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3 years ago

13

What potential difference is dropped over the 60 ohms resistor

1 answer:

timurjin [86]3 years ago

7 0

Voltage = (current) x (resistance).

The current through the 60 ohms resistor is 0.10 A.

V = (0.10 A) x (60 ohms)

V = 6 volts

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Jamal is playing on the swings in the playground. At what point does he have the greatest potential energy?

denis23 [38]

Answer:

D

Explanation:

8 0

4 years ago

A satellite is in a circular Earth orbit of radius r. The area A enclosed by the orbit depends on r2 because A = πr2. Determine

Salsk061 [2.6K]

Answer:

a. $T=r^{3/2}$

b. $K=\frac{1}{r}$

c. $v=\frac{1}{\sqrt{r}}$

d. $v=\sqrt{r}$

Explanation:

To make analysis about the satellite circular earth the depends or r and A

$T^2=\frac{4\pi}{GM}*r^3$

$K=\frac{GM*m}{2*r}$

a.

$T^2=r^3$

$T=r^{3/2}$

b.

$K=\frac{GM*m}{2}*\frac{1}{r}$

$K=\frac{1}{r}$

c.

$K=\frac{1}{2}*m*v^2$

$v^2=\frac{2*K}{m}=\frac{2*Gm*m}{2*m*r}$

$v=\frac{1}{\sqrt{r}}$

d.

$v=\sqrt{2*GMr}$

$v=\sqrt{r}$

3 0

3 years ago

When a comet is within the inner solar system, its visible tails point __________.

astraxan [27]

away from the sun

hope it helps...!!!!

6 0

2 years ago

(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force

balandron [24]

Answer:

The force is $F = 1164.6\ lbf$

The time is $\Delta t = 2.44 \ s$

Explanation:

From the question we are told that

The mass of the car is $m = 2500 \ lbm$

The initial velocity of the car is $u = 25 \ mi/hr$

The final velocity of the car is $v = 50 \ mi/hr$

The acceleration is $a = 15 ft/s^2 = \frac{15 * 3600^2}{ 5280} = 36818.2 \ mi/h^2$

Generally the acceleration is mathematically represented as

$a = \frac{v-u}{\Delta t}$

=> $36818.2 = \frac{50 - 25 }{ \Delta t}$

=> $t = 0.000679 \ hr$

converting to seconds

$\Delta t = 0.0000679 * 3600$

=> $\Delta t = 2.44 \ s$

Generally the force is mathematically represented as

$F = m * a$

=> $F = 2500 * 15$

=> $F = 37500 \ \frac{lbm * ft}{s^2}$

Now converting to foot-pound-second we have

$F = \frac{37500}{32.2}$

=> $F = 1164.6\ lbf$

7 0

3 years ago

What is the strength of an electric field that will balance the weight of a 3.3 g plastic sphere that has been charged to -7.6 n

sp2606 [1]

As the plastic sphere is charged, therefore it experience an electric force when placed in an electric fields and also experiences gravitational force acts downward so the electric force must act upward.

Let $F_{E}$ is electric force and $F_{G}$ is gravitational force.

If these forces are balanced, therefore $F_{E} = F_{G}$

or $\left | q \right |E=mg\\\\\ E=\frac{mg}{\left | q \right |}$

Given, $q=-7.6 nC=-7.6\times10^{-9} C$ and $m=3.3 g= 3.3\times10^{-3} kg$ .

Substituting these values in above equation we get,

$E=\frac{3.3\times10^{-3} kg\times9.8 m/s^{2} }{7.6\times10^{-9} C}\\\\E=4.2\times10^{6} N/C$

Thus, the magnitude of electric field is $4.2\times10^{6} N/C$ .

As the charge is negative, the electric field at the location of the plastic sphere must be pointing downward.

8 0

3 years ago