All categories

3 years ago

A weightlifter raises a 200-kg barbell through a height of 2 m in 2.2 s. the average power he develops during the lift is

2 answers:

4 0

Power is calculated as work per unit time, and work in turn is calculated as force multiplied by distance. In this case, the force required is equivalent to the weight of the barbell multiplied by acceleration due to gravity.
P = W/t = Fd/t = mgd/t = (200 kg)(9.81 m/s^2)(2 m)/2.2 s = 1783.64 Watts.

Liula [17]3 years ago

4 0

Answer:

Power, P = 1781.81 watts

Explanation:

It is given that,

Mass of the barbell, m = 200 kg

It is lifted to a height of, h = 2 m

Time taken, t = 2.2 s

We need to find the develops during the lift. We know that the pwer developed by an object is equal to the rate of doing work. Mathematically, it is given by :

$P=\dfrac{W}{t}$

$P=\dfrac{mgh}{t}$

$P=\dfrac{200\times 9.8\times 2}{2.2}$

P = 1781.81 watts

So, the average power developed during the lift is 1781.81 watts. Hence, this is the required solution.

You might be interested in

A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h

Elina [12.6K]

Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: $F_{net}=m.a$

Acceleration is the change of velocity in a period of time: $a=\frac{\Delta v}{\Delta t}$

Velocity of the car is in km/h. Transforming it in m/s:

$v=\frac{47.10^{3}}{36.10^{2}}$

v = 13 m/s

At the moment the car decelerates, acceleration is

$a=\frac{13}{0.2}$

a = 65 m/s²

Then, force will be

$F_{net}=19(65)$

$F_{net}$ = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

5 0

2 years ago

The current supplied by a battery as a function of time is ) -(0.64 A)e-/(6.0 hr). What is the total number of electrons transpo

Paha777 [63]

Answer:

The total number of electrons is $8.6\times10^{22}$

(3) is correct option.

Explanation:

Given that,

The current equation is

$I=0.64 e^{\dfrac{-t}{6.0 hr}}$

We know that,

The formula of charge

$q=\int_{0}^{\infty}{I dt}$

$q=\int_{0}^{\infty}{0.64 e^{\dfrac{-t}{6.0 hr}}dt$

$q=0.64\int_{0}^{\infty}{e^{\dfrac{-t}{6.0 hr}}dt$

$q=0.64\int_{0}^{\infty}{e^{\dfrac{-t}{21600}}dt$

$q=0.64(21600e^{\dfrac{-t}{21600}})_{0}^{\infty}$

$q=0.64(0-21600)$

$q=0.64\times21600$

$q=13824\ C$

We need to calculate the number of electron

Using formula of charge

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{13824}{1.6\times10^{-19}}$

$n=8.6\times10^{22}$

Hence, The total number of electrons is $8.6\times10^{22}$

7 0

3 years ago

Matt and Anna Killian are frequent fliers on Fast-n-Go Airlines. They often fly between two cities that are a distance of 1575

marin [14]

Answer:

Speed of wind = 50mi/hr, Speed of plane in still air = 400mi/hr

Explanation:

Let the speed of the wind = Vw,

Speed of the plane in still air = Vsa,

The first trip the average speed of the plane = 1575mi/4.5hours = 350mi/hr

The coming trip the wind behind = 1575mi/3.5hrs = 450

Write the motion in equation form

First trip ( the plane flew into the wind)

Vaverage = Vsa - Vw

350 = Vsa - Vw

Second trip the wind was behind

450 = Vsa +Vw

Adding the two equation

800 = 2Vas

Vas = 800/2 = 400mi/hr

Substitute for Vas into equation 1

350mi/hr = 400mi/hr - Vw

Vw = 400-350 = 50mi/hr

6 0

2 years ago

If Jill starts out at 20 m/s, and in 10 s speeds up to 40 m/s, what is her acceleration?

Iteru [2.4K]

Answer:

$2 m/s^2$

Explanation:

The acceleration of an object (or a person, as in this case) is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 20 m/s

v = 40 m/s

t = 10 s

Therefore Jill's acceleration is

$a=\frac{40-20}{10}=2 m/s^2$

6 0

3 years ago

Read 2 more answers

Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

melisa1 [442]

Answer:

$H_2 = 91.55 km$

Explanation:

Gravity on the surface of planet is given as

$g = \frac{GM}{R^2}$

as we know that

$M = 8.93 \times 10^{22} kg$

$R = 1821 km$

now gravity on the planet is

$g = \frac{(6.67 \times 10^{-11})(8.93 \times 10^{22})}{(1821 \times 10^3)^}$

so we have

$g = 1.8 m/s^2$

now we know that

$H_{max} = \frac{v^2}{2g}$

so we will say

$\frac{H_1}{H_2} = \frac{g_2}{g_1}$

$\frac{500}{H_2} = \frac{9.81}{1.8}$

$H_2 = 91.55 km$

5 0

3 years ago