Answer:
The gravitational potential energy equals the work needed to lift the object.
Explanation:
here we know that
work done is given as
Potential energy is given as
force due to gravity is given as
now here if we plug in the value of distance and force in the formula of work done then we will have
so here we got
so we can concluded that
The gravitational potential energy equals the work needed to lift the object.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
Newton's third law says:
"<span>For every action, there is an equal and opposite reaction. ".
So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>
<span>where the label "t" means "on Tom", while the label "s" means "on the sister".
From Newton's second law, we also know
</span>
where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>
<span>And find Tom's acceleration:
</span>
<span>
</span>
Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
- When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
- While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
- And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.
Answer with Explanation:
We are given Avogadro's constant =
There are eight significant figures.
We have to round off.
1.If we round off to four significant figures
The ten thousandth place of Avogadro's constant is less than five therefore, digits on left side of ten thousandth place remains same and digits on right side of ten thousandth place and ten thousandth place replace by zero.
Then ,Avogadro's constant can be written as
If we round off to 2 significant figures
Hundredth place of given number is less than 5 therefore, digits on left side of hundredth place remains same and digits on right side of hundredth place and hundredth place replace by zero.
Then,Avogadro's constant can be written as
If we round off six significant figures
6 is greater than 5 therefore, 1 will be added to 3 and digits on right side of 6 and 6 replace by zero and digits on left side of 6 remains same except 3.
Then, the Avogadro's constant can be written as
