Answer:
μ = 0.725
Explanation:
This problem refers to Newton's second law.
F = ma
Let's write the equations on each axis
Y Axis
N-W = 0
N = W
N = mg
X axis
F-fr = ma
With the body not started moving its acceleration is zero
F-fr = 0
F = fr
The friction force equation is
fr = μ N
fr = μ m g
Let's replace and calculate
F = μ m g
μ = F / mg
μ = 321 /45.2 9.8
μ = 0.725
Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
Answer
for every meters it will go up 15 so if it was 2 secoonds it woudl be 30 and if it was 3 seconds it would be 45 meters
Explanation:
Answer:
false is the answer because it was around in the 1880's
Explanation:
I hope this helped
To answer this question, we must use the equation for the volumetric expansion of gases at constant pressure. This equation is given by:
We know:
is the initial volume 
ΔT is the temperature change = 45 ° -20 ° = 25 °
is the coefficient of gas expansion and is equal to 1/273
Then the final volume of the gas is:
