\frac{-(ΔKE+ΔU)}{h}[/tex] = $\frac{-(392 + (-2940))}{12}$

= $\frac{(-392 + 2940)}{12}$ = 212.33 N

5 0

3 years ago

A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera

Fantom [35]

In order to accelerate the dragster at a speed $v_f = 100 m/s$ , its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$ :
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$

3 0

3 years ago

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