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Lemur [1.5K]
2 years ago
11

Why do astronauts feel weightless in space?

Physics
1 answer:
JulsSmile [24]2 years ago
6 0
Weight = mass × gravitational field strength
At space gravitational field strength is very low so its"weightless".
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Between which two points of a concave mirror should an object be placed to obtain a magnificati of -3​
kobusy [5.1K]

Answer:

When object is placed between the focus (F) and pole (P) of a concave mirror, magnified and erect image of the object is formed on the back of the mirror.

When object is placed between the centre of curvature and the principal focus of a concave mirror, magnified and inverted image is formed in front of the mirror.

Explanation:

8 0
3 years ago
2. Use the diagram below to answer this question. As the ball moves from point A
Thepotemich [5.8K]

Answer:

at point A the ball possess pontetial energy , point B kinetic energy then point C pontetial energy

5 0
3 years ago
the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m
olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

x = \frac {-nmg}{k}

Substituting into the formula, we have;

x = \frac {-49*0.002*9.8}{24}

x = \frac {-0.9604}{24}

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

3 0
2 years ago
Help! <br><br><br>I'm not sure my answer is correct.<br><br><br>Problem is attached!​
kaheart [24]
  • Let G be xN

\\ \sf\longmapsto x-20=30

\\ \sf\longmapsto x=30+20

\\ \sf\longmapsto x=50N

Option B

3 0
2 years ago
Read 2 more answers
Help!!! I need it today <br> Thank you in advance
svlad2 [7]

Answer:

 F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.

Explanation:

The student wants to prove hooke's law which has the form

          F = - k (x-xo)

To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.

Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,

we must be careful when hanging the weights so as not to create oscillations in the spring

we look for the mass of each weight

         W = mg

          m = W / g

and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.

The fact of obtaining a line already proves Hooke's law.

5 0
2 years ago
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