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Aleksandr-060686 [28]

3 years ago

14

You are in a car and go around a corner very fast. What happens to you?

1 answer:

Fiesta28 [93]3 years ago

3 0

Answer:

You may tip the car over or crash.

Explanation:

Going really fast at high speeds and turn a corner might make the car crash into an object or fall over.

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An RC circuit consists of a resistor with resistance 1.0 kΩ, a 120-V battery, and two capacitors, C1 and C2, with capacitances o

Serhud [2]

Answer:

$Q_t= 8.3 * 10^3 C$

Explanation:

From the question we are told that:

Resistor $R=1000ohms$

Voltage $v=120_V$

Capacitance of c_1 $c_1=20 \mu F$

Capacitance of c_2 $c_2=60 \mu F$

Time $t=0$

Generally the equation for charges is mathematically given by

$For C_1\\Charge\ on\ C_1 = CV = 20*120 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2.4 x 10^-3 C\\$

$ForC_2\\Charge on C_2 = 60*120 =7200 μC = 7.2 x 10^-3\\Charge on C_2 = 7.2 x 10^-3$

Generally the equation for voltage across capacitors is mathematically given by

$V_c(t)=V(1-e^{-t/RC})$

$C=C_1+C_2=80 \mu f\\t=2RC=>160000s$

$V_c(t)=120(1-e^{-(160000)/1000*(80)})$

$V_c(t)=103.7598$

Generally the equation for charges is mathematically given by

$Q1(t) = C1Vc(t)\\Q1(t) = 20*103.7598\\Q1(t) = 2075.196\\\\Q2(t) = 60*103.7598\\Q2(t) = 6225.6\\$

Generally the equation for total charges $Q_t$ is mathematically given by

$Q_t=Q1(t)+Q2(t)$

$Q_t= 8.3 * 10^3 C$

5 0

3 years ago

There are stars located in the center bulge of the Milky Way and the spiral arms of the Milky Way. What is the difference betwee

nadya68 [22]

Answer:

The stars at the center bulge are bigger and brighter than the stars in the arms.

Explanation:

5 0

3 years ago

A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is th

IceJOKER [234]

Answer: $E=75\ N/m$

Explanation:

Given

First charge of $q_1=15\ nC$ is placed at $x=1.5\ m$

Second charge $q_2=-20\ nC$ is placed at $y=-2\ m$

Electric field is given by

$E=\frac{kq}{r^2}$

Electric field due to $q_1$ is away from it

$E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}$

$E_1=60\ N/m$

Electric field due to $q_2$

$E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}$

$E_2=45\ N/m$

Net electric field will be vector addition of two

$\vec{E_{net}}=\vec{E_1}+\vec{E_2}$

$\vec{E_{net}}=-60\hat{i}-45\hat{j}$

Magnitude of Electric field is

$E=\sqrt{60^2+45^2}$

$E=75\ N/m$

8 0

3 years ago

Five race cars speed toward the finish line at the Jasper County Speedway. The table lists each car’s speed in meters/second. If

nekit [7.7K]

We have: v = d/t
From the expression, we conclude speed is indirectly proportional to speed.
So, the car which will take longer time must have the smallest speed. Among all the options Car C has the smallest speed. So, it would be your answer.

In short, Your Answer would be Option C) Car C will take longer time than any other.

Hope this helps!

6 0

3 years ago

Read 2 more answers

Black holes must have a mass _____ or more times the mass of the Sun.

statuscvo [17]

<span>Black holes must have a mass THREE or more times the mass of the Sun.</span>

8 0

4 years ago