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Bezzdna [24]
3 years ago
12

Salt is now added to the water in the bucket, increasing the density of the liquid. what happens to the tension in the string ?

Physics
1 answer:
AlexFokin [52]3 years ago
5 0

The question involves a ping-pong ball that is held submerged in a bucket by a string attached to the bottom of the bucket.

The answer is the tension of the string will increase. This is because making the water salty increases its density, and consequently, increases its buoyancy. This is why sea water is more buoyant than fresh water. Therefore the ping pong is pushed more upwards by the water when salt is added than initially. This gives the string more tension.


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What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to
Vadim26 [7]

Answer:

The correct answer to the question is

B. It always decreases

Explanation:

To solve the question, we note that the foce of gravity is given by

F_G=\frac{Gm_1m_2}{r^2} where

G= Gravitational constant

m₁ = mass of first object

m₂ = mass of second object

r = the distance between both objects

If the mass of one object remains unchanged while the distance to the second object and the second object’s mass are both doubled, we have

F_{G2} =\frac{Gm_1(2m_2)}{(2r)^2} = \frac{2}{4} \frac{Gm_1m_2}{r^2}

Therefore the gravitational force is halved. That is it will always decrease

4 0
2 years ago
alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;
zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0
3 years ago
Two common terms for a decrease in velocity are
Colt1911 [192]

deceleration or rėtardation i’m pretty sure (it won’t let me say the second word but it’s correct)

6 0
3 years ago
Read 2 more answers
Never mind i dont need help anymore
Usimov [2.4K]
Yay good for you!!!!
8 0
2 years ago
A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at
frutty [35]

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s

Therefore, the speed of the ball is 10 m/s

6 0
3 years ago
Read 2 more answers
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