Given:
Amount of heat produced = 100 kcal per hour
Let's find the rate of energy production in joules.
We know that:
1 calorie = 4.184 Joules
1 kcal = 4.184 Joules
To find the rate of energy production in Joules, we have:
Therefore, the rate of energy production in joules is 418.4 kJ/h which is equivalent to 418400 Joules
ANSWER:
418.4 kJ/h
Answer:
The answer to your question is: t = 2.5 s
Explanation:
Data
vo = 30 m/s
a = -12 m/s2
t = ?
vf = 0 m/s
Formula
vf = vo + at
Substitution
0 m/s = 30 + (-12)t
Solve it for t -30 = -12t
t = -30 / -12
t = 30/12 = 15/6 = 5/2
t = 2.5 s
Answer:
v = 5.7554 m/s
Explanation:
First of all we need to know if the angle of the vine is measured in the horizontal or vertical.
To do this easier, let's assume the angle is measured with the horizontal. In this case, the innitial height of the monkey will be:
h₀ = h sinα
h₀ = 5.32 sin43° = 3.6282 m
As the monkey is dropping from the innitial point which is the suspension point, is also dropping from 5.32. Then the actual height of the monkey will be:
Δh = 5.32 - 3.63 = 1.69 m
In order to calculate the speed of the monkey we need to understand that the monkey has a potential energy. This energy, because of the gravity, is converted in kinetic energy, and the value will be the same. Therefore we can say that:
Ep = Ek
From here, we can calculate the speed of the monkey.
Ep = mgΔH
Ek = 1/2 mv²
The potential energy is:
Ep = 16.9 * 9.8 * 1.69 = 279.9
Now with the kinetic energy:
1/2 * (16.9) * v² = 279.9
v² = (279.9) * 2 / 16.9
v² = 33.12
v = √33.12
<h2>
v = 5.7554 m/s</h2>
Hope this helps
Newton's second law states that the product between the mass and the acceleration of an object is equal to the force applied:
from which we find an expression for the acceleration:
(1)
Both objects are moving by uniformly accelerated motion (because the force applied is constant), so we can also using the following relationship
(2)
where
is the final speed of the object
is the initial speed
S is the distance covered
By substituting (1) into (2), and by removing
(since the final velocity of the two objects is zero), we find
where we can ignore the negative sign (because the force F will bring another negative sign).
For the first object, we have
![S= \frac{(2.0 m/s)^2 (4.0 kg)}{2F} = \frac{8}{F} [m]](https://tex.z-dn.net/?f=S%3D%20%5Cfrac%7B%282.0%20m%2Fs%29%5E2%20%284.0%20kg%29%7D%7B2F%7D%20%3D%20%20%5Cfrac%7B8%7D%7BF%7D%20%5Bm%5D%20)
And for the second object we have
![S= \frac{(4.0 m/s)^2 (1.0 kg)}{2F} = \frac{8}{F} [m]](https://tex.z-dn.net/?f=S%3D%20%5Cfrac%7B%284.0%20m%2Fs%29%5E2%20%281.0%20kg%29%7D%7B2F%7D%20%3D%20%5Cfrac%7B8%7D%7BF%7D%20%5Bm%5D%20)
And since the braking force applied to the two objects is the same, the two objects cover the same distance.
C. It pulls them together
