Select three possible applications of a capacitor. Select all that apply.

A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels

Aleksandr [31]

Answer:

<em>(a) The average velocity is 16 m/s</em>

<em>(b) The acceleration is 0.4 m/s^2</em>

<em>(c) The final velocity is 24 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

$v_f=v_o+at\qquad\qquad [1]$

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

$\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16$

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

(c) From [2] we can solve for a:

$\displaystyle a= 2\frac{x-v_ot}{t^2}$

Since vo=8 m/s, x=640 m, t=40 s:

$\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4$

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

$v_f=8+0.4\cdot 40$

$v_f=8+16=24$

The final velocity is 24 m/s

3 0

3 years ago

4-08 t 1 - / S P = 1 / P = S 1 S= P Def. of Speed

nordsb [41]

Answer:

what do I answer? not enough information

8 0

2 years ago

What is the momentum of a 200 kg football player running west at a speed of 8m/s

lutik1710 [3]

Answer:

<h3>The answer is 1600 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 200 kg

velocity / speed = 8m/s

We have

momentum = 200 × 8

We have the final answer as

Hope this helps you

7 0

3 years ago

How are weather satellites and weather stations similar?

alexandr402 [8]

<span>Weather satellites and weather stations are similar, because they both have the same purpose. They are used to help predict future weather as well as as current conditions. The satellites are viewing the weather from a distance at a large scope, but stations are using data they collect on earth to help with the same task.</span>

6 0

3 years ago

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$

5,061.96 J = 21.5 kg × v₂² + 53. $\overline 3$ kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0

3 years ago