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ss7ja [257]
3 years ago
13

How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?

Physics
1 answer:
4vir4ik [10]3 years ago
5 0

Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

W=(20kg)(9.81m/s²)(4.0m)

W=784.8J

P=W/∆t

P=784.8J/2.5s

P=313.92 watts

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the best way to prevent cancer is to eat right

Explanation:

4 0
3 years ago
The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and
Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

d=\sqrt{(x-x_o)^2+(y-y_o)^2}   (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m

hence, the displacement of the airplane is 3.45*10^4 m

6 0
3 years ago
A centrifuge in a forensics laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates 46.0 times be
ra1l [238]

Answer:

Explanation:

Given,

initial angular speed, ω = 3,700 rev/min

                                      = 3700\times \dfrac{2\pi}{60}=387.27\ rad/s

final angular speed = 0 rad/s

Number of time it rotates= 46 times

angular displacement, θ = 2π x 46 = 92 π

Angular acceleration

\alpha = \dfrac{\omega_f^2 - \omega^2}{2\theta}

\alpha = \dfrac{0 - 387.27^2}{2\times 92\ pi}

\alpha = -259.28 rad/s^2

3 0
3 years ago
6. A person lifts a package weighing 75 N. If she lifts it 1.2 m off the floor,
balandron [24]

Answer:

90 Joules

Explanation:

75×1.2=90.

i believe it is this

7 0
3 years ago
An archer pulls her bowstring back 0.410 m by exerting a force that increases uniformly from zero to 270 N. How much work is don
Nitella [24]

Answer:

<em>110.7Joules</em>

Explanation:

<em>Work is said to be done when the force applied to a body cause the body to move through a distance.</em> Mathematically:

Work done = Force * Distance

Given the following

Force = 270 -0  = 270N

Distance moved = 0.410m

Required

The work done

Substitute the given parameters into the formula

Workdone = 270 * 0.41

Workdone = 110.7Joules

<em>Hence the work done in pulling the bow is 110.7Joules</em>

6 0
3 years ago
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