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nadya68 [22]
2 years ago
14

The metal wire in an incandescent lightbulb glows when the lights is switch on and stops glowing when switched off. This simple

process is which kind of a change?
Physics
2 answers:
Radda [10]2 years ago
7 0

Opening or closing a circuit.

lutik1710 [3]2 years ago
4 0

Answer:

When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat.  

Explanation:

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A golf club with mass 0.5 kg is swung at 23 m/s toward a stationary golf ball on a tee. The club barely slows down to 17 m/s, bu
mylen [45]

Answer:

I gotchu fam. 0.0435 kg

Explanation:

m1v1+m2v2=m1v1f+m2v2f

(0.5)(23)+(m2)(0)=(0.5)(17)+(m2)(69)

11.5=8.5+69m2

3=69m2

0.0435 kg

5 0
2 years ago
How do determine which is Y2, Y1, X2, and X1 on a graph. <br><br> And how do find rise over run.
fredd [130]
One point will be X1,Y1 and the other will be X2,Y2. It does not matter which is which except that X1 and Y1 have to be the same point and X2 and Y2 have to be the same point. For example, let's say you were given (2,3) and (6,8). No matter which point is X1,Y1 and the other is X2,Y2, the slope will still be 5/4. 

The rise is the change in y from one point to the other. The run would be the change in x from one point to the other.
7 0
3 years ago
A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b)
Alexeev081 [22]

Answer:

a) 0.01232 A

b) 0.00075 s = 0.75 ms

c) 0.0045323 A = 4.532 mA

d) 3.894 V

Explanation:

R = 500 Ω

V = 6.16 V

C = 1.50 μF

Let Vs be the voltage of the emf source

Let Vc be the voltage across the capacitor at any time

a) Current flows as a result of potential difference between two points. So, the current flows according to difference in voltage between the emf source and the capacitor.

At time t = 0,

There is no voltage on the capacitor; Vc = 0 V

Current in the circuit is given by

I = (Vs - Vc)/R

I = (6.16 - 0)/500

I = 0.01232 A

b) Time constant for an RC circuit is given by τ

τ = RC = (500) (1.5 × 10⁻⁶) = 0.00075 s

c) The current decay in an RC circuit (called decay because the current in the circuit starts to fall as the capacitor's voltage rises as the capacitor charges) is given by

I = I₀ e⁻ᵏᵗ

where k = (1/τ)

I₀ = Current in the circuit at t = 0 s; I₀ = 0.01232 A

At t = τ = 0.00075 s, kt = (τ/τ) = 1

I = 0.01232 e⁻¹ = 0.0045323 A = 4.532 mA

d) The voltage for a charging capacitor is given by

Vc = Vs (1 - e⁻ᵏᵗ)

where k = (1/τ)

At t = τ = 0.00075 s, Vc = ?, Vs = 6.16 V, kt = 1

Vc = 6.16 (1 - e⁻¹) = 6.16 (0.6321)

Vc = 3.894 V

4 0
3 years ago
Read 2 more answers
Calculate the monentun pf 75 kg bicycle and boy who has ghe velocitg of 3m/s
Grace [21]

Explanation:

n=75 kg

v=3m/s

m=n×v

m=75x3

m= 225n

6 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.
BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

y_{i}=0m

v_i=80\frac{m}{s}

a=4\frac{m}{s^2}

y_{e}=1000m

g=9.8\frac{m}{s^2}

From Kinematics, the position y as a function of time when the engine still works will be:

y(t)=v_it+\frac{1}{2}at^2

At what time the altitud will be y_{e}=1000m?

v_it+\frac{1}{2}at^2=y_{e} ⇒ \frac{1}{2}at^2+v_it-y_{e}=0

Using the quadratic formula: t_1=10s.

How much time does it take for the rocket to touch the ground? No the function of position is:

y(t)=y_{e}+v_et-\frac{1}{2}gt^2

Where our new initial position is y_{max}, the velocity when the engine breaks is v_e=v_i+at=120\frac{m}{s} and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

y_{e}+v_et-\frac{1}{2}gt^2=0

Again, using the quadratic ecuation:

t_2=24.49s

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

t_T=t_1+t_2=34.49s.

6 0
3 years ago
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