Answer:
50%
Explanation:
Efficiency = work out / work in
e = Fd / W
e = (2000 N) (2 m) / (8000 J)
e = 0.5
Answer:
a. 8.33 x 10 ⁻⁶ Pa
b. 8.19 x 10 ⁻¹¹ atm
c. 1.65 x 10 ⁻¹⁰ atm
d. 2.778 x 10 ⁻¹⁴ kg / m²
Explanation:
Given:
a.
I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s
P rad = I / us
P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s
P rad = 8.33 x 10 ⁻⁶ Pa
b.
P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]
P rad = 8.19 x 10 ⁻¹¹ atm
c.
P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]
P rad = 1.67 x 10 ⁻⁵ Pa
P₁ = 1.013 x 10 ⁵ Pa /atm
P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm
d.
P rad = I / us
ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²
ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²
Answer:
4.17 m/s
Explanation:
To solve this problem, let's start by analyzing the vertical motion of the pea.
The initial vertical velocity of the pea is
Now we can solve the problem by applying the suvat equation:
where
is the vertical velocity when the pea hits the ceiling
is the acceleration of gravity
s = 1.90 is the distance from the ceiling
Solving for ,
Instead, the horizontal velocity remains constant during the whole motion, and it is given by
Therefore, the speed of the pea when it hits the ceiling is
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.
Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
The radius of the circular path is 1.5 m.
The circumference is then
The ball moves 3π m every 2.2 s, so the speed is
