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3 years ago

If a 2 kg ball is traveling at a speed of 4 m/s, what is its kinetic energy?

Physics

2 answers:

Kay [80]3 years ago

8 0

Answer: 16J

Explanation:

K.E = $\frac{1}{2}$ m $v^{2}$

K.E = $\frac{1}{2}$ X 2 X 16

K.E = 16

Art [367]3 years ago

4 0

The kinetic energy of the ball is 16 J.

Answer: Option D

<u>Explanation: </u>

Kinetic energy is the form of energy exhibited by a body when it is in motion. The kinetic energy will be directly proportional to the mass of the body as well as to the square of the speed of the body at which it is moving. So the mathematical representation of kinetic energy of any object is

$\text {Kinetic energy}=\frac{1}{2} \times m \times(v)^{2}$

In the present case, the ball is the object with mass of around 2 kg and it is moving with the speed of about 4 m/s. So, the kinetic energy exhibited by the ball during its motion is

$\text {Kinetic energy of the ball}=\frac{1}{2} \times 2 \times(4)^{2}=16 \mathrm{J}$

So, the kinetic energy exhibited by the ball is 16 J.

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A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin

Leya [2.2K]

Answer:

v = 13.19 m / s

Explanation:

This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

X axis

N = m a

Centripetal acceleration is

a = v² / r

Y Axis

fr -W = 0

fr = W

The force of friction is

fr = μ N

Let's calculate

μ (m v² / r) = mg

μ v² / r = g

v² = g r / μ

v = √ (g r /μ)

v = √ (9.8 11 / 0.62)

v = 13.19 m / s

7 0

3 years ago

Determine the kinetic energy of a 2000 kg roller coaster car that is moving at the speed of 10 ms

kondaur [170]

Answer:

$\boxed {\boxed {\sf 100,000 \ Joules}}$

Explanation:

Kinetic energy is energy due to motion. The formula is half the product of mass and velocity squared.

$E_k= \frac{1}{2} mv^2$

The mass of the roller coaster car is 2000 kilograms and the car is moving 10 meters per second.

m= 2000 kg
s= 10 m/s

Substitute these values into the formula.

$E_k= \frac{1}{2} (2000 \ kg ) \times (10 \ m/s)^2$

Solve the exponent.

(10 m/s)²= 10 m/s * 10 m/s= 100 m²/s²

$E_k= \frac{1}{2} (2000 \ kg ) \times (100 \ m^2/s^2)$

Multiply the first two numbers together.

$E_k= 1000 \ kg \times (100 \ m^2/s^2)$

Multiply again.

$E_k= 100,000 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer of 100,000 kg*m²/s² is equal to 100,000 Joules.

$E_k= 100,000 \ J$

The roller coaster car has <u>100,000 Joules</u> of kinetic energy.

3 0

3 years ago

A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i

nexus9112 [7]

Answer:

E = k q / a² (1.3535) (- i ^ + j ^)

E = k q / a² 1.914 , θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

E₁₂ = k q / a²

On the y axis

E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

d = √ (a² + a²) = a √2

let's look for the field

E₁₃ = k q / d²

E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

cos 45 = E₁₃ₓ / E₁₃

E₁₃ₓ = E₁₃ cos 45

sin 45 = E_{13y} / E₁₃

E_{13y} = E₁₃ sin 45

E₁₃ₓ = k q / 2a² cos 45

E_{13y} = k q / 2a² sin 45

let's find each component of the electric field

X axis

Eₓ = -E₁₂ - E₁₃ₓ

Eₓ = - k q / a² - k q / 2a² cos 45

Eₓ = - k q / a² (1 + cos 45/2)

cos 45 = sin 45 = 0.707

Eₓ = - k q / a² (1 + 0.707 / 2)

Eₓ = - k q / a² (1.3535)

Y axis

E_y = E₁₄ + E_{13y}

E_y = k q / a² + k q / 2a² sin 45

E_y = k q / a² (1 + sin 45/2)

E_y = k q / a² (1.3535)

we can give the results in two ways

E = k q / a² (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

E = √ (Eₓ² + E_y²)

E = k q / a² 1.3535 √2

E = k q / a² 1.914

we use trigonometry for the angle

tan θ = E_y / Eₓ

θ = tan⁻¹ (E_y / Eₓ)

θ = tan⁻¹ (1 / -1)

θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

θ‘= 90 + 45

θ’= 135

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Answer:

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