Which of the following is true about ALL energy conversions?

HELP ASAP! <br>why is rolling friction much smaller than sliding friction? Also give an example.

IceJOKER [234]

Answer:

Rolling friction is much smaller than sliding friction because Rolling friction is considerably less than sliding friction as there is no work done against the body that is rolling by the force of friction. For a body to start rolling a small amount of friction is required at the point where it rests on the other surface, else it would slide instead of roll.

Rolling Friction example: Anything with weels (cars,skateboards) or a ball rooling.

Sliding Friction example: Bicycle brakes,skinning your knee walking,writing.

6 0

3 years ago

An electromagnet is a solenoid with a piece of ferromagnetic material within it.

Vlada [557]

ANSWER:
The answer will be OT

3 0

3 years ago

A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa

Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

$Q=P\tan\phi$ ...(I)

We need to calculate the $\phi$

Using formula of $\phi$

$\phi=\cos^{-1}(Power\ factor)$

Put the value into the formula

$\phi=\cos^{-1}(0.6)$

$\phi=53.13^{\circ}$

Put the value of Φ in equation (I)

$Q=600\tan(53.13)$

$Q=799.99\ kVAR$

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

$Q'=600\tan(0.95)$

$Q'=9.94\ KVAR$

We need to calculate the difference between Q and Q'

$Q''=Q-Q'$

Put the value into the formula

$Q''=799.99-9.94$

$Q''=790.05\ KVAR$

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0

3 years ago

A proton in a particle accelerator is traveling at a speed of 0.99c.(a) If you use the approximate nonrelativistic equation for

____ [38]

Answer:

a) p = 4.96 10⁻¹⁹ kg m / s , b) p = 35 .18 10⁻¹⁹ kg m / s ,

c) p_correst / p_approximate = 7.09

Explanation:

a) The moment is defined in classical mechanics as

p = m v

Let's calculate its value

p = 1.67 10⁻²⁷ 0.99 3. 10⁸

p = 4.96 10⁻¹⁹ kg m / s

b) in special relativity the moment is defined as

p = m v / √(1 –v² / c²)

Let's calculate

p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)

p = 4.96 10⁻¹⁹ / 0.141

p = 35 .18 10⁻¹⁹ kg m / s

c) the relationship between the two values is

p_correst / p_approximate = 35.18 / 4.96

p_correst / p_approximate = 7.09

4 0

3 years ago

The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo

ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

$(V_{f})^{2}=(V_{o})^{2}+2aS\\ So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}$

As we find acceleration.Now we need to find time

So

$V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s$

Now for distance

So

$Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m$

7 0

3 years ago