The astronaut's weight on the Earth's surface can be determined from
<span>F = m g = 579.2 N subsituting </span><span> mass equal to 59.1 kg and acceleration due to gravity equal to 9.8 m/s². When the variables are mass of the earth and the radius of the earth, </span>F = k m / r². Thus, doubling the mass of the earth would double his weight and doubling the radius would decrease the original weight by 1/4. Hence, <span>579.2 N* 2* 1/4 equal to 290 N. Answer is B.</span>
Matter is a lot like energy.
<em>It can't be created or destroyed</em>. It can only be moved around, shifted, rearranged, transformed, disguised, etc. Never created or destroyed.
If you suddenly see some matter that wasn't there a moment ago, it had to come from somewhere. It couldn't just appear out of nothing.
If you suddenly STOP seeing some matter that was there a moment ago, it had to go somewhere. It couldn't just disappear into nothing.
Answer:
D. accelerated
Explanation:
According to Newton's first law of motion, an object always remains in its state of rest or in uniform motion until an external unbalanced force is acted upon the object. This means if an external unbalanced force is acted on the object, it will come to a non-uniform motion i.e., an accelerated motion.
Let me first get you clear that uniform motion is determined by a constant velocity and a state of rest is considered by a zero velocity or speed. But, here we have an unbalanced force acting on the object. This means the object will change its velocity and hence, it has an accelerated motion.
Answer:
(a) Angular acceleration will be ![\alpha =18.66rev/min^2](https://tex.z-dn.net/?f=%5Calpha%20%3D18.66rev%2Fmin%5E2)
(B) Final angular velocity will be 28 rev/min
Explanation:
We have given time t = 1.5 min
Angular displacement ![\Theta =21rev](https://tex.z-dn.net/?f=%5CTheta%20%3D21rev)
(a) Initial angular speed ![\omega _i=0rad/sec](https://tex.z-dn.net/?f=%5Comega%20_i%3D0rad%2Fsec)
From second equation of motion we know that ![\Theta =\omega _it+\frac{1}{2}\alpha t^2](https://tex.z-dn.net/?f=%5CTheta%20%3D%5Comega%20_it%2B%5Cfrac%7B1%7D%7B2%7D%5Calpha%20t%5E2)
So ![21 =0\times 1.5+\frac{1}{2}\times \alpha\times 1.5^2](https://tex.z-dn.net/?f=21%20%3D0%5Ctimes%201.5%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Calpha%5Ctimes%20%201.5%5E2)
![42 =\alpha\times 1.5^2](https://tex.z-dn.net/?f=42%20%3D%5Calpha%5Ctimes%20%201.5%5E2)
![\alpha =18.66rev/min^2](https://tex.z-dn.net/?f=%5Calpha%20%3D18.66rev%2Fmin%5E2)
(b) Now from first equation of motion
![\omega _f=\omega _i+\alpha t=0+18.66\times 1.5=28rev/min](https://tex.z-dn.net/?f=%5Comega%20_f%3D%5Comega%20_i%2B%5Calpha%20t%3D0%2B18.66%5Ctimes%201.5%3D28rev%2Fmin)
Answer: D
Explanation: In the energy conversion of a nuclear power plant, uranium acts as nuclear fuel, so, Intranuclear Potential Energy (uranium) is converted to thermal energy which is Kinetic Energy (high-pressure steam) which is converted to mechanical energy that is,
Kinetic Energy (spinning turbine) then to Kinetic Energy (generator) and finally Kinetic Energy (electricity in
transmission lines)
Therefore, the best option is D which is
D. III, I, II, V, IV