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1. 12m/s and here’s how. The equation for calculating velocity of a wave is lambda x frequency. So 2m x 6Hz. Hz can be 1/s so the only unit of measurement we get is meters and seconds. So the answer will be 12m/s.
2. Frequency=2Hz period= 0.5 seconds
Equation for frequency is velocity/wavelength. 10m/s divides by 5m = 2Hz.
Equation for period is 1 / frequency. 1 divided by 2 = 0.5 seconds
3. The answer is the picture of you can’t read my hand writing comment and say “ I need the info about the wave because I can’t read your hand writing “ thank you and I hoped I helped
2. Frequency=2Hz period= 0.5 seconds
Equation for frequency is velocity/wavelength. 10m/s divides by 5m = 2Hz.
Equation for period is 1 / frequency. 1 divided by 2 = 0.5 seconds
3. The answer is the picture of you can’t read my hand writing comment and say “ I need the info about the wave because I can’t read your hand writing “ thank you and I hoped I helped
Answer:
a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis
0.7 kg is 109.4 ms on the x axis
b) Y = 109.3 m
Explanation:
This is a moment and projectile launch exercise.
a) Let's start by finding the initial velocity of the projectile
sin 40 = voy / v₀
= v₀ sin 40
= 50.0 sin40
= 32.14 m / s
cos 40 = v₀ₓ / V₀
v₀ₓ = v₀ cos 40
v₀ₓ = 50.0 cos 40
v₀ₓ = 38.3 m / s
Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis
Let's write the amounts
Initial mass of the projectile M = 2.0 kg
Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and = -10.0 m / s
Fragment mass 2 m₂ = 0.7 kg moves in the x direction
Fragment mass 3 m₃ = 0.3 kg moves up (y axis)
Moment before the break
X axis
p₀ₓ = m v₀ₓ
Y Axis y
= 0
After the break
X axis
= m₂ v₂
Axis y
= m₁ v₁ + m₃ v₃
Let's write the conservation of the moment and calculate
Y Axis
0 = m₁ v₁ + m₃ v₃
Let's clear the speed of fragment 3
v₃ = - m₁ v₁ / m₃
v₃ = - (-10) 1 / 0.3
v₃ = 33.3 m / s
X axis
M v₀ₓ = m₂ v₂
v₂ = v₀ₓ M / m₂
v₂ = 38.3 2 / 0.7
v₂ = 109.4 m / s
The fragment speeds of 0.3 kg is 33.3 m / s on the y axis
0.7 kg is 109.4 ms on the x axis
b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics
² =
² - 2 gy
0 = ²-2gy
y = ² / 2g
y = 32.14² / 2 9.8
y = 52.7 m
This is the height where the break occurs, which is the initial height for body movement of 0.3 kg
² =
² - 2 g y₂
0 = ² - 2 g y₂
y₂ = ² / 2g
y₂ = 33.3²/2 9.8
y₂ = 56.58 m
Total body height is
Y = y + y₂
Y = 52.7 + 56.58
Y = 109.3 m