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3 years ago

Why is it that the weight of an object weighing 1N air, weighs more when immersed in water ?

1 answer:

4 0

There is no "why", because that's not what happens. The truth is
exactly the opposite.

Whatever the weight of a solid object is in air, that weight will appear
to be LESS when the object is immersed in water.

The object is lifted by a force equal to the weight of the fluid it displaces.
It displaces the same amount of air or water, and any amount of water
weighs more than the same amount of air. So the force that lifts the
object in water is greater than the force that lifts it in air, and the object
appears to weigh less in the water.

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It is now 9:11 a.m. but when the bell rings at 9:12 a.m. Susie will be late for Mrs. Garner's U.S. History class for the 3rd tim

GaryK [48]

Answer:

3.1 m/s

Explanation:

The total distance she has to run is the addition of the three lengths:

47 + 63 + 76 = 186 meters.

She needs to cover it one minute (60 seconds). Therefore her speed must be:

186 m / 60 s = 3.1 m/s

6 0

3 years ago

If a sprinter accelerates from rest to 12 m/s north , what is their change in velocity ?

natima [27]

Answer: It would be 12 m/s.

Explanation: It would be this because If you go from rest to sprint it would be 12 m/s. Also, I did this the other day.

5 0

3 years ago

Read 2 more answers

Someone pls help me with this!

Lina20 [59]

Answer:

F=8.87 N

Explanation:

Coulomb's Law

The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge in coulomb

d= The distance between the objects in meters

Object 1 has a charge of

$q_1=-2.3\cdot 10^{-6}\ c$

Object 2 has a charge of

$q_2=-4.2\cdot 10^{-6}\ c$

They are separated by a distance of

d = 0.099 m

Calculate the force:

$\displaystyle F=9\cdot 10^9\frac{2.3\cdot 10^{-6}*4.2\cdot 10^{-6}}{0.099^2}$

F=8.87 N

5 0

2 years ago

A 200. N wagon is to be pulled up a 30 degree incline at constant speed. How large a force parallel to the incline force is need

Sever21 [200]

For the question above, here is the equation to follow:
F = mgsinα = Wsinα
=200 x 0.5 = 100 N
OR

Sin30 * 200N = 100 N

The asnwer is 100N. I hope this answer helps.

3 0

3 years ago

Read 2 more answers

Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2, đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g,

dusya [7]

Answer:

The angle is 18.3 degree.

Explanation:

A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?

surface charge density, σ = 4 x 10^-5 C/m^2

Charge, q = 10^-8 C

mass, m = 0.008 kg

Let the angle is A and the tension in the string is T.

The electric field due to a plane is

$E =\frac{\varepsilon \sigma }{2\varepsilon o}\\\\E =\frac{4\times 10^{-5}}{2\times 8.85\times 10^{-12}}\\\\E = 2.26\times 10^6 V/m \\$

Now equate the forces,

$T sin A = q E.... (1)\\\\T cos A = m g ..... (2)\\\\divide (1) by (2)\\\\tan A = \frac{10^{-8}\times 2.6\times 10^6}{0.008\times 9.8}\\\\tan A = 0.33\\\\ A = 18.3 degree$

5 0

3 years ago