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3 years ago

Whats the answer to this question show in the picture 2 questions

Physics

2 answers:

astraxan [27]3 years ago

3 0

1) D

2) I would say A, but not 100%, its the only one that makes sense tho

almond37 [142]3 years ago

3 0

The first one would be KE=0.5m(v to the 2nd power)

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A fixed 15.3-cm-diameter wire coil is perpendicular to a magnetic field 0.77 T pointing up. In 0.20 s , the field is changed to

maksim [4K]

Answer:

Induced emf will be 0.468 volt

Explanation:

We have given diameter of wire d = 15.3 cm

So radius $r=\frac{d}{2}=\frac{15.3}{2}=7.65cm$

So area $A=\pi r^2=3.14\times (7.65)^2=183.76\times 10^{-4}m^2$

Change in magnetic field dB = 0.26 - 0.77 = -0.51 T

Time for change in magnetic field dt = 0.26 sec

We know that emf is given by $e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-183.76\times 10^{-4}\times \frac{-0.51}{0.2}=0.0468volt$

5 0

3 years ago

Calculate the angle of refraction of 30.0° light shines from water into ice. The indices of refraction for water and ice are 1.3

WITCHER [35]

Answer:

The angle of refraction Ø2 equals 62.95° ≈ 63°

Explanation:

The relationship between the angles of incidence and

refraction , with respect to light or other waves passing through two different substances or media, such as glass, water or air is given by Snell's Law.

Snell's Law states that the when light travels from one medium to another, it generally refracts.

It is given by the mathematical expression;

[SinØ1°/SinØ2°] = [n2/n1]

Cross multiplying, we have;

n1 × SinØ1° = n2 × SinØ2°

where, n is the indices of refraction of each substance

Ø is the angle between the ray and the line normal to the surface.

Given the following values;

n1 = 1.36 n2 = 1.31 Ø1 = 30° Ø = ?

n1 × SinØ1° = n1 × SinØ2°

SinØ2° = [n1 × SinØ1°]/n2

substituting the values respectively;

SinØ2° = [1.36 × Sin30°]/1.31

SinØ2° = [1.36 × 0.5]/1.31

SinØ2° = 0.68 × 1.31

SinØ2° = 0.8906

Ø2° = Sin–¹(0.8906)

Ø2° = 62.95° ≈ 63°

4 0

3 years ago

the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible. what are the bes

slamgirl [31]

Answer:

1.) Micrometres screw gauge

2.) Tape rule.

Explanation:

Given that the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible.

what are the best instruments to use ?

To measure the diameter of a thin wire, the best instrument to use is known as micrometres screw gauge.

And to measure the length of a thin wire up to 1 m, the measuring device can be tape rule or long metre rule.

5 0

2 years ago

A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th

forsale [732]

Answer:

$F=-18412.9N$ , where the minus indicates the direction is opposite to that of the throw.

Explanation:

Since MKS stands for meter-kilogram-second and we know that:

$1\ hour = 3600\ seconds$

$1\ mile = 1600\ meters$

$1000g = 1kg$

We can write that:

$\frac{1\ hour}{3600\ seconds}=1$

$\frac{1600\ meters}{1\ mile}=1$

$\frac{1kg}{1000g}=1$

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

$90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s$

$110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s$

$145 g=145 g(\frac{1kg}{1000g})=0.145kg$

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is $a=\frac{\Delta v}{\Delta t}$ , so we have:

$F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}$

Taking the throw direction as the positive one, for our values we have:

$F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N$

4 0

3 years ago

How do you find the oscillation period in seconds for different pendulum lengths?

GalinKa [24]

To find:

The equation to find the period of oscillation.

Explanation:

The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.

Thus the period of a pendulum is given by the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

Where L is the length of the pendulum and g is the acceleration due to gravity.

On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.

Final answer:

The period of oscillation of a pendulum can be calculated using the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

3 0

9 months ago