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True [87]
3 years ago
11

A roller coaster car rapidly picks up speed as it rolls down a slope as it starts down the slope its speed is 4m/s but 3 seconds

later at the bottom of the slope its speed is 22m/s what is its average acceleration
Physics
1 answer:
Gwar [14]3 years ago
3 0

Answer:

The acceleration is 6 [m/s^2]

Explanation:

We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.

v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\

Now replacing the values we have:

a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]

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Marta, Cato, and Juan plan a skit to illustrate one of the methods of charging. Marta and Cato are walking together. Each has on
sveta [45]

Answer:

Option D.

Explanation:

Methods of charging:

1. By friction: when two materials of different electro-negativities rub against each, electrons transfer from one material to another.

2. By induction: When a charged body is brought near an uncharged body, the later gets polarized.

3. By conduction: when electrons from one body to another uncharged while contact, it is known as charging by conduction.

Here, a flower is analogous to an electron. A box of candy is positive charge. A bouquet of flowers is like a charged body which is brought near an uncharged body. It induces charge on ti it by causing it to transfer electron to another.

Thus, option D is correct.

8 0
3 years ago
An adiabatic nozzle is used to accelerate 6000 kg/hour of CO2 to 450 m/s. CO2 enters the nozzle at 1000 kPa and 500 C. The inlet
den301095 [7]

Answer:

a. inlet velocity = 60.8m/s

b. exit temperature = 686k

Explanation:

7 0
3 years ago
1. A pumpkin with a mass of 2 kg accelerates 2 m/s/s when an unknown force is applied to it. What is the amount of the force? ​
Andreyy89

Answer:

<h3>The answer is 4 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 2 × 2

We have the final answer as

<h3>4 N</h3>

Hope this helps you

8 0
3 years ago
If the equipotential surfaces due to some charge distribution are vertical planes, what can you say about the electric field dir
Elina [12.6K]

Answer:

The correct option is

(e)either (c) or (d) could be correct.

Explanation:

The electric field of a charge radiates out in all directions and the intensity of the electric field strength given by E = F/q₀, diminishes as the lines of force moves further away from the source. The direction of F and E is in the line of potential motion of the source charge in the field.

Equipotential surfaces are locations in the radiated electric that have the same field strength or electric potential. The work done in moving within an equipotential surface is zero and as such since

Work = Force × distance = 0 where distance ≠ 0.

The force acting between two points on an equipotential surface is also zero or the component of the force within an equipotential surface is zero and since there is a force in the electric field, it is acting at right angles to the equipotential surface which could be horizontally to the left or right directions where the equipotential surfaces due to the charge distribution are in the vertical plane.

Therefore it is either horizontally to the left, or horizontally to the right.

7 0
3 years ago
In a World Cup soccer match, Juan is running due north toward the goal with a speed of 7.60 m/s relative to the ground. A teamma
spayn [35]

Answer:

Part a)

v_{bj} = 11.03 m/s

Part b)

\theta = 4.57 degree East of South

Explanation:

Part a)

Velocity of Juan is given as

v_1 = 7.60 m/s \hat j

velocity of the ball is given as

v_2 = 12.9(cos31.4 \hat i + sin31.4\hat j)

now we have

v_2 = 11\hat i + 6.72\hat j

Part a)

We need to find velocity of ball with respect to Juan

so it is given as

v_{bJ} = \vec v_b - \vec v_j

v_{bj} = 11\hat i + 6.72 \hat j - 7.6\hat j

v_{bj} = 11\hat i - 0.88\hat j

magnitude of the speed is given as

v_{bj} = \sqrt{11^2 + 0.88^2}

v_{bj} = 11.03 m/s

Part b)

direction of velocity of the ball

tan\theta = \frac{v_y}{v_x}

tan\theta = \frac{-0.88}{11}

\theta = 4.57 degree East of South

3 0
3 years ago
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