Hi there!
We can use the following (derived) equation to solve for the final velocity given height:
vf = √2gh
We can rearrange to solve for height:
vf² = 2gh
vf²/2g = h
Plug in the given values (g = 9.81 m/s²)
(13)²/2(9.81) = 8.614 m
We can calculate time using the equation:
vf = vi + at, where:
vi = initial velocity (since dropped from rest, = 0 m/s)
a = acceleration (in this instance, due to gravity)
Plug in values:
13 = at
13/a = t
13/9.81 = 1.325 sec
Answer:
Time = t = 6.62 s
Explanation:
Given data:
Height = h = 215 m
Initial velocity = 
= 0 m/s
gravitational acceleration = g = 9.8 m/s²
Time = t = ?
According to second equation of motion
As initial velocity is zero, So the first term of right hand side of above equation equal to zero.
t² = 
t =
t = 
t = 6.62 s
Answer:
Explanation:
Given that, current generated from lightning range from
10⁴ A < I < 10^5 A
We know that,
The magnetic force is given as
F = iLB
The magnetic field on the earth surface is
B = 10^-5 T
So, let assume the worst case of a 15m flag pole
L = 15m
Then,
F = iLB
F = 10^5 × 10 × 10^-5
F = 15 N
Therefore, 15N is fairly strong so it will come to the material that was use for the material of the flag pole.
Therefore, it is possible that the student is right depending on the material of the flag pole.
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
