Answer:
T = 76.39°C
Explanation:
given,
coffee cup temperature = 95°C
Room temperature= 20°C
expression

temperature at t = 0

T(0) = 95°C
temperature after half hour of cooling

t = 30 minutes


T(30) = 61.16° C
average of first half hour will be equal to

![T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B1%7D%7B30%7D%5B%2820t%20-%20%5Cdfrac%7B75%20e%5E%7B%5Cdfrac%7B-t%7D%7B50%7D%7D%7D%7B%5Cdfrac%7B1%7D%7B50%7D%7D%29%5D_0%5E30)
![T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B1%7D%7B30%7D%5B%2820t%20-%203750e%5E%7B%5Cdfrac%7B-t%7D%7B50%7D%7D%5D_0%5E30)
![T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B1%7D%7B30%7D%5B%2820%5Ctimes%2030%20-%203750%20e%5E%7B%5Cdfrac%7B-30%7D%7B50%7D%7D%20%2B%203750%5D)
![T = \dfrac{1}{30}[600 - 2058.04 + 3750]](https://tex.z-dn.net/?f=T%20%3D%20%5Cdfrac%7B1%7D%7B30%7D%5B600%20-%202058.04%20%2B%203750%5D)
T = 76.39°C
The force exerted on the box is 56 N
Explanation:
The work done by a force on an object is given by

where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
For the box in this problem, we have:
W = 2240 J is the work done
d = 40 m is the displacement of the box
Assuming that the force is parallel to the displacement, 
Solving the equation for F, we find the force exerted on the box:

Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
#LearnwithBrainly
Answer:
20m/s
Explanation:
it covers 20 metres in a second
Answer:
F = 2.26 × 10⁻³ N
Explanation:
given,
length of rod = 11 cm
charge = 19 nC
linear charge density = 3.9 x 10⁻⁷ C/m
electric force at 2 cm away.

F = E q

integrating from 0.02 to 0.02 + L
![F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2K%5Clambda%5C%20q%7D%7BL%7D%5Bln%280.02%2BL%29-ln%280.002%29%5D)
![F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]](https://tex.z-dn.net/?f=F%3D%20%5Cdfrac%7B2%5Ctimes%209%20%5Ctimes%2010%5E9%5Ctimes%203.9%5Ctimes%2010%5E%7B-7%7D%5Ctimes%2019%20%5Ctimes%2010%5E%7B-9%7D%7D%7B0.11%7D%5Bln%280.02%2B0.11%29-ln%280.002%29%5D)
F = 2.26 × 10⁻³ N