Answer:
Δt = 5.29 x 10⁻⁴ s = 0.529 ms
Explanation:
The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.
where,
Δt = required time interval = ?
Δs = distance between ears = 18 cm = 0.18 m
v = speed of sound = 340 m/s
Therefore,
<u>Δt = 5.29 x 10⁻⁴ s = 0.529 ms</u>
Answer: increases
Explanation: As speed and kinetic energy increase, the gravitational potential energy decreases.
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
Answer: Yes
Explanation: The Sun like all other stars will eventually go through nuclear fusion. But for a very long time (about 5 billion years) When it does it is more than likely that it will consume the earth as the sun will expand very larger during the process and become a red giant.
B)
To calculate the values of for acceleration, it is helpful to remember what acceleration actually is.
Acceleration is the change in speed over time, with units
. The y-axis of this graph gives speed, and the x-axis gives time, so we find the point where the car is gaining speed, and we find the rate at which it gained speed.
i) We do this by calculating the slope. We obtain the points where the car starts and stops accelerating,
, and find the average rate of change (slope) between these two points.
ii) We do the same for the deceleration. We choose
.
c) We can most easily do this by calculating the distance for each segment of the graph.
For the first 20 seconds, the car moves at an average speed of 20 m/s.
For the middle 20 seconds, the car moves at 40 m/s.
For the final 10 seconds, the car moves at an average speed of 20 m/s.
You can also do this by finding the area under the graph. You don't need Calculus, because this isn't a curve!