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Deffense [45]
3 years ago
8

Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Strings of holiday lights can be designed in one of two ways. In so

me strings of lights, each light is connected to the others along a single wire (in series). In others, each light is attached to its own wire (in parallel). Suppose a single light bulb burns out. How do you think this will affect lights that are strung along a single wire
Physics
1 answer:
IgorLugansk [536]3 years ago
8 0

Answer:

They would go out

Explanation:

This is because, in a series connection, the same current passes through each light. Since the current is the same, if one light burns out, it cuts off the rest of the other lights and thus, no current flows in the string again.

Whereas, in a parallel connection, each light is attached to its own wire and thus has a different current flowing through it than the rest of the other wires.  If one of the lights goes out, current stops flowing through it but, it doesn't affect the other lights.

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A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d
prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance  

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

V_{t}= V_{o}e^{\frac{-t}{R*C} }

Voltage in this case is the energy dissipated so

E_{t}= E_{o}e^{\frac{-t}{R*C} }

\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }

\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }

Using the equation to find capacitance

ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }

C= 5.32x10^{-6} F

C= 5.32 uF because u is the symbol for micro that is equal to 10^{-6}

8 0
3 years ago
What is your initial speed if you accelerate at 5.8 m/s/s for 3.0 seconds and achieve a final speed of 45 m/s?
Natali5045456 [20]

Answer:

27.6 m/s

Explanation:

hopefully it makes sense and is visible

:)

8 0
2 years ago
Which action involves the most efficient conversion of electrical energy into thermal energy?
azamat

Answer:

the answer is option d

because in oven which works through electricity ,it will bake the cake (heat energy is produced)

7 0
3 years ago
Let the displacement function of a particle is x(t)=(20t^2-15t+200). Find the total displacement, instantaneous velocity and ins
irga5000 [103]

Answer:

A.) 39.5 m

B.) 0

C.) 60m/s^2

Explanation:

Given that a displacement function of a particle is x(t)=(20t^2-15t+200).

To Find the total displacement,

Reduce everything by dividing them by 5

X(t) = 4t^2 - 3t + 40 ...... (1)

For instantaneous velocity, differentiate x(t). That is,

dy/dt = 60t - 15 ...... (2)

But dy/dt = velocity.

If dy/dt = 0, then

60t - 15 = 0

60t = 15

t = 15/60

t = 0.25s

Substitutes t in equation (1)

Total displacement will be

X(t) = 4(0.25)^2 - 3(0.25) + 40

X(t) = 0.25 - 0.75 + 40

Total displacement = 39.5 m

To calculate instantaneous velocity, substitute t into equation (2)

V = 60 (0.25) - 15

V = 0.

and to find instantaneous acceleration, differentiate dv/dt

dv/dt = 60

Therefore, acceleration = 60 m/s^2

4 0
2 years ago
Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

  • V_r = resultant velocity of the plane
  • \theta = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

5 0
3 years ago
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