Answer:
Eastward, at 11 m/s^2
Explanation:
64N-31N=unbalanced force of 33N
F=ma
33N=(3kg)a
a=11m/s^2 to the East
Answer:
the propagation velocity of the wave is 274.2 m/s
Explanation:
Given;
length of the string, L = 1.5 m
mass of the string, m = 0.002 kg
Tension of the string, T = 100 N
wavelength, λ = 1.5 m
The propagation velocity of the wave is calculated as;

Therefore, the propagation velocity of the wave is 274.2 m/s
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.
It must be a virtual image, because this is the only kind of image it can produce.
Answer:
The intensity at 10° from the center is 3.06 × 10⁻⁴I₀
Explanation:
The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ
I₀ = maximum intensity of light
a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m
θ = angle at intensity point = 10°
λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m
α = πasinθ/λ
= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m
= 1.0911/650 × 10³
= 0.001679 × 10³
= 1.679
Now, the intensity I is
I = I₀(sinα/α)²
= I₀(sin1.679/1.679)²
= I₀(0.0293/1.679)²
= 0.0175²I₀
= 0.0003063I₀
= 3.06 × 10⁻⁴I₀
So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀