Answer:
Math Skills Calculating Kinetic Energy and Potential Energy ... moving object depends on its mass and speed. ... represents the object's speed, measured in meters per second (m/s). ... Example 1: A bicycle and rider with a combined mass of ... the bicycle and rider? Given: m = 110 kg v = 8 m/s. Unknown: Kinetic energy (KE).
I think
Explanation:
Speed has now direction while velocity does. For example, if i say that im running at 10 mph, i have given you my speed<span>. If i say that i am running at 10 mph north, then i haven given you my velocity</span><span> Acceleration is the rate of change in velocity. </span>
Answer: You do not specify what is being asked for. ∆E? ∆H?
∆E = (430 - 238) J = 192 J
∆H = 430 J
Explanation:
If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.
Therefore ∆H = 430 J
If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w
The question states that 238 J of work are done AND the system expanded
(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)
Therefore, ∆E = (430 - 238) J = 192 J
a) Speed of the Moon: 1025 m/s
The speed of the moon is equal to the ratio between the circumference of its orbit and the orbital period:
![v=\frac{2 \pi r}{T}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%20%5Cpi%20r%7D%7BT%7D)
where
is the radius of the orbit of the Moon
is the orbital period
Substituting into the formula, we find
![v=\frac{2 \pi (3.85\cdot 10^8 m)}{2.36\cdot 10^6 s)}=1025 m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%20%5Cpi%20%283.85%5Ccdot%2010%5E8%20m%29%7D%7B2.36%5Ccdot%2010%5E6%20s%29%7D%3D1025%20m%2Fs)
b) Centripetal force: ![2.0 \cdot 10^{20} N](https://tex.z-dn.net/?f=2.0%20%5Ccdot%2010%5E%7B20%7D%20N)
The centripetal force acting on the Moon is given by:
![F=m\frac{v^2}{r}](https://tex.z-dn.net/?f=F%3Dm%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
is the mass of the Moon
is its orbital speed
is the radius of the orbit
Substituting into the formula, we find
![F=(7.35 \cdot 10^{22} kg) \frac{(1025 m/s)^2}{3.85\cdot 10^8 m}=2.0 \cdot 10^{20} N](https://tex.z-dn.net/?f=F%3D%287.35%20%5Ccdot%2010%5E%7B22%7D%20kg%29%20%5Cfrac%7B%281025%20m%2Fs%29%5E2%7D%7B3.85%5Ccdot%2010%5E8%20m%7D%3D2.0%20%5Ccdot%2010%5E%7B20%7D%20N)
c) Gravitational force
The only relevant force that acts on the Moon, and that keeps the Moon in circular motion around the Earth, is the gravitational force exerted by the Earth on the Moon. In fact, this force "pulls" the Moon towards the Earth, so towards the centre of the orbit of the Moon, therefore it acts as source of centripetal force for the Moon.
Answer:
42.69 N and 18.07 N
Explanation:
We are given that
Mass of ladder=6.2 kg
Length of ladder=1.97 m
Distance of Sawhorse A from one end=0.64 m
Distance of sawhorse B from other end=0.17 m
Let center of Ladder=![\frac{1.97}{2}=0.985 m](https://tex.z-dn.net/?f=%5Cfrac%7B1.97%7D%7B2%7D%3D0.985%20m)
Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m
Distance of sawhorse B from center of ladder=0.985-0.17=0.815 m
Force one ladder due to gravity=mg=![6.2\times 9.8=60.76N](https://tex.z-dn.net/?f=6.2%5Ctimes%209.8%3D60.76N)
Where ![g=9.8 m/s^2](https://tex.z-dn.net/?f=g%3D9.8%20m%2Fs%5E2)
Torque applied on Sawhorse A=![0.345F_a](https://tex.z-dn.net/?f=0.345F_a)
Torque applied on Sawhorse B=![0.815F_b](https://tex.z-dn.net/?f=0.815F_b)
In equilibrium
![0.345F_a=0.815F_b](https://tex.z-dn.net/?f=0.345F_a%3D0.815F_b)
![F_b=\frac{0.345}{0.815}F_a](https://tex.z-dn.net/?f=F_b%3D%5Cfrac%7B0.345%7D%7B0.815%7DF_a)
Total force=![F_a+F_b](https://tex.z-dn.net/?f=F_a%2BF_b)
![F_a+\frac{0.345}{0.815}F_a=60.76](https://tex.z-dn.net/?f=F_a%2B%5Cfrac%7B0.345%7D%7B0.815%7DF_a%3D60.76)
![\frac{0.815F_a+0.345F_a}{0.815}=60.76](https://tex.z-dn.net/?f=%5Cfrac%7B0.815F_a%2B0.345F_a%7D%7B0.815%7D%3D60.76)
![\frac{1.16}{0.815}F_a=60.76](https://tex.z-dn.net/?f=%5Cfrac%7B1.16%7D%7B0.815%7DF_a%3D60.76)
![F_a=\frac{60.76\times 0.815}{1.16}=42.69 N](https://tex.z-dn.net/?f=F_a%3D%5Cfrac%7B60.76%5Ctimes%200.815%7D%7B1.16%7D%3D42.69%20N)
![F_b=\frac{0.345}{0.815}\times 42.69=18.07 N](https://tex.z-dn.net/?f=F_b%3D%5Cfrac%7B0.345%7D%7B0.815%7D%5Ctimes%2042.69%3D18.07%20N)