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snow_tiger [21]
3 years ago
9

What are two things that might be found at the site of excavation?

Physics
1 answer:
denis23 [38]3 years ago
3 0
A fire or carbon dioxide ora main break
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How does Kepler's first law refine the Copernican model?
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Copernicus's model states that the sun is in the center, and that the planets move around it in a circle. Kepler's first law of planetary motion says that they move around the sun in an ellipse.
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3 years ago
What is the<br> cancer treatments
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One is chemo. Chemo is a special magnetic field like to treat cancer


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3 years ago
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A 950-kg car strikes a huge spring at a speed of 22m/s (fig. 11-54), compressing the spring 5.0m. (a) what is the spring stiffne
alukav5142 [94]

(a) The spring stiffness constant of the spring is 18,392 N/m.

(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.

<h3>Kinetic energy of the car</h3>

The kinetic energy of the car is calculated as follows;

K.E = ¹/₂mv²

K.E = ¹/₂ x 950 x 22²

K.E = 229,900 J

<h3>Stiffness constant of the spring</h3>

The stiffness constant of the spring is calculated as follows;

K.E =  U = ¹/₂kx²

k = 2U/x²

k = (2 x 229,900)/(5)²

k = 18,392 N/m

<h3>Force exerted on the spring</h3>

F = kx

F = 18,392 x 5

F = 91,960 N

<h3>Time of impact</h3>

F = mv/t

t = mv/F

t = (950 x 22)/(91960)

t = 0.23 s

Learn more about spring constant here: brainly.com/question/1968517

#SPJ4

3 0
1 year ago
A car speeds up from 18.54 m/s to<br> 29.52 m/s in 13.84 s.<br> The acceleration of the car is:
valkas [14]

Answer:

.7934m/s^{2}

Explanation:

Acceleration = change in velocity / change in time

A = 10.98m/s / 13.84s

A = .7934m/s^{2}

5 0
3 years ago
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How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t
Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

W = U_f - U_i

we know that

U_f = -\frac{GMm}{6r}

U_i = -\frac{GMm}{r}

now from above formula

W = -\frac{GMm}{6r} + \frac{GMm}{r}

W = \frac{5GMm}{6r}

so above is the work done to move the mass from surface to given altitude

7 0
3 years ago
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