The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body.
P = W = mass x acceleration due to gravity
P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N
Solving for the static friction force (F),
F = P x c
F = (2.94 N) x 0.6 = 1.794 N
Therefore, the maximum force of static friction is 1.794 N.
Answer:
T=280.41 °C
Explanation:
Given that
At T= 24°C Resistance =Ro
Lets take at temperature T resistance is 2Ro
We know that resistance R given as
R= Ro(1+αΔT)
R-Ro=Ro αΔT
For copper wire
α(coefficient of Resistance) = 3.9 x 10⁻³ /°C
Given that at temperature T
R= 2Ro
Now by putting the values
R-Ro=Ro αΔT
2Ro-Ro=Ro αΔT
1 = αΔT
1 = 3.9 x 10⁻³ x ΔT
ΔT = 256.41 °C
T- 24 = 256.41 °C
T=280.41 °C
So the final temperature is 280.41 °C.
120n
since the speed is doubled, her force is doubled
This question is poorly stated, but I assume you mean what conditions are needed. It would have to be cold outside, correct?
Answer:
Net pull = 110 N to the left
Explanation:
Group the different pulls according to the direction (right or left)
2 pull 196 N each to the right
4 pull 98 N each to the left
5 pull 62 N each to the left
3 pull 150 N each to the right
1 pull 250 N to the left
Since positive direction is to the right, the pulls to the left will have a minus (-)
The resulting force is negative, meaning the direction is to the left
