D) does the growth of human settlements affect where tigers can live?
Answer:
k = 9.6 x 10^5 N/m or 9.6 kN/m
Explanation:
First, we need to use the expression to calculate the spring constant which is:
w² = k/m
Solving for k:
k = w²*m
To get the angular velocity:
w = 2πf
The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:
f = V/x
f = 5.7 / 4.9 = 1.16 Hz
Now the angular velocity:
w = 2π*1.16
w = 7.29 rad/s
Finally, solving for k:
k = (7.29)² * 1800
k = 95,659.38 N/m
In two significant figures it'll ve 9.6 kN/m
You would need to ear defenders (I'm assuming these are ear protectors for use with loud sounds) while firing at a gun range. You could use it for loud construction areas.
Answer:
induced emf = 28.65 mV
Explanation:
given data
diameter = 7.3 cm
magnetic field = 0.61
time period = 0.13 s
to find out
magnitude of the induced emf
solution
we know radius is diameter / 2
radius = 7.3 / 2
radius = 3.65 m
so induced emf is dπ/dt = Adb/dt
induced emf = A × ΔB / Δt
induced emf = πr² × ΔB / Δt
induced emf = π (0..65)² × ( 0.61 - (-0.28)) / 0.13
induced emf = 0.0286538 V
so induced emf = 28.65 mV
Answer:
v = 47.85 m / s
, θ = 64.7º
Explanation:
This is a missile throwing exercise.
Let's find the speed to reach the maximum height, at this point the vertical speed is zero
= v_{oy}^{2} - 2 g y
0 = v_{oy}^{2} - 2gy
v_{oy} = √2gy
let's calculate
v_{oy} = √ (2 9.8 21.3)
v_{oy} = 20.43 m / s
now we can calculate the time it takes to get to this point
vy = v_{oy} - g t
t = v_{oy} / g
t = 20.43 / 9.8
t = 2.08 s
in projectile launching, the time it takes for the body to rise is the same as the time it takes to go down, so the total launch time is
= 2 t
t_{v} = 2 2.08 = 4.16 s
let's use the horizontal throw ratio
x = v₀ₓ t_{v}
v₀ₓ = x / t_{v}
v₀ₓ = 180 / 4.16
v₀ₓ = 43.27 m / s
initial velocity is
v = √ (v₀ₓ² + v_{oy}^{2})
v = √ (20.43² + 43.27²)
v = 47.85 m / s
with an angle of
tan θ = I go / vox
θ = tan⁻¹ (43.27 / 20.43)
θ = 64.7º