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2 years ago

13

Write down the relation between energy and power

2 answers:

gizmo_the_mogwai [7]2 years ago

6 0

Answer:

Energy is what makes change happen and can be transferred form one object to another. Energy can also be transformed from one form to another. Power is the rate at which energy is transferred. It is not energy but is often confused with energy.

Explanation:

Nookie1986 [14]2 years ago

3 0

Answer:

when u have energy u will have power n u can work as power is the capacity to do the work

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Prisons are typically under whose control?

sineoko [7]

The wardens or the po's

6 0

3 years ago

A water skier lets go of the tow rope upon leaving the end ofa

ANTONII [103]

Answer:

1.35m

Explanation:

At the highest point of the jump, the vertical speed of the skier should be 0. So the 13m/s speed is horizontal, this speed stays the same from the jumping point to the highest point. The 14m/s speed at jumping point is the combination of both vertical and horizontal speeds.

The vertical speed at the jumping point can be computed:

$v_v^2 + v_h^2 = v^2$

$v_v^2 + 13^2 = 14^2$

$v_v^2 = 196 - 169 = 27$

$v_v = \sqrt{27} = 5.2 m/s$

When the skier jumps to the its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv_v^2/2$

where m is the skier mass and h is the vertical distance traveled, $v_v$ is the vertical velocity at jumping point, and h is the highest point.

Let g = 10m/s2

We can divide both sides of the equation by m:

$gh = v_v^2/2$

$h = \frac{v_v^2}{2g} = \frac{27}{2*10} = 1.35 m$

3 0

3 years ago

The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re

mihalych1998 [28]

Answer:

Explanation:

Given

$\omega =13-\frac{1}{2}\cdot t^2$

Motor reverse its direction when \omega =0

$13-0.5t^2=0$

$26=t^2$

$t=\sqrt{26}=5.099\approx 5.1 s$

(b)

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

$\int d\theta =\int_{0}^{5.1}\omega dt$

$\int d\theta =\int_{0}^{t}(13-.05t^2)dt$

$\theta =(13t-0.1667\times t^3)_0^{5.1}$

$\theta =44.192^{\circ}$

4 0

2 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

$\tau = 798.525N.m$

3 0

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Marizza181 [45]

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

<h3>c) 2.02 x 10^16 nuclei</h3>

7 0

2 years ago