All categories

2 years ago

How high off the ground must you be if you mass is 75 kg and you have 11,025 J of energy stored inside?

Physics

1 answer:

pashok25 [27]2 years ago

3 0

Answer:

15 m

Explanation:

Given,

Mass ( m ) = 75 kg

Potential energy ( P.E ) = 11,025 J

To find : Hight ( h )

Formula : -

P.E = mgh

[ Note : The value of g = 9.8 m/s² ]

h = P.E / mg

= 11,025 / ( 75 x 9.8 )

= 11,025 / 735

h = 15 m

Hence,

15 m is the high off the ground must you be if you mass is 75 kg and you have 11,025 J of energy stored inside.

You might be interested in

video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It

natka813 [3]

Answer:

$\omega_f = 0.4\ rad/s$

Explanation:

given,

width of door dimension = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

$I_{total} =I_{door} + I_{bullet}$

$I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2$

a) from conservation of angular momentum

$L_i = L_f$

$mv\dfrac{W}{2} = I_{total}\omega_f$

$mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f$

$\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}$

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}$

$\omega_f = 0.4\ rad/s$

8 0

3 years ago

It took a bulldozer 62,000 J of work to move a rock 30 m. It took 5 minutes. How much force did the bulldozer have to apply?

NeX [460]

Answer:

A (2066,6 N)

Explanation:

Use the Work formula

62.000J = F . 30

62.000/30 = 2066,6 N

The amout of time it took to move the rock doesn´t matter at all.

It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.

5 0

3 years ago

A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant

atroni [7]

Answer:

The time constant is $\tau = 1.265 s$

Explanation:

From the question we are told that

the time take to charge is $t = 2.4 \ s$

The mathematically representation for voltage potential of a capacitor at different time is

$V = V_o - e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant

$V_o$ is the potential of the capacitor when it is full

So the capacitor potential will be 100% when it is full thus $V_o =$ 100% = 1

and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

V = 0.85

Hence

$0.85 = 1 - e^{-\frac{2.4}{\tau } }$

$- {\frac{2.4}{\tau } } = ln0.15$

$\tau = 1.265 s$

7 0

3 years ago

a 46 kilogram student climbs 11 meter up a rope at a constant speed if the student power output is 230 watts how long in seconds

Bezzdna [24]

230×46=10580÷11=961 second

3 0

3 years ago

Select the correct answer.

spin [16.1K]

Answer:

There isnt enough in your question to answer the question bro, like we need a picture or something bro.

Explanation:

7 0

3 years ago