Here's a quick way to find out. Pick up your glasses, bifocals work best, and find the focal length with a flashlight against a book. If I remember right, the object should be magnified and upside down. So, A.
Answer:
t = 1.4[s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.
![P=m*v\\or\\P=F*t](https://tex.z-dn.net/?f=P%3Dm%2Av%5C%5Cor%5C%5CP%3DF%2At)
where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 50 [kg]
v = velocity [m/s]
F = force = 200[N]
t = time = [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.
![(m_{1}*v_{1})-F*t=(m_{1}*v_{2})](https://tex.z-dn.net/?f=%28m_%7B1%7D%2Av_%7B1%7D%29-F%2At%3D%28m_%7B1%7D%2Av_%7B2%7D%29)
where:
m₁ = mass of the object = 50 [kg]
v₁ = velocity of the object before the impulse = 18.2 [m/s]
v₂ = velocity of the object after the impulse = 12.6 [m/s]
![(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]](https://tex.z-dn.net/?f=%2850%2A18.2%29-200%2At%3D50%2A12.6%5C%5C910-200%2At%3D630%5C%5C200%2At%3D910-630%5C%5C200%2At%3D280%5C%5Ct%3D1.4%5Bs%5D)
Answer:
404.4 m
Explanation:
Converting the initial speed from km/h to m/s then
![140\times \frac {1000m}{3600s}=38.88888889 m/s \approx 38.89 m/s](https://tex.z-dn.net/?f=140%5Ctimes%20%5Cfrac%20%7B1000m%7D%7B3600s%7D%3D38.88888889%20m%2Fs%20%5Capprox%2038.89%20m%2Fs%3C%2Fp%3E%3Cp%3E)
The acceleration is resolved as shown in the figure hence
deceleration of the truck along the inclined plane will be
where g is acceleration due to gravity
Substituting g with
then
![a=-9.81 m/s^{2} sin 11^{\circ}=-1.871836245\approx -1.87 m/s^{2}](https://tex.z-dn.net/?f=a%3D-9.81%20m%2Fs%5E%7B2%7D%20sin%2011%5E%7B%5Ccirc%7D%3D-1.871836245%5Capprox%20-1.87%20m%2Fs%5E%7B2%7D%3C%2Fp%3E%3Cp%3E)
Using kinematic equation
and making s the subject then
where v and u are final and initial velocities respectively
Substituting 0 for v, 38.89 m/s for u and
then
![s=\frac {0^{2}-38.89^{2}}{2\times -1.87}=404.3936096 m\approx 404.4 m](https://tex.z-dn.net/?f=s%3D%5Cfrac%20%7B0%5E%7B2%7D-38.89%5E%7B2%7D%7D%7B2%5Ctimes%20-1.87%7D%3D404.3936096%20m%5Capprox%20404.4%20m)
Use this equation
Acceleration = frequency x wavelength
You have the wavelength and frequency so multiply them.
3.0 x .5 = 1.5
I don't remeber if any conversions need to be made
To find out scientific notation, you want to make sure that number is less than 10. So do 5.000000, you don't rally need the zeros but I just want to make my point. So use 10^x meaning ten the whatever power adds zeros like 5.000000x10^6 meaning it is increasing it by six zeros moving it out of the decimals and letting become 5,000,000.