Explain how the distance to a lightning bolt (Fig. CQ17.5) can be determined by counting the seconds between the flash and the s
1 answer:
The distance to a lightning bolt can be determined by counting the seconds between the flash and the sound of thunder due to the speed of light is greater than the speed of sound
We must consider that the speed of light (3.00× 10⁸ m/s) is greater than the speed of sound (340 m/s), this is why we can first see the lightning and after seeing it we can hear it.
In order to calculate the distance we must count the time since the lightning bolt and use the speed of sound, applying the following formula:
x = v * t
Where:
- x = distance
- t = time
- v = velocity
It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).
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Answer:
skid distance = 180 m
Explanation:
given,
mass of the car = 1000 Kg
speed of the car = 40 Km/hr
distance to stop = 20 m
if speed is now = 120 Km/he
distance to stop = ?
from given question we can state that,
Work done by the friction is equal to the change in kinetic energy of the car.
............(1)
And we also know that work is directly proportional to displacement
W = F × x...........................(2)
from the equation (1) and (2)
x ∝ v²
now,
here velocity is increased from 40 km/h to 120 km/h means velocity is increase 3 times
so displacement will be equal to 3² or 9 times
hence, skid distance will be equal to 9 x 20 = 180 m
skid distance = 180 m
Answer:
The speed will be "1.06 m/s".
Explanation:
The given values are:
Momentum,
m1 = 244 g
m2 = 45.2 g
On applying momentum conservation ,
Let v2 become the final golf's speed.
From Momentum Conservation
⇒
⇒
On putting the estimated values, we get
⇒
⇒
⇒
⇒
⇒
(a) 4.06 cm
In a simple harmonic motion, the displacement is written as
(1)
where
A is the amplitude
is the angular frequency
is the phase
t is the time
The displacement of the piston in the problem is given by
(2)
By putting t=0 in the formula, we find the position of the piston at t=0:
(b) -14.69 cm/s
In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:
(3)
Differentiating eq.(2), we find
And substituting t=0, we find the velocity at time t=0:
(c) -101.13 cm/s^2
In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:
Differentiating eq.(3), we find
And substituting t=0, we find the acceleration at time t=0:
(d) 5.00 cm, 1.26 s
By comparing eq.(1) and (2), we notice immediately that the amplitude is
A = 5.00 cm
For the period, we have to start from the relationship between angular frequency and period T:
Using and solving for T, we find