A 20 kg box is being pushed with a force of 60 N while friction is 20 N. What is the net force on the

Describe what happens to the particles of water when it boils.

Nezavi [6.7K]

Answer:

The liquid turns to a gas.

Explanation:

If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. Particles in the middle of the liquid form bubbles of gas in the liquid.

8 0

3 years ago

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lyudmila [28]

Answer: Answer to the I'm Tall when I'm Young, and I'm Short when I'm Old. What am I? Riddle is a candle

Hope this helps

7 0

3 years ago

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A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the

Reika [66]

Answer:

0.25m/s

Explanation:

Given parameters

m₁ = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

m₁v₁ + m₂v₂ = v(m₁ + m₂)

Insert parameters and solve for v;

5 x 1 + 15 x 0 = v (5 + 15 )

5 = 20v

v = $\frac{5}{20}$ = 0.25m/s

5 0

3 years ago

Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp

In-s [12.5K]

Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

8 0

3 years ago

A test charge of -1.4 x 10-7 coulombs experiences a force of 5.4 x 10-1 newtons. Calculate the magnitude of the electric field c

VARVARA [1.3K]

Answer:

3.86×10⁶ Newton/coulombs

Explaination:

Applying,

E = F/q....................... Equation 1

Where E = Electric Field, F = Force, q = charge.

From the question,

Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs

Substitute these values into equation 1

E = 5.4×10⁻¹/ -1.4×10⁻⁷

E = -3.86×10⁶ Newtons/coulombs

Hence the magnitude of the electric field created by the

negative test charge is 3.86×10⁶ Newton/coulombs

5 0

3 years ago