Ask question

Ask question

All categories

xxTIMURxx [149]

3 years ago

8

Alien A lifts a 500-newton child from the floor to a height of 0.40 meters in 2 seconds

1 answer:

vivado [14]3 years ago

4 0

Strong alien you got there good luck bud you never asked a question

You might be interested in

Is james west still alive?

mafiozo [28]

Greetings

and james west is ALIVE

8 0

3 years ago

Read 2 more answers

If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0

Ilya [14]

Work of the force = 10 N

Time required for the work = 50 sec

Height = 7 m

We are given with the value of work and time in the question.

Substitute the values in the formula of power and then you'll get the power required.

We know that,

w = Work

p = Power

t = Time

By the formula,

Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,

p = 70/50

p = 1.4 watts

Therefore, the power required is 1.4 watts.

Hope it helps!

3 0

3 years ago

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

$Mm=44.0 g/mol$ is the molar mass of the carbon dioxide

So, the number of moles of the gas is

$n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol$

Now we can find the pressure of the gas by using the ideal gas equation:

$pV=nRT$

where

p is the pressure

$V=1.00 L = 0.001 m^3$ is the volume

n = 0.023 mol is the number of moles

$R=8.314 J/mol K$ is the gas constant

$T=25.0^{\circ}+273=298 K$ is the temperature of the gas

Solving the equation for p, we find

$p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa$

And since we have

$1 atm = 1.01\cdot 10^5 Pa$

the pressure in atmospheres is

$p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm$

5 0

3 years ago

[25 points] Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. T

evablogger [386]

6 meters is left because you subtract 12 meters from 6

5 0

3 years ago

A certain light truck can go around a flat curve having a radius of 150 m with a maximum spee dof 32.0 m/s. With what maximum sp

Dvinal [7]

Answer

Maximum speed at 75 m radius will be 22.625 m /sec

Explanation:

We have given radius of the curve r = 150 m

Maximum speed $v_{max}=32m/sec$

Coefficient of friction $\mu =\frac{v_{max}^2}{rg}=\frac{32^2}{150\times 9.8}=0.6965$

Now new radius r = 75 m

So maximum speed at new radius $v_{max}=\sqrt{\mu rg}=\sqrt{0.6965\times 75\times 9.8}=22.625m/sec$

7 0

3 years ago