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Sveta_85 [38]
3 years ago
15

When using a simple machine, the benefit of using less force to lift an object is offset by the need to push for a larger ____ s

o that energy can be conserved.
Physics
2 answers:
Alecsey [184]3 years ago
7 0

Answer:

distance

Explanation:

WITCHER [35]3 years ago
5 0

When using a simple machine, the benefit of using less force to lift an object is offset by the need to push for a larger distance so that energy can be conserved.

<u>Explanation: </u>

A simple machine utilized for completing the works easily. So in order for easy and quick completion of works, these machines may increase the amount of force acting on the object by increasing the velocity or speed of the machine. As velocity term is present so the machines may also change the directions of force acting on the object of concern to do the work soon just like pulley.

Also other way of completing the work with less input force requirement is by increasing the distance or area of action for the force acting on it. As work done is a measure of acting force on a region multiplied with the displacement occurred with that force.

It can be known that force seems to be inversely proportional to distance or area of action. So if we need to use less force to lift an object, it is offset by the need to push for a larger distance in order to conserve the energy.

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A person of mass m will bungee-jump from a bridge over a river where the height from the bridge to the river is h. The bungee co
Likurg_2 [28]

Answer:k_{min}=\frac{18mg}{h}

Explanation:

Given

mass of person is m

Distance between bridge and river is h

chord has an un-stretched length of \frac{2h}{3}

Let spring constant be k

Person will just stop before hitting the river

Conserve energy i.e. Potential Energy of Person is converted in to elastic energy of chord

mgh=\frac{kx^2}{2}

x=h-\frac{2h}{3}=\frac{h}{3}

mgh=\frac{kh^2}{18}

k=\frac{18mg}{h}

Thus k_{min}=\frac{18mg}{h}

5 0
4 years ago
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t
Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = 854\times 10^{3}\ Hz

Power, P = 50 kW = 50\times 10^{3}\ W

I_{p} = 1\ photon/s/m^{2}

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s

Area of the sphere, A = 4\pi r^{2}

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is 1\ photon/s/m^{2}

I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}

1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}

r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m

Now,

We know that:

1 ly = 9.4607\times 10^{15}\ m

Thus

r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly

5 0
3 years ago
Mars had a mass of 6.38* 10^23 kg and a radius of 3,390,000meter what is the force of gravity experienced by a mars rover with a
d1i1m1o1n [39]

I hope its helps you . you can also write 1.44×10^3 it will be better

7 0
2 years ago
A ski jumper has a mass of 59.6 kg. She is moving with a speed of 23.4 m/s at a height of 44.6 meters above the ground. Determin
Daniel [21]

Hello!

Use the formula:

M = k * p

Data:

M = Mechanic energy

k = Kinetic energy

p = Potencial energy

Descomposing:

M = (0,5*mv²) + (mgh)

Replacing:

M = (0,5 * 59,6 kg * (23,4 m/s)²) + (59,6 kg * 9,81 m/s² * 44,6 m)

M = 16317,28 J + 26076,54 J

M = 42393,82 J

The mechanic energy is <u>42393,82 Joules.</u>

7 0
3 years ago
An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
solong [7]

Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
  • T_3 = The third temperature of air inside the tire = 85^\circ C =(273+85)\ K=358 \ K
  • V_1 = The first volume of air inside the tire
  • V_2 = The second volume of air inside the tire = 30\% V_1 = 0.3V_1
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)

Part (a):

Using the above equation for this part of compression in the air, we have

\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa

Hence, the pressure in the tire after the compression is 3.81\times 10^5\ Pa.

Part (b):

Again using the equation for this part for the air, we have

\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

8 0
3 years ago
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