Answer:
F=507.7N
Explanation:
According to Newton's second law:
![\sum F=m.a](https://tex.z-dn.net/?f=%5Csum%20F%3Dm.a)
in this case, the football players need to drag the coach at a constant velocity, thus means with no acceleration, so:
![\sum F=0\\\\-F_f+F_p*cos(\theta)=0\\\\](https://tex.z-dn.net/?f=%5Csum%20F%3D0%5C%5C%5C%5C-F_f%2BF_p%2Acos%28%5Ctheta%29%3D0%5C%5C%5C%5C)
there are 20 degrees between the two ropes that means each player exerts a force 10 degrees from the zero reference.
![F_p=2*F\\\\F=\frac{1000N}{2*cos(10)}\\\\F=507.7N](https://tex.z-dn.net/?f=F_p%3D2%2AF%5C%5C%5C%5CF%3D%5Cfrac%7B1000N%7D%7B2%2Acos%2810%29%7D%5C%5C%5C%5CF%3D507.7N)
Answer:
1) joule
2) ![kgm^{2}/s^{2}](https://tex.z-dn.net/?f=kgm%5E%7B2%7D%2Fs%5E%7B2%7D)
3) ![10\%](https://tex.z-dn.net/?f=10%5C%25)
Explanation:
1) Luminosity is the <u>amount of light emitted</u> (measured in Joule) by an object in a unit of<u> time</u> (measured in seconds). Hence in SI units luminosity is expressed as joules per second (
), which is equal to Watts (
).
This amount of light emitted is also called radiated electromagnetic power, and when this is measured in relation with time, the result is also called radiant power emitted by a light-emitting object.
Therefore, if we want to calculate luminosity the Joule as a unit will be used.
2) Work
is expressed as force
multiplied by the distane
:
Where force has units of
and distance units of
.
If we input the units we will have:
This is 1Joule (
) in the SI system, which is also equal to ![1 Nm](https://tex.z-dn.net/?f=1%20Nm)
3) The formula to calculate the percent error is:
![\% error=\frac{|V_{exp}-V_{acc}|}{V_{acc}} 100\%](https://tex.z-dn.net/?f=%5C%25%20error%3D%5Cfrac%7B%7CV_%7Bexp%7D-V_%7Bacc%7D%7C%7D%7BV_%7Bacc%7D%7D%20100%5C%25)
Where:
is the experimental value
is the accepted value
![\% error=\frac{|7.34 (10)^{-11} Nm^{2}/kg^{2}-6.67 (10)^{-11} Nm^{2}/kg^{2}|}{6.67 (10)^{-11} Nm^{2}/kg^{2}} 100\%](https://tex.z-dn.net/?f=%5C%25%20error%3D%5Cfrac%7B%7C7.34%20%2810%29%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D-6.67%20%2810%29%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D%7C%7D%7B6.67%20%2810%29%5E%7B-11%7D%20Nm%5E%7B2%7D%2Fkg%5E%7B2%7D%7D%20100%5C%25)
This is the percent error
Explanation:
Given:
![r_a = 3.570R_E](https://tex.z-dn.net/?f=r_a%20%3D%203.570R_E)
![R_E = 1.499×10^{11}\:\text{m}](https://tex.z-dn.net/?f=R_E%20%3D%201.499%C3%9710%5E%7B11%7D%5C%3A%5Ctext%7Bm%7D)
![M_S = 1.989×10^{30}\:\text{kg}](https://tex.z-dn.net/?f=M_S%20%3D%201.989%C3%9710%5E%7B30%7D%5C%3A%5Ctext%7Bkg%7D)
![G = 6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2](https://tex.z-dn.net/?f=G%20%3D%206.674%C3%9710%5E%7B-11%7D%5C%3A%5Ctext%7BN-m%7D%5E2%5Ctext%7B%2Fkg%7D%5E2)
Let
= mass of the asteroid and
= orbital radius of the asteroid around the sun. The centripetal force
is equal to the gravitational force ![F_G:](https://tex.z-dn.net/?f=F_G%3A)
![F_c = F_G \Rightarrow m_a\dfrac{v_a^2}{r_a} = G\dfrac{m_aM_S}{r_a^2}](https://tex.z-dn.net/?f=F_c%20%3D%20F_G%20%5CRightarrow%20m_a%5Cdfrac%7Bv_a%5E2%7D%7Br_a%7D%20%3D%20G%5Cdfrac%7Bm_aM_S%7D%7Br_a%5E2%7D)
or
![\dfrac{4\pi^2 r_a}{T^2} = G\dfrac{M_S}{r_a^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Cpi%5E2%20r_a%7D%7BT%5E2%7D%20%3D%20G%5Cdfrac%7BM_S%7D%7Br_a%5E2%7D)
where
![v = \dfrac{2\pi r_a}{T}](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7B2%5Cpi%20r_a%7D%7BT%7D)
with T = period of orbit. Rearranging the variables, we get
![T^2 = \dfrac{4\pi^2 r_a^3}{GM_S}](https://tex.z-dn.net/?f=T%5E2%20%3D%20%5Cdfrac%7B4%5Cpi%5E2%20r_a%5E3%7D%7BGM_S%7D)
Taking the square root,
![T = 2\pi \sqrt{\dfrac{r_a^3}{GM_S}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cdfrac%7Br_a%5E3%7D%7BGM_S%7D%7D)
![\:\:\:\:=2\pi \sqrt{\dfrac{(3.57(1.499×10^{11}\:\text{m}))^3}{(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)(1.989×10^{30}\:\text{kg})}}](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5C%3A%5C%3A%3D2%5Cpi%20%5Csqrt%7B%5Cdfrac%7B%283.57%281.499%C3%9710%5E%7B11%7D%5C%3A%5Ctext%7Bm%7D%29%29%5E3%7D%7B%286.674%C3%9710%5E%7B-11%7D%5C%3A%5Ctext%7BN-m%7D%5E2%5Ctext%7B%2Fkg%7D%5E2%29%281.989%C3%9710%5E%7B30%7D%5C%3A%5Ctext%7Bkg%7D%29%7D%7D)
![\:\:\:\:= 2.13×10^8\:\text{s} = 6.75\:\text{years}](https://tex.z-dn.net/?f=%5C%3A%5C%3A%5C%3A%5C%3A%3D%202.13%C3%9710%5E8%5C%3A%5Ctext%7Bs%7D%20%3D%206.75%5C%3A%5Ctext%7Byears%7D)
Answer:
The change in internal energy is zero.
(a) is correct option
Explanation:
Given that,
Work done W = -400 J
The First law of thermodynamics is law conservation of energy
Law of conservation energy is defined the total amount of energy is constant in isolated system.
The energy can not be created and destroyed but the energy can be changed from one form to other form.
The First law of thermodynamics is given by
Where,
W = work done by the system or on the system
= Total internal energy
Q = heat
In isothermal process, the temperature is constant
The energy can not be change and the internal energy is the function of temperature.
So, The internal energy will be zero .
Hence, The change in internal energy is zero.