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kumpel [21]
3 years ago
9

Using the data table to the right, determine the

Physics
2 answers:
SVEN [57.7K]3 years ago
8 0
C I hope it helps I think
andreyandreev [35.5K]3 years ago
6 0
Silver sable is in spiderman lol
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A woman is driving her car due east at a velocity of 10 m/s. If the woman has a mass of 50 kg and her car has a mass of 1.000 kg
brilliants [131]

Answer: 10,500 kg m/s

Explanation: (1,000 + 50)(10)

7 0
3 years ago
The angle of refraction rounded to the nearest whole number
Alika [10]

Answer:

15^{\circ}

Explanation:

Using Snell's law which is represented by

n_1sin\theta_1 = n_2sin\theta_2

Making \theta_2 the subject of the formula then

\theta_2=sin^{-1}(\frac {n_1sin\theta_1}{n_2})

Here \theta_1 and \theta_2 are the angles of incidence and refraction in water and air respectively

n_1 and n_2 are refraction index

Substituting 1.0003 for n_1 and 1.33 for n_2 then 20^{\circ} for \theta 1 we obtain

\theta_2=sin^{-1}(\frac {1.0003\times sin 20^{\circ}}{1.33})=14.90606875^{\circ}\approx 15^{\circ}

6 0
3 years ago
Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is
Xelga [282]

Answer:

7.3114*10^{-5}V

Explanation:

To give a solution to the exercise, it is necessary to consider the concepts related to magnetic flux and Faraday's law of induction. Faraday's law states that the voltage induced in a closed circuit is directly proportional to the speed with which the magnetic flux that crosses any surface with the circuit as an edge changes over time.

It is represented under the equation,

\epsilon = N \frac{\Delta\Phi}{\Delta t}

Where,

\epsilonis the induced electromotive force

N = Number of loops

\Delta t= Time

\Delta\Phi= Magnetic Flux

For definition the change in magnetic flux is:

\Delta \Phi = \Delta B A cos\phi

Where,

B= Magnetic field

Substituting at the first equation we have

\epsilon = N \frac{\Delta B A Cos\phi}{\Delta t}

\epsilon = N \frac{(B_2-B_1) (\pi r^2) Cos\phi}{\Delta t}

Our values are given by,

N = 1 turn

B_2 = 1T

B_1 = 0T

r = 1.6mm

\phi = 0\°

\Delta t = 100ms

Replacing,

\epsilon = (1) \frac{(1-0) (\pi (1.6*10^{-3})^2) Cos(0)}{110*10^{-3}}

\epsilon = 7.311*10^{-5}V

<em>Therefore the magnitud of the induced emf around a horizontal circle of tissue is 7.3114*10^{-5}V</em>

8 0
4 years ago
Please can somebody help me
Airida [17]
Possibly, what’s the question?
8 0
4 years ago
Read 2 more answers
A body of mass 20kg initially at rest is subjected to a force of 40m for 15sec. Calculate the change in k.e
Ede4ka [16]

Explanation:

mass(m)=20kg

velocity(v)=d/t=2.67

k.e=?

now,

k.e=1/2mv^2

=1/2*20*(2.67)^2

=71J

3 0
2 years ago
Read 2 more answers
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