Answer:
21 psi
Explanation:
The weight of the car is:
W = mg
W = 1000 kg * 9.8 m/s²
W = 9800 N
Divided by 4 tires, each tire supports:
F = W/4
F = 9800 N / 4
F = 2450 N
Pressure is force divided by area, so:
P = F / A
P = (2450 N) / (0.13 m × 0.13 m)
P ≈ 145,000 Pa
101,325 Pa is the same as 14.7 psi, so:
P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)
P ≈ 21 psi
Answer:
P = 180.81 J
Explanation:
Given that,
Mass of a object, m = 4.1 kg
It is lifted to a height of 4.5 m
We need to find the potential energy of the object due to gravity. It is given by the formula as follows :
P = mgh Where g is acceleration due to gravity
P = 4.1 kg × 9.8 m/s² × 4.5 m
P = 180.81 J
Hence, the potential energy is 180.81 J.
Answer:
80.386 degrees
Explanation:
We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse
The adjacent side = 699ft
The hypotenuse = 1034ft
using cos∅ = Adjacent/hypotenuse
where ∅ is the unknown angle
cos ∅ = 699/1034 = 0.167
∅ = arccos 0.167 = 80.368°
As easy as one can imagine
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
