Answer:
The last option is the only correct one if you like to multiply
The second last option is good if you like to divide.
Explanation:
Each fraction in the last two options has a value of 1
example
dividing by 1
15 cm /(100 cm/ 1 m) = 0.15 m 0.15 m / (1000 m/ 1km) = 0.00015 km
and
multiplying by 1
15 cm(1 m / 100cm) = 0.15 m 0.15m(1 km/1000m) = 0.00015 km
only one of the two fractions in each of the top two options has a value of 1.
Answer:
Explanation:
The equivalent resistance for three resistors connected in parallel is given as
(1/R)=(1/R₁)+(1/R₂)+(1/R₃)
now we.need to.insert the value of 3 resistances but only 2 are given in the question.
To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force
![F_p = 6*10^{22}N](https://tex.z-dn.net/?f=F_p%20%3D%206%2A10%5E%7B22%7DN)
![F_p = \frac{GMm}{R^2}](https://tex.z-dn.net/?f=F_p%20%3D%20%5Cfrac%7BGMm%7D%7BR%5E2%7D)
![F_p = 9*10^{22}N](https://tex.z-dn.net/?f=F_p%20%3D%209%2A10%5E%7B22%7DN)
Distance between planet and star
![r = \frac{R}{2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7BR%7D%7B2%7D)
Gravitational force is
![F = \frac{GMm}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BGMm%7D%7Br%5E2%7D)
Applying the new distance,
![F = \frac{GMm}{(\frac{R}{2})^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BGMm%7D%7B%28%5Cfrac%7BR%7D%7B2%7D%29%5E2%7D)
![F = 4\frac{GMm}{R^2}](https://tex.z-dn.net/?f=F%20%3D%20%204%5Cfrac%7BGMm%7D%7BR%5E2%7D)
Replacing with the previous force,
![F = 4F_p](https://tex.z-dn.net/?f=F%20%3D%204F_p)
Replacing our values
![F= 4(9*10^{22}N)](https://tex.z-dn.net/?f=F%3D%204%289%2A10%5E%7B22%7DN%29)
![F = 36*10^{22}N](https://tex.z-dn.net/?f=F%20%3D%2036%2A10%5E%7B22%7DN)
Therefore the magnitude of the force on the star due to the planet is ![36*10^{22}N](https://tex.z-dn.net/?f=36%2A10%5E%7B22%7DN)
Explanation:
We have,
Length of a metal rod is 55 cm or 0.55 m
Change in length is 0.2 cm or 0.002 m
It is required to find the change in temperature of a metal rod. The coefficient of linear expansion is given by :
![\alpha =\dfrac{\Delta L}{L_0\Delta T}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7B%5CDelta%20L%7D%7BL_0%5CDelta%20T%7D)
is the change in temperature
![\Delta T =\dfrac{\Delta L}{L_0\alpha }\\\\\Delta T =\dfrac{0.002}{0.55\times 12\times 10^{-6}}\\\\\Delta T= 303.03^{\circ} C](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%5Cdfrac%7B%5CDelta%20L%7D%7BL_0%5Calpha%20%7D%5C%5C%5C%5C%5CDelta%20T%20%3D%5Cdfrac%7B0.002%7D%7B0.55%5Ctimes%2012%5Ctimes%2010%5E%7B-6%7D%7D%5C%5C%5C%5C%5CDelta%20T%3D%09303.03%5E%7B%5Ccirc%7D%20C)
So, the change in temperature is 303.03 degrees Celsius.
Answer:
2.12 J
Explanation:
Initial kinetic energy = final elastic energy + work by friction
KE = EE + W
KE = ½ kx² + W
5 J = ½ (1600 N/m) (0.06 m)² + W
W = 2.12 J