Answer
given,
mass of block (m)= 6.4 Kg
spring is stretched to distance, x = 0.28 m
initial velocity = 5.1 m/s
a) computing weight of spring
k x = m g
![k = \dfrac{mg}{x}](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7Bmg%7D%7Bx%7D)
![k = \dfrac{6.4 \times 9.8}{0.28}](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7B6.4%20%5Ctimes%209.8%7D%7B0.28%7D)
k = 224 N/m
b) ![f = \dfrac{\omega}{2\pi}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B%5Comega%7D%7B2%5Cpi%7D)
![\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%7D%3D%20%5Csqrt%7B%5Cdfrac%7B224%7D%7B6.4%7D%7D%20%3D%205.92%20%5C%20rad%2Fs)
![f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cdfrac%7Bk%7D%7Bm%7D%7D)
![f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cdfrac%7B224%7D%7B6.4%7D%7D)
![f =0.94\ Hz](https://tex.z-dn.net/?f=f%20%3D0.94%5C%20Hz)
c) ![v_b = -v cos \omega t](https://tex.z-dn.net/?f=v_b%20%3D%20-v%20cos%20%5Comega%20t)
![v_b = -5.1 \times cos (5.92 \times 0.42)](https://tex.z-dn.net/?f=v_b%20%3D%20-5.1%20%5Ctimes%20cos%20%285.92%20%5Ctimes%200.42%29)
![v_b = 4.04\ m/s](https://tex.z-dn.net/?f=v_b%20%3D%204.04%5C%20m%2Fs)
d) ![a_{max} = v \omega](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%20v%20%5Comega)
![a_{max} = 4.04 \times 5.92](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%204.04%20%5Ctimes%205.92)
![a_{max} =23.94\ m/s^2](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D23.94%5C%20m%2Fs%5E2)
e)
![A = \dfrac{v}{\omega}](https://tex.z-dn.net/?f=A%20%3D%20%5Cdfrac%7Bv%7D%7B%5Comega%7D)
![A = \dfrac{4.04}{5.92}](https://tex.z-dn.net/?f=A%20%3D%20%5Cdfrac%7B4.04%7D%7B5.92%7D)
A = 0.682 m
Force =![m \omega^2 |Y|](https://tex.z-dn.net/?f=m%20%5Comega%5E2%20%7CY%7C)
=![6.4 \times 5.92^2\times 0.42](https://tex.z-dn.net/?f=6.4%20%5Ctimes%205.92%5E2%5Ctimes%200.42)
F = 94.20 N
The answer would be:
Precipitation sometimes occurs when the horizontal winds move air against mountain ranges, forcing air to rise as it passes over the mountains.
This happens when the air is forced to move from low elevation to high elevation due to rising terrain. This causes the air to cool adiabatically. This increases the relative humidity and causes clouds to form, under certain conditions it can also create precipitation.
Answer:
C
Explanation:
It has to travel 600 light years before we would be able to observe the explosion.
Answer:
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
c) True. Information is missing to perform the calculation
Explanation:
Let's consider solving this exercise before seeing the final statements.
We use Newton's second law Rotational
τ = I α
T r = I α
T gR = I α
Alf = T R / I (1)
T = α I / R
Now let's use Newton's second law in the mass that descends
W- T = m a
a = (m g -T) / m
The two accelerations need related
a = R α
α = a / R
a = (m g - α I / R) / m
R α = g - α I /m R
α (R + I / mR) = g
α = g / R (1 + I / mR²)
We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant
Let's review the claims
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
b) False. Missing data for calculation
c) True. Information is missing to perform the calculation
d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases
Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by
![P = VI](https://tex.z-dn.net/?f=P%20%3D%20VI)
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,
![I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BP%7D%7BV%7D%20%20%5C%5C%5C%5CI%20%3D%20%5Cfrac%7B100%7D%7B220%7D%20%5C%5C%5C%5CI%20%3D%200.4545%20%5C%3A%20A%20%5C%5C%5C%5C)
Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by
![V = IR](https://tex.z-dn.net/?f=V%20%3D%20IR)
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,
![R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7BV%7D%7BI%7D%20%5C%5C%5C%5CR%20%3D%20%5Cfrac%7B220%7D%7B0.4545%7D%20%5C%5C%5C%5CR%20%3D%20484%20%5C%3A%20%5COmega)
Therefore, the resistance of the bulb is 484 Ω