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emmainna [20.7K]
2 years ago
13

Huge loops of gas that link different parts of sunspot regions are called.

Physics
1 answer:
lapo4ka [179]2 years ago
3 0

Answer:

Prominences

Explanation:

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Which diagram shows diffraction as light passes through an opening?
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A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
A 9.0-V battery moves 20 mC of charge through a circuit running from its positive terminal to its negative terminal. How much en
dem82 [27]

Answer:

E = 0.18 J

Explanation:

given,

Potential of the battery,V = 9 V

Charge on the circuit, Q = 20 m C

                                        = 20 x 10⁻³ C

energy delivered in the circuit

E = Q V

E = 20 x 10⁻³ x 9

E =  180 x 10⁻³

E = 0.18 J

Energy delivered in the circuit is equal to E = 0.18 J

7 0
3 years ago
If anyone here can help me with physics please comment. its about coulombs law
Minchanka [31]
I may be able to help you with This subject
3 0
3 years ago
A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th
vladimir1956 [14]

Answer:

Force, F=2.27\times 10^4\ N

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

F.d=\dfrac{1}{2}m(v^2-u^2)

Finally, it stops, v = 0

F.d=-\dfrac{1}{2}m(u^2)

F=\dfrac{-mu^2}{2d}

F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}

F = -22713.92 N

F=2.27\times 10^4\ N

So, the magnitude of the force that stops the bullet is 2.27\times 10^4\ N

7 0
3 years ago
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