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katrin2010 [14]
2 years ago
15

if you push a box a distance of 2000 meters with a force of 1 newton, how many calories have you used

Physics
1 answer:
kap26 [50]2 years ago
4 0
Note that
1 J = 0.239 cal

By definition,
Work = Force x Distance

Therefore work done is
W = (1 N)*(2000 m) = 2000 J

In calories,
W = (2000 J)*(0.239 cal/J) = 478 cal

Answer: 478 calories

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s344n2d4d5 [400]
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5 yes it is The water on our Earth today is the same water that’s been here for nearly 5 billion years. Only a tiny bit of it has escaped out into space. As far as we know, new water hasn’t formed either.

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7 0
3 years ago
. Rohit had only one light source in his room. He was told by his friend that he can make the room more lighted by using mirror.
LUCKY_DIMON [66]

Answer:

very smooth and polished glass mirror

Explanation:

The best type of mirror for this would be a very smooth and polished glass mirror. A metal mirror will also work but it would need to be extremely polished. This would allow light to efficiently be reflected across the room, thus effectively adding an "extra" light source to the room. If the mirror is not smooth then the light may become distorted and not reflect properly as it is supposed to. This may defeat the entire purpose of the mirror for this scenario.

3 0
2 years ago
Make the following conversion. 0.0097 mg = _____ g
Cloud [144]
Your answer should be 9.7 :)
8 0
3 years ago
Read 2 more answers
When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

3 0
2 years ago
The Cosmological Argument for the existence of God intellectually follows cause-and-effect back in time -- in effect asking "Whe
ankoles [38]

Answer:

First uncaused cause

Explanation:

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2 years ago
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