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Norma-Jean [14]

3 years ago

7

A car accident occurs around midnight on the night of a full moon. The driver at fault claims he was blinded momentarily by the

Moon rising on the eastern horizon. Should the police believe him?

2 answers:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer:

No

Explanation:

It is well explained in the question itself that the accident took place at mid night on a full moon.

At mid night , the moon will be high on sky, it will not be rising.

So the statement of driver is false as there will not be any moon rising on eastern horizon when moon is already at the highest on sky.

This shows he is certainly lying and police should not believe him.

Flura [38]3 years ago

5 0

At full moon the moon is directly opposite the sun as viewed from earth.

That is, when the sun is setting the moon is rising. At midnight a full

moon would be directly overhead.

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A 12 kg<br> mass is lifted to a height of 2 m. What is its potential energy<br> at this position?

Romashka-Z-Leto [24]

Answer:

Explanation:

Potential energy is the energy stored within an object, due to the object's position, arrangement or state

4 0

3 years ago

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The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use

saul85 [17]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

$M = \frac{Nl}{f_ef_0}$

Where

N = Near point

l = distance between the object lens and eye lens

$f_0$ = Focal length

$f_e$ = Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

$M = \frac{25*10}{3*5}$

$M = 16.67\approx 17\\$

Therefore the correct answer is C.

3 0

3 years ago

Please help me solve all of them ( a, b, c and d ) thankiew !! <br> I’m also kind of in a rush

Sergio039 [100]

Answer:

a-

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

b

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

c

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

d

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

V = IR

12V = I × 5Ohm

I=2.4 A

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2 years ago

Calculate the speed of the car at each checkpoint by

lana66690 [7]

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Explanation: I did the assignment and got it correct :)

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3 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

3 years ago