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sleet_krkn [62]
3 years ago
8

Suppose you increase your walking speed from 7 m/s to 13 m/s in a period of 3 s. What is your acceleration? i need the answer an

d the units
Physics
1 answer:
Alekssandra [29.7K]3 years ago
5 0
2 m/s per second is the answer
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It takes 300 newtons of force and a distance of 20 meters for a moving car to come to stop
Andrei [34K]
The answer is 15 I think.
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3 years ago
Which statement describes the speed of electromagnetic waves?
goldfiish [28.3K]

Answer:

Their speed in a vacuum is a constant value.

Explanation:

Electromagnetic waves consits of oscillations of electric field and magnetic field. The oscillations of these fields occur in a direction perpendicular to the direction of propagation of the waves, so they are transverse wave. Electromagnetic waves, contrary to mechanical waves, do not need a medium to propagate, so they can also travel through a vacuum. In a vacuum, their speed is constant and has always the same value, the speed of light:

c=3\cdot 10^8 m/s

7 0
2 years ago
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What is the refractive index of air if light travels through it at 3.0 108 m/s?
olchik [2.2K]

Answer:

n = c/v = (3.00 x 108 m/s)/(2.76 x 108 m/s) = 1.09. This does not equal any of the indices of refraction listed in the table.

4 0
2 years ago
-2 m
Nana76 [90]
Report this clown who put the first answer he’s trying to get your ip
4 0
2 years ago
A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th
fiasKO [112]
Let's use the mirror equation to solve the problem:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}
where f is the focal length of the mirror, d_o the distance of the object from the mirror, and d_i the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for d_i by using the numbers given in the text of the problem:
\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}
\frac{1}{d_i}= -\frac{7}{60 cm}
d_i = -8.6 cm
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.
6 0
3 years ago
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