Explanation:
KE = 1/2mv^2
6,250 J =1/2(250kg) v ^2
<u>6250 J </u>= <u>125kgv^2</u>
125kg. 125kg
v =
In case of an object sitting at rest on another base, there are two equal and opposite forces – Normal force and the gravity.
Answer: Option A
<u>Explanation:
</u>
When an object is placed at rest position on another object, there is a force exerted by the surfaces of the two contact objects. This force is denoted as Normal Force.
When an object such as a box is placed on a shelf, its surface exerts a contact force on the base of the shelf- The Normal force directed upward. Meanwhile, the gravity stays at its action and tries to pull the box towards itself.
Both of these forces however are equal and opposite and therefore, there is zero net force on the box. That's why it remains at rest, holding on Newton's third law.
Answer:
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
On Earth, the acceleration of gravity is 9.8 m/s² downward.
So any object with only gravity acting on it gains 9.8 m/s of
downward speed every second.
If the rock starts out moving upward at 10 m/s, then it will
continue upward for only (10/9.8) = 1.02 second, before
it stops rising and starts falling.
Its average speed during that time is (1/2) (10 + 0) = 5 m/s .
At an average speed of 5 m/s for 1.02 sec,
the rock rises
(5 m/s) x (1.02 sec) = 5.102 meters .
Answer:
The latent heat of vaporization of water is 2.4 kJ/g
Explanation:
The given readings are;
The first (mass) balance reading (of the water) in grams, m₁ = 581 g
The second (mass) balance reading (of the water) in grams, m₂ = 526 g
The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ
The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ
The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature
Based on the measurements, we have;
The latent heat of vaporization = ΔQ/Δm
∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g
The latent heat of vaporization of water = 2.4 kJ/g
