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riadik2000 [5.3K]
3 years ago
10

Al is floating freely in her spacecraft, and you are accelerating away from her with an acceleration of 1g. 5) How will you feel

in your spacecraft? A) You will be floating weightlessly. B) You will feel weight, but less than on Earth. C) You will feel weight, but more than on Earth. D) You will feel the same weight as you do on Earth. E) You will feel yourself pressed against the back of your spaceship with great force, making it difficult to move.
Physics
1 answer:
wariber [46]3 years ago
5 0

Answer:

D. You will feel the same weight as you do on Earth

Explanation:

In free space, she is suppose to be weightless.

Free fall can be described as body in motion where the body is under the effect of acceleration due to gravity only and no other acceleration..

Since I am accelerating away from her at an acceleration of 1g

Then,

F=ma, where a=g

Then F=mg

Since my weight on earth is W=mg

This is equals to my weight in the spaceship, then I will feel the same weights as I do on earth.

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Which type radiation can be observed well from Earth's surface?
umka2103 [35]

Answer:

Eletromagnetic radiation which is also known as visible light.

Explanation:

4 0
3 years ago
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Belly-flop Bernie dives from atop a tall flagpole into a swimming pool below. His potential energy at the top is 7000 J (relativ
elena55 [62]

Answer:

KE₂ = 6000 J

Explanation:

Given that

Potential energy at top U₁= 7000 J

Potential energy at bottom U₂= 1000 J

The kinetic energy at top ,KE₁= 0 J

Lets take kinetic energy at bottom level =  KE₂

Now from energy conservation

U₁+ KE₁= U₂+ KE₂

Now by putting the values

U₁+ KE₁= U₂+ KE₂

7000+ 0 = 1000+ KE₂

KE₂ = 7000 - 1000 J

KE₂ = 6000 J

Therefore the kinetic energy at bottom is 6000 J.

5 0
2 years ago
A glass capillary tube with a diameter of 8.5 mm and length 8 cm is filled with a salt solution with a resistivity of 2.5 ?m. Wh
MA_775_DIABLO [31]

Answer:

The resistance is 3.5\times10^{-4}\ \Omega

Explanation:

Given that,

Diameter of tube = 8.5 mm

Length = 8 cm

Resistivity = 2.5 m

We need to calculate the resistance

The resistance is equal to the product of the resistivity and length divided by the area of cross section .

In mathematical form,

R = \dfrac{\rho\times l}{A}

Where, \rho=resistivity

l = length

A = area of cross section

Put the value into the formula

R = \dfrac{2.5\times8\times10^{-2}}{3.14\times(\dfrac{8.5}{2}\times10^{-3})^2}

R=3526.32\ \Omega

R=3.5\times10^{-4}\ \Omega

Hence, The resistance is 3.5\times10^{-4}\ \Omega

6 0
3 years ago
What will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? Assume the system is big enough to co
MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

F = qE

If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.

4 0
2 years ago
A spider begins to spin a web by first hanging from a ceiling by his fine, silk fiber. He has a mass of 0.025 kg and a charge of
Rasek [7]

Answer:

a) (5.59 × 10³) N/C

b) 0.226 N directed away from the spider.

Explanation:

a) Electric field, E, felt as a result of point charge, Q, at a distance, d away is given by

E = kQ/d²

So, magnitude of the electric field due to the charge on the second spider at the position of the first spider

Q = 4.2 µC = 4.2 × 10⁻⁶ C

k = Coulomb's constant = 8.99 × 10⁹ Nm²/C

d = 2.6 m

E = (8.99 × 10⁹ × 4.2 × 10⁻⁶)/2.6²

E = 5.59 × 10³ N/C

b) Tension in the silk fiber above the spider is the net force due to the weight of spider one and the force of repulsion due the two charges.

Force due to the two charges = Eq

where q now represents the charge of the first spider at the first point, feeling the electric field calculated in (a)

F = 5.59 × 10³ × 3.4 × 10⁻⁶ = 0.01901 N directed upwards. (That is, F = + 0.019 N)

Weight of the spider = mg = 0.025 × 9.8 = 0.245 N directed downwards. (That is, W = -0.245 N)

Net force, T = mg - F = 0.245 - 0.019 = 0.226 N (that is, 0.226 N, directed upwards, away from the spider).

8 0
3 years ago
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