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4 years ago

While online last week, you saw the following advertisement:

Physics

2 answers:

tankabanditka [31]4 years ago

8 0

The answer for Apex is B.

Karolina [17]4 years ago

3 0

No scientific testing has been made to check for ion transfer, and the claims are purely empirical. Also, nine out of ten people is hardly a representative sample, and the people can claim whatever they want since "feeling" is subjective. This is most likely a pseudoscientific claim, made to sound legitimate to consumers. The best answer is choice D.

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If you throw an object straight up into the air with an initial velocity of 42m/s. What is it’s velocity at the peak of its fl

solniwko [45]

0 m/s. It’s trivial

7 0

3 years ago

If a car accelerates at 6.4 m/s?

lawyer [7]

initial velocity=u=0
Acceleration=a=6.4m/s
Time=t=9s

Final velocity be v

According to first equation of kinematics

v=u+at
v=0+6.4(6)
v=57.6m/s

6 0

3 years ago

An object moving at 30 m/s takes 5s to come to a stop. what is the object's acceleration?

Taya2010 [7]

Use SUVAT where:
S = distance
U = initial velocity = 30m/s
V = end velocity = 0 m/s
A = acceleration = ?
T = time = 5 secs

then pick out the equation v = u + at

so:
0 = 30 + 5a
-30 / 5 = a
a = -6 m/s^2

4 0

3 years ago

A scientist fixes a test charge q′ to point P and then measures the electrostatic force it experiences there. She then calculate

nordsb [41]

Answer:

<em>Correct answer: B.</em>

Explanation:

Electrostatic Field

It measures the electric effect of a charge distribution in its surroundings. If we wanted to test or measure the electric field by using a point-charge of value q', then the electric field is

$\displaystyle E=\frac{F}{q'}$

The electrostatic force between two point charges is

$\displaystyle F=\frac{k\ q_1\ q'}{r^2}$

So, the electric field is

$\displaystyle E=\frac{\frac{k\ q_1\ q'}{r^2}}{q'}$

$\displaystyle E=\frac{k\ q_1}{r^2}$

The theory shows that the electric field doesn't depend on the test charge used, i.e., if we now use q'=-q', the electric force will be of the same magnitude but in the opposite direction, thus the electric field will be the same. Let's recall the formula is used to compute the scalar value of the field, the direction must be studied separately.

Now, if we changed the test charge to another value, say 2q, the measured force will be

$\displaystyle F=\frac{k\ q_1\ 2q}{r^2}$

And the new electric field is

$\displaystyle E=\frac{k\ q_1}{r^2}$

We can see the electric field is not affected by the value of the test charge.

Finally, if we move the test charge to another location and keep the same charge, the electric force will vary and the electric field will be different.

Correct answer: B.

7 0

3 years ago

Helppp pls yes or no question

svp [43]

Answer:

yes, should be

Explanation:

This is a hard yes or no question becuase the amplitudes are the same height but in different beating orders.

7 0

3 years ago

Read 2 more answers