Answer:
<em>11.06m/s²</em>
Explanation:
According to Newtons second law of motion
![\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\](https://tex.z-dn.net/?f=%5Csm%20F_x%20%3D%20ma_x%5C%5CF_m%20-%20F_f%20%3D%20ma_x%5C%5Cmgsin%20%5Ctheta%20-%20%5Cmu%20R%20mgcos%20%5Ctheta%20%3D%20ma_x%5C%5C)
Given
Mass m = 17kg
Fm = 208N
theta = 36 degrees
g = 9.8m/s²
a is the acceleration
Substitute
208 - 0.148(17)(9.8)cos 36 = 17a
208 - 24.6568cos36 = 17a
208 - 19.9478 = 17a
188.05 = 17a
a = 188.05/17
a = 11.06m/s²
<em>Hence the the magnitude of the resulting acceleration is 11.06m/s²</em>
Explanation:
Given that,
Rate of cooling of air
Initial temperature= 80°C
Final temperature = 5°C
We need to calculate
Using newton's law of cooling
![\dfrac{dT}{dt}=c(T-T_{0})](https://tex.z-dn.net/?f=%5Cdfrac%7BdT%7D%7Bdt%7D%3Dc%28T-T_%7B0%7D%29)
![\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})](https://tex.z-dn.net/?f=%5Cdfrac%7BdT%7D%7Bdt%7D%3Dc%28%5Cdfrac%7BT_%7B1%7D%2BT_%7B2%7D%7D%7B2%7D-T_%7B0%7D%29)
Where, ![dT=T_{1}-T_{2}](https://tex.z-dn.net/?f=dT%3DT_%7B1%7D-T_%7B2%7D)
Here,
= 25°C (surrounding temperature)
dt = 1 minute
![\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})](https://tex.z-dn.net/?f=%5Cdfrac%7BdT%7D%7Bdt%7D%3Dc%28%5Cdfrac%7BT_%7B1%7D%2BT_%7B2%7D%7D%7B2%7D-T_%7B0%7D%29)
Put the value into the formula
![\dfrac{80-5}{1}=c(\dfrac{85}{2}-25)](https://tex.z-dn.net/?f=%5Cdfrac%7B80-5%7D%7B1%7D%3Dc%28%5Cdfrac%7B85%7D%7B2%7D-25%29)
![c=\dfrac{75}{17.5}](https://tex.z-dn.net/?f=c%3D%5Cdfrac%7B75%7D%7B17.5%7D)
![c=4.285\ cubic\ meter/minute](https://tex.z-dn.net/?f=c%3D4.285%5C%20cubic%5C%20meter%2Fminute)
Hence, This is the required answer.
The magnitude of the electric field at the proton's location is 10,437.5 N/C.
<h3>What the magnitude of the
electric field?</h3>
The size of the electric field is basically characterized as the power per charge on the test charge. On the off chance that the electric field strength is meant by the image E. Very much like gravity, electric fields work the same way. In any case, while gravity generally draws in, an electric field, then again, can either rebuff or draw in. By and large, the Electric Field submits to the super-position guideline. the all out Electric Field from various charges is equivalent to the amount of the electric fields from each charge separately. An electric field is the actual field that encompasses electrically charged particles and applies force on any remaining charged particles in the field, either drawing in or repulsing them.
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