Answer:
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g
U is the speed = 300m/s
H is the maximum height = 78.4m
g is the acceleration due to gravity = 9.8m/s²
Substitute into the fromula;
R = 300√2(78.4)/9.8
R = 300 √(16)
R = 300*4
R = 1200m
Hence the projectile travelled 1200m before hitting the ground
Answer:
9.34 N
Explanation:
First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:
where
f is the frequency of the wave
is the wavelength
For the waves in this string we have:
, since it completes 625 cycles per second
is the wavelength
So the speed of the wave is
The speed of the waves in a string is related to the tension in the string by
(1)
where
T is the tension in the string
is the linear density
In this problem:
is the mass of the string
L = 0.75 m is the its length
Solving the equation (1) for T, we find the tension:
Answer:
Answer is D. 8.04 x 10^4 J
Explanation:
1. D
2. A
3. D
4. B
5. C
6. B
7. D
8. C
9. B
10. D
All correct i promise you that
Answer:
Intensity of the light (first polarizer) (I₁) = 425 W/m²
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²
Explanation:
Given:
Unpolarized light of intensity (I₀) = 950 W/m²
θ = 65°
Find:
a. Intensity of the light (first polarizer)
b. Intensity of the light (second polarizer)
Computation:
a. Intensity of the light (first polarizer)
Intensity of the light (first polarizer) (I₁) = I₀ / 2
Intensity of the light (first polarizer) (I₁) = 950 / 2
Intensity of the light (first polarizer) (I₁) = 425 W/m²
b. Intensity of the light (second polarizer)
Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ
Intensity of the light (second polarizer) (I₂) = (425)(0.1786)
Intensity of the light (second polarizer) (I₂) = 75.905 W/m²