Ask question

Ask question

All categories

3 years ago

8

2. You decide to lift a 25 kg box to a height of 3 meters. How much work will you do while moving

1 answer:

nadya68 [22]3 years ago

4 0

Answer:

W = 735.75[J]

Explanation:

Work is defined as the product of force by distance. Therefore we can use the following equation.

$W=F*d$

where:

W = work [J] (units of Joules)

F = force [N] (units of Newtons)

d = distance = 3 [m]

But first, we must determine the force that is equal to the product of mass by gravity (weight of the body).

$F=m*g\\F=25*9.81\\F=245.25[N]$

$W=F*d\\W=245.25*3\\W=735.75[J]$

You might be interested in

Which of these materials is permeable?

kogti [31]

Correct answer is letter B. sandstone

8 0

2 years ago

A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of

scoray [572]

Answer:1200

Explanation:

Given data

Upper Temprature $\left ( T_H\right )=240^{\circ}\approx 513$

Lower Temprature $\left ( T_L\right )=20^{\circ}\approx 293$

Engine power ouput $\left ( W\right )=910 W$

Efficiency of carnot cycle is given by

$\eta =1-\frac{T_L}{T_H}$

$\eta =\frac{W_s}{Q_s}$

$1-\frac{293}{513}=\frac{910}{Q_s}$

$Q_s=2121.954 W$

$Q_r=1211.954 W$

rounding off to two significant figures

$Q_r=1200 W$

5 0

3 years ago

You want to make a ride so you do not want to exceed 1.1g’s, if the radius of the turns are 10m, then what is the maximum speed

Citrus2011 [14]

The maximum speed is 10.4 m/s

Explanation:

For a body in uniform circular motion, the centripetal acceleration is given by:

$a=\frac{v^2}{r}$

where

v is the linear speed

r is the radius of the circular path

In this problem, we have the following data:

- The maximum centripetal acceleration must be

$a=1.1 g$

where $g=9.8 m/s^2$ is the acceleration of gravity. Substituting,

$a=(1.1)(9.8)=10.8 m/s^2$

- The radius of the turn is

r = 10 m

Therefore, we can re-arrange the equation to solve for v, to find the maximum speed the ride can go at:

$v=\sqrt{ar}=\sqrt{(10.8)(10)}=10.4 m/s$

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

8 0

3 years ago

What is the Potential Energy of ball 1??

Dvinal [7]

GPE= height x mass x gravitational field strength
5 x 10 x 9,8=490J

7 0

3 years ago

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

belka [17]

Given that,

Energy $H=2.7\times10^{31}\ W$

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

$H=Ae\sigma T^4$

$A=\dfrac{H}{e\sigma T^4}$

$4\pi R^2=\dfrac{H}{e\sigma T^4}$

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}$

$R=5.0\times10^{10}\ m$

(b). Given that,

Radiates energy $H=2.1\times10^{23}\ W$

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}$

$R=5.42\times10^{6}\ m$

Hence, (a). The radius of the star is $5.0\times10^{10}\ m$

(b). The radius of the star is $5.42\times10^{6}\ m$

8 0

3 years ago