Answer:
50 The emissivity of galvanized steel sheet, a common roofing material, is s = 0.13 at temperatures around 300 K, while its abso
1 answer:
You might be interested in
Answer:
Explanation:
given,
weight of swimmer = 510 N
length of ledge, L = 1.75 m
vertical height of the cliff, h = 9 m
speed of the swimmer = ?
horizontal velocity of the swimmer should be that much it can cross the wedge.
distance = speed x time
d = v_x × t
1.75 = v_x × t ........(1)
now,time taken by the swimmer to cover 9 m
initial vertical velocity of the swimmer is zero.
using equation of motion for time calculation
t² = 1.938
t = 1.39 s
same time will be taken to cover horizontal distance.
now, from equation 1
1.75 = v_x × 1.39
horizontal speed of the swimmer is equal to 1.26 m/s
Answer:
The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s
Explanation:
The given parameters of the motion of the branch are;
The height from which the branch is thrown = 3.00 m
The horizontal distance the branch lands from where it was thrown, x = 8.00 m
The direction in which the branch is thrown = Horizontally
Therefore, the initial vertical velocity of the branch, = 0 m/s
The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;
s = ·t + 1/2·g·t²
Where;
= 0 m/s
s = The initial height of the object = 3.00 m
g = The acceleration due to gravity = 9.8 m/s²
∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²
t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246
The horizontal distance covered before the branch touches the ground, x = 8.00 m
Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;
uₓ = x/t = 8.00 m/((√30)/7 s)
Using a graphing calculator, we get;
uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s
The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.
Answer:
The spring constant = 104.82 N/m
The angular velocity of the bar when θ = 32° is 1.70 rad/s
Explanation:
From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:
Also;
Thus;
where;
= deflection in the spring
k = spring constant
b = remaining length in the rod
m = mass of the slender bar
g = acceleration due to gravity
Thus; the spring constant = 104.82 N/m
b
The angular velocity can be calculated by also using the conservation of energy;
Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s
