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Lubov Fominskaja [6]
3 years ago
7

prc-2) A soccer ball accelerates from rest and rolls 6.5m down a hill for 3.1 seconds. It is then stopped by a tree. Find the fi

nal velocity.
Physics
1 answer:
anastassius [24]3 years ago
7 0

Answer:

15?6 and 17 plus 18 equals e b d f g so correct

Explanation:

yws you're welcome and thank for your help with this and I will be there is it ok if I get a and I can

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A bicycle slows down when the rider applies the brakes.
aniked [119]

This is kinetic energy to heat energy


4 0
3 years ago
Read 2 more answers
When a body moves from point A to point B the distance travelled in 500 meters with in a given time of 0.6 hours and find out th
Sav [38]

Answer

Explanation:

Convert the time to seconds = 0.6 × 60 × 60

= 2160seconds

Velocity = distance ÷ time

Velocity = 500 ÷ 2160

Velocity = 0.23meters per seconds(m/s)

Acceleration = Velocity ÷ time

Acceleration = 0.23 ÷ 2160

Acceleration = 0.000106meters per seconds ²(m/s²)

6 0
3 years ago
In an RC-circuit, a resistance of R=1.0 "Giga Ohms" is connected to an air-filled circular-parallel-plate capacitor of diameter
solniwko [45]

Answer:

\tau = 1\ ms

Explanation:

First we need to find the capacitance of the capacitor.

The capacitance is given by:

C = \epsilon_0 * area / distance

Where \epsilon_0 is the air permittivity, which is approximately 8.85 * 10^(-12)

The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:

Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2

So the capacitance is:

C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}

C = 10^{-12}\ F = 1\ pF

The time constant of a rc-circuit is given by:

\tau = RC

So we have that:

\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms

7 0
3 years ago
A 5-N force is applied to a 3-kg ball to change its velocity from 9 m/s to 3 m/s. The impulse is encountered by the ball for a t
anyanavicka [17]

Answer:

3.6 s.

Explanation:

Impulse: This can be defined as the product of force and time. The S.I unit of impulse is N.s.

From newton's second law of motion,

The expression for impulse is given as,

I = Ft = m(u-v) ................. Equation 1

make t the subject of the equation

t = m(v-u)/F................... Equation 2

Where I = impulse, v = final velocity, u = initial velocity, m = mass of the ball, F = force, t = time

Given: m = 3 kg, F = 5 N, u = 9 m/s, v = 3 m/s

Substitute into equation 2

t = 3(9-3)/5

t = 3(6)/5

t = 3.6 s.

4 0
4 years ago
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
Illusion [34]

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

6 0
3 years ago
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