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ExtremeBDS [4]
3 years ago
11

Can anyone help with this physics question?

Physics
1 answer:
lys-0071 [83]3 years ago
7 0
You can't tell without knowing the values of the resistors. Whichever resistor has less resistance (less ohms) will have more current flowing through it.
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What will occur if a resistor with high resistance is inserted into a circuit?
Anton [14]
If the resistor is in series with the rest of the circuit then a is the correct answer. The voltage across the resistor in series with another resistor is
V \frac{R}{R+r}
where R is the big resistor and r is the small one and V is the total voltage drop across both.  This is called a voltage divider
5 0
3 years ago
Read 2 more answers
4. What is the acceleration of a plane that changes velocity from 75 m/s to
kompoz [17]

Answer:

a =  \frac{v - u}{t}  \\ s =  \frac{ {v}^{2} -  {u}^{2}  }{2a}

Explanation:

Apply SUVAT

4 0
3 years ago
Please answer this. Science 7th grade.
nikdorinn [45]

Answer:

the answer would be 2

Explanation:

it would be 2 because if u look at the diagram the darkest arrow is pointsin towards earth and the moon and when the moon is infront of the sun it cause's an eclispe

6 0
3 years ago
The world's most powerful laser is the LFEX laser in Japan. It can produce a 2 petawatt(2×10^15W) laser pulse that last for 1 ps
kogti [31]

Answer:

2.82942\times 10^{24}\ W\m^2

Explanation:

d = Diameter of spot = 30 μm

r = Radius of spot = \frac{d}{2}=\frac{30}{2}=15\ mu m

P = Power of the laser = 2\times 10^{15}\ W

A = Area = \pi r^2

Intensity is given by

I=\frac{P}{A}\\\Rightarrow I=\frac{2\times 10^{15}}{\pi\times (15\times 10^{-6})^2}\\\Rightarrow I=2.82942\times 10^{24}\ W/m^2

The light intensity within this spot is 2.82942\times 10^{24}\ W/m^2

8 0
3 years ago
A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap
viktelen [127]

Answer:

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

V_r + V_L + V_c = V_{net}

now we will have

iR + L\frac{di}{dt} + \frac{q}{C} = 12 V

now we have

1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12

So we will have

q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}

at t = 0 we have

q = 0

0 = 3\times 10^{-6} + c_1  + c_2

also we know that

at t = 0 i = 0

0 = -4000 c_1 - 1000c_2

c_2 = -4c_1

c_1 = 1 \times 10^{-6}

c_2 = -4 \times 10^{-6}

so we have

q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}

at t = 0.001 we have

q = 1.55 \times 10^{-6} C

at t = 0.01

q = 2.99 \times 10^{-6} C

at t = infinity

q = 3 \times 10^{-6} C

5 0
3 years ago
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