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3 years ago

Look at the circuit below. What is the voltage between points C and D?

Physics

2 answers:

stepan [7]3 years ago

7 0

Option c) 1.5 V

Explanation:

<em>As the circuit is build in series first we will find the current passing through the complete circuit. Current stays the same in each element is the series cirucuit, however, the voltage is different.</em>

Voltage is given by the following formula:

V = IR

<em>Because we have to find current through whole circuit, we will first find resistance of the whole circuit.</em>

Equivalent Resistance R(eq): R1 + R2 = 60 + 60 = 120 ohm

Current passing through whole circuit be:

$I = \frac{V}{R(eq)}\\I = \frac{3}{120}$ = 0.025

Now we will find out the voltage between C and D:

Current stays the same in series circuit: I = 0.025 c

Resistance between C and D is, R = 60 ohm

Voltage becomes, V = IR = 0.025 * 60 = 1.5 V

andre [41]3 years ago

3 0

Answer:

Explanation:

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How is a crowbar used to remove the top of a box change the direction of the force

Lemur [1.5K]

The crowbar is placed under a small edge under the top of the box. The crowbar will send the box lid flying from the opposite direction that the crowbar is placed. Please give Brainlest if you understand.

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3 years ago

Monochromatic light from a distant source is incident on a slit 0.705 mm wide. On a screen 2.13 m away, the distance from the ce

vampirchik [111]

Answer:

492.183 nm

Explanation:

x = Distance from the central maximum to the first minimum = 1.35 mm

l = Distance of screen = 2.13 m

d = Distance of gap = 0.705 mm

m = Order = 1

We have the relation

$tan\theta=\dfrac{x}{l}\\\Rightarrow \theta=tan^{-1}\dfrac{1.35\times 10^{-3}}{2.13}\\\Rightarrow \theta=0.04^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow \lambda=\dfrac{dsin\theta}{m}\\\Rightarrow \lambda=\dfrac{0.705\times 10^{-3}\times sin0.04}{1}\\\Rightarrow \lambda=4.92183\times 10^{-7}=492.183\ nm$

The wavelength of the light is 492.185 nm

5 0

3 years ago

Read 2 more answers

The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach to

irga5000 [103]

(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration $a_1$ for a total time $t_1$ During this part of the motion, he covers a distance equal to $s_1 = 45 m$ , until he finally reaches a velocity of $v_1 = u + a_1t_1$ . We can use the following suvat equation:

$s_1 = u t_1 + \frac{1}{2}a_1t_1^2$

which reduces to

$s_1 = \frac{1}{2}a_1 t_1^2$ (1)

since u = 0.

- In the second part, he continues with constant speed $v_1 = a_1 t_1$ , covering a distance of $d_2 = 55 m$ in a time $t_2$ . This part of the motion is a uniform motion, so we can use the equation

$s_2 = v_1 t_2 = a_1 t_1 t_2$ (2)

We also know that the total time is 10.0 s, so

$t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)$

Therefore substituting into the 2nd equation

$s_2 = a_1 t_1 (10-t_1)$

From eq.(1) we find

$a_1 = \frac{2s_1}{t_1^2}$ (3)

And substituting into (2)

$s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1$

Solving for t,

$s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s$

So from (3) we find the acceleration in the first phase:

$a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2$

And so the average force exerted on the sprinter is

$F=ma=(66 kg)(2.34 m/s^2)=154.5 N$

b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

$v_1 = u +a_1 t_1$

where we have

u = 0

$a_1 =2.34 m/s^2$ is the acceleration

$t_1 = 6.2 s$ is the time of the first part

Solving the equation,

$v_1 = 0 +(2.34)(6.2)=14.5 m/s$

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3 years ago

A clock on a moving train runs __________ an identical clock at rest.

katovenus [111]

It depends who is watching it.

-- A passenger on the train says that his clock is normal, but the clock at the station he just passed is running slow.

-- A ticket agent at the station says that his clock is normal, but the clock on the train that just rolled through is running slow.

-- They're both correct, but ...

-- The difference is too small for either of them to notice, unless the train is going faster than about half the speed of light.

-- It isn't.

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3 years ago

4. How far apart are a proton and electron if they exert an attractive force of 3 N on one another?

nikitadnepr [17]

Answer:

For the force due to electromagnetism we use Coulomb’s inverse square law:

F=keq1q2r2

(where F is the electrostatic force, q1 & q2 are the two charges, r is the distance between them, and Coulomb’s Constant ke=14πϵ0=8.99×109Nm2C−2, where ϵ0 is the permittivity of free space)

r=keq1q2F−−−−−√

r=8.99×109Nm2C−2−1.60×10−19C×−1.60×10−19C1.0N−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

r=1.52×10−14m

Explanation:

7 0

3 years ago