Answer:
1058.78 ft/sec
Explanation:
Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis
The horizontal velocity of a body can be calculated as shown below.\
Vh = Vcos∅.......................... Equation 1
Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.
Given: V = 1178 ft/sec, ∅ = 26°
Substitute into equation 1
Vh = 1178cos26
Vh = 1178(0.8988)
Vh = 1058.78 ft/sec
Hence the horizontal component of the velocity = 1058.78 ft/sec
Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
Answer:
Compression is the answer
Explanation:
got it right
Answer:
A. The wavelengths of the new sound waves are longer
Explanation:
This is the Doppler effect which can be best illustraded for the case of a siren of an ambulance approaching us having a greater frequency and getting lower in frequency and deeper as the ambulance passes us.
Since the wavelength is inversely proportional to the frequency it follows the wavelengths are longer when the frequency decreases lowering its pitch and getting deeper.
<span>1.0x10^3 Joules
The kinetic energy a body has is expressed as the equation
E = 0.5 M V^2
where
E = Energy
M = Mass
V = Velocity
Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion
E = 0.5 * 7.2 kg * (17 m/s)^2
E = 3.6 kg * 289 m^2/s^2
E = 1040.4 kg*m^2/s^2
E = 1040.4 J
So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>