The key results indicate that lightning contains electricity.
Answer:
none of the answers is correct, the time is the same t₁ = t₂ = 0.600 s
Explanation:
This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)
let's find the time to hit the ground
y = y₀ + I go t - ½ g t²
0 = y₀ - ½ g t²
t = √ 2y₀ / g
with the data from the first launch
y₀i = ½ g t²
y₀ = ½ 9.8 0.6²
y₀ = 1,764 m
with this is the same height the time to descend in the second case is the same
t₂ = 0.600 s
this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis
Therefore, none of the answers is correct, the time is the same
t₁ = t₂ = 0.600 s
Answer:
The frequency of oscillations is 7 Hz
Explanation:
Given;
mass of car, = 1700 kg
mass of driver, = 66 kg
compression of the spring, x = 5mm = 0.005 m
The frequency of the oscillation is given as;
![F = \frac{1}{2 \pi } \sqrt{\frac{k}{m} }](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B1%7D%7B2%20%5Cpi%20%7D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D)
where;
k is force constant
m is the total mass of the car and the driver
m = 1700 kg + 66 kg = 1766 kg
Weight of the car and the driver;
W = mg
W = 1766 x 9.8
W = 17306.8 N
Apply hook's law, to determine the force constant;
F = kx
W = F
Thus, k = W/x
k = 17306.8 / 0.005
k = 3461360 N/m
Now, calculate the frequency
![F = \frac{1}{2\pi}\sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi}\sqrt{\frac{3461360}{1766} }\\\\F = 7 \ Hz](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D%20%20%5C%5C%5C%5CF%20%3D%20%20%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cfrac%7B3461360%7D%7B1766%7D%20%7D%5C%5C%5C%5CF%20%3D%207%20%5C%20Hz)
Therefore, the frequency of oscillations is 7 Hz
Well, first of all, you really shouldn't use ' W ' for the unit when you
talk about resistors.
You may have seen the resistors written as 6ω, 12ω, and 2ω in your
book or on the homework sheet. But that little symbol ' ω ' is not a ' w '.
It's the small Greek letter 'omega'. The CAPITAL omega is ' Ω '. It's used
to label resistors because it's short for "ohms". So the resistors in this
problem have resistances of 6Ω, 12Ω, and 2Ω, and we have to do some
manipulating of the individual resistors to find out what resistance the
battery actually sees.
The parallel combination of the first two resistors looks like a single
resistor, whose value is
1 / (1/6 + 1/12)
= 1 / (2/12 + 1/12)
= 1 / (3/12)
= 12/3 = 4Ω .
Now, that parallel combination is connected in series with 2Ω .
All three resistors together look like a single resistor of
4Ω + 2Ω = 6Ω .
So the battery thinks there's a single resistor connected to it,
with 6Ω of resistance. The current out of the battery is
I = V / R = (24v) / (6Ω) = 4 Amperes.
That 4 Amperes of current will split between the parallel resistors,
but it will ALL flow through the series 2Ω resistor because there's
no other path through that part of the circuit.
So the current through the 2Ω resistor is 4 Amperes. (B).
Note:
The POWER dissipated by the 2Ω resistor is
P = I² R = (4A)² · (2Ω) = 32 watts .
This is a fair amount of heat, so you'll need to provide some way
to remove the heat from the resistor, otherwise it'll burn or crack.
Answer:
Maximum force will be equal to 720 N
Explanation:
We have given that spring constant ![k=1.2\times 10^4N/m](https://tex.z-dn.net/?f=k%3D1.2%5Ctimes%2010%5E4N%2Fm)
Maximum stretch of the spring x = 6 cm = 0.06 m
We have to find the maximum force on the spring
We know that spring force is given by
![F=kx=1.2\times 10^4\times 0.06=720N](https://tex.z-dn.net/?f=F%3Dkx%3D1.2%5Ctimes%2010%5E4%5Ctimes%200.06%3D720N)
So the maximum force which is necessary to relaxed the spring will be eqaul to 720 N