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3 years ago

Which of the following ways is usable energy lost?

Physics

2 answers:

Vsevolod [243]3 years ago

5 0

The answer to your question is D all of the above

tamaranim1 [39]3 years ago

4 0

A. Friction or all of the above

You might be interested in

If 2 objects have the same momentum which statement is true

Vika [28.1K]

C is the correct option

As momentum = (mass) (velocity)

P=mv

So the object having greater mass has less velocity and the object having smaller mass will move faster. But product of mass and velocity is same to each other.

5 0

3 years ago

An earthquake produces longitudinal P waves that travel outward at 8000 m/s and transverse S waves that move at 4500 m/s. A seis

vivado [14]

Answer:

1234285.7 m or 1234.3 km

Explanation:

Let the distance be $d$ , the time taken by P waves be $t_P$ and the time taken by the S waves be $t_S$ .

$\text{Velocity}\dfrac{\text{Distance}}{\text{Time}}$

$\text{Time}\dfrac{\text{Distance}}{\text{Velocity}}$

For the P waves,

$t_P=\dfrac{d}{8000}$

$d=8000t_P$

For the S waves,

$t_S=\dfrac{d}{4500}$

$d=4500t_S$

Equating the $d$ ,

$8000t_P=4500t_S$

Divide both sides of the equation by 500 to reduce the terms.

$16t_P=9t_S$

Since S waves arrive 2 minutes (= 120 seconds) after P waves,

$t_S-t_P=120$

$t_S=120+t_P$

Substitute this in the equation of the distance.

$16t_P=9(t_P+120)$

$16t_P=9t_P+1080$

$7t_P=1080$

$t_P=\dfrac{1080}{7}$

Substitute this in the equation for $d$ involving $t_P$ .

$d=8000t_P$

$d=8000\times\dfrac{1080}{7}$

$d=1234285.7 \text{ m }= 1234.3 \text{ km}$

4 0

3 years ago

What would happen if you use a thicker wire around the iron nail of an electromagnet? (thats the whole question)

puteri [66]

Answer:

When we have a current I, we will have a magnetic field perpendicular to this current.

Then if we have a wire in a "spring" form. then we will have a magnetic field along the center of this "spring".

Now suppose we put an iron object in the middle (where the magnetic field is) then we will magnetize the iron object.

Of course, the intensity of the magnetic field is proportional to the current, given by:

B = (μ*I)/(2*π*r)

Where:

μ is a constant, I is the current and r is the distance between to the current.

Now remember that for a resistor:

R = ρ*L/A

R is the resistance, ρ is the resistivity, which depends on the material of the wire, L is the length of the wire, and A is the cross-section of the wire.

If we increase the area of the wire (if we use a thicker wire).

And the relation between resistance and current is:

I = V/R

Where V is the voltaje.

Now, if we use a thicker wire, then the cross-section area of the wire increases.

Notice in the resistance equation, that the cross-section area is on the denominator, then if we increase the area A, the resistance decreases.

And the resistance is on the denominator of the current equation, then if we decrease R, the current increases.

If the current increases, the magnetic field increases, which means that we will have a stronger electromagnet.

3 0

3 years ago

Urgent<br> Please write the complete answer

Crazy boy [7]

Multiply field strength (N/kg) by mass (kg) to get weight (N)

At the start, the car is carrying

4.7 kg * (9.8 N/kg) = 46.06 N

of fuel.

At the end, it is carrying

3.0 kg * (9.8 N/kg) = 29.4 N.

Assuming the car remains completely intact, its weight was reduced by

46.06 N - 29.4 N = 16.66 N.

5 0

3 years ago

A man works on a striaght road from his home to market 2.5km away dwith a speed of 5km/h.Finding the market closed, he instantly

umka21 [38]

Answer:

5.625km/h

Explanation:

We are given that

Distance between home and market, d=2.5 km

Speed, v1=5km/h

Speed, v2=7.5 km/h

We have to find the average speed of the man over the interval of time 0 to 40min.

Time, $t=\frac{distance}{speed}$

Using the formula

$t_1=\frac{2.5}{5}=0.5 h=0.5\times 60=30 min$

1 hour =60 min

$t_2=\frac{2.5}{7.5}=\frac{1}{3}hour=\frac{60}{3}=20min$

Distance traveled by man in 10 min with speed 7.5 km/h= $\frac{2.5}{20}\times 10=2.5/2km$

Therefore,

Total distance covered=2.5+2.5/2=3.75 km

Time=40 min=40/60=2/3 hour

Average speed= $\frac{total\;distance}{total\;time}$

Average speed= $\frac{3.75}{2/3}=5.625km/h$

7 0

3 years ago