Answer:
The bumper will be able to move by 0.01155m.
Explanation:
The magnitude of deceleration of the car in the front end collision.
![a = \frac{F_m}{m} \\](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BF_m%7D%7Bm%7D%20%5C%5C)
![a = \frac{80000}{1500} \\](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B80000%7D%7B1500%7D%20%5C%5C)
![a = 53.33](https://tex.z-dn.net/?f=a%20%3D%2053.33)
This is the deceleration of the car that is generated to stop due to a front end collision.
4 km/h = 1.11 m/s
Now, the initial speed of the bumper in the relation of car, Vi = 0
Now, the initial speed of the bumper in the relation of car, Vf = 1.11 m/s
Use the below equation:
![s = \frac{(Intitial \ speed)^2 – (Final \ speed)^2}{2a} \\](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%28Intitial%20%5C%20speed%29%5E2%20%E2%80%93%20%28Final%20%5C%20speed%29%5E2%7D%7B2a%7D%20%5C%5C)
![s = \frac{(1.11)^2 – (0)}{2 \times 53.33} \\](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%281.11%29%5E2%20%E2%80%93%20%280%29%7D%7B2%20%5Ctimes%2053.33%7D%20%5C%5C)
![s = 0.01155 \\](https://tex.z-dn.net/?f=s%20%3D%200.01155%20%5C%5C)
Thus, the bumper can move relative to the car is 0.01155 m .
Answer:
He barely makes it but hit his leg on the pool edge.
Explanation:
Using kinematic formula: Δx=![v_{0}t+(1/2)at^2](https://tex.z-dn.net/?f=v_%7B0%7Dt%2B%281%2F2%29at%5E2)
We can plug in delta x, v zero, and acceleration.
delta x = 8 meters, v0 = 0. a = 9.8
8=(1/2)9.8*t^2
Around 1.278 seconds of airtime.
4 * 1.278 = 5.112 meters horizontally assuming no air resistance
That was close!
Answer:
The original volume of the first bar is half of the original volume of the second bar.
Explanation:
The coefficient of cubic expansivity of substances is given by;
γ = ΔV ÷ (
Δθ)
Given: two metal bars with equal change in volume, equal change in temperature.
Let the volume of the first metal bar be represented by
, and that of the second by
.
Since they have equal change in volume,
Δ
= Δ
= ΔV
For the first metal bar,
2γ = ΔV ÷ (
Δθ)
⇒ Δθ = ΔV ÷ (2γ
)
For the second metal bar,
γ = ΔV ÷ (
Δθ)
⇒ Δθ = ΔV ÷ (
γ)
Since they have equal change in temperature,
Δθ of first bar = Δθ of the second bar
ΔV ÷ (2γ
) = ΔV ÷ (
γ)
So that;
(1 ÷ 2
) = (1 ÷
)
2
= ![V_{2}](https://tex.z-dn.net/?f=V_%7B2%7D)
= ![\frac{V_{2} }{2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B2%7D%20%7D%7B2%7D)
Thus, original volume of the first bar is half of the original volume of the second bar.
Answer:
20 N up
Explanation:
Since the object on the floor is sitting still. Based on Newton's law of force. Objects must have equal reaction and force against each other for still objects. So the answer is 20 N up. Since the force are equal on both sides making the object not moving.
A pressure antinode in a sound wave is not a region of high pressure, while a pressure node is not a region of low pressure.
The answer is false