Answer:
Incomplete question: "Each block has a mass of 0.2 kg"
The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s
Explanation:
Given data:
θ = angle of the surface = 37°
m = mass of each block = 0.2 kg
v = speed = 0.35 m/s
t = time to collision = 0.5 s
Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?
Change in momentum:
![delta(P)=F*delta(t)](https://tex.z-dn.net/?f=delta%28P%29%3DF%2Adelta%28t%29)
![P_{f} -P_{i}=F*delta(t)](https://tex.z-dn.net/?f=P_%7Bf%7D%20-P_%7Bi%7D%3DF%2Adelta%28t%29)
![2m(v_{f} -v_{i})=F*delta(t)](https://tex.z-dn.net/?f=2m%28v_%7Bf%7D%20-v_%7Bi%7D%29%3DF%2Adelta%28t%29)
![v_{i} =0.35-0.35=0](https://tex.z-dn.net/?f=v_%7Bi%7D%20%3D0.35-0.35%3D0)
It is neccesary calculate the force:
![F=(m+m)*g*sin\theta](https://tex.z-dn.net/?f=F%3D%28m%2Bm%29%2Ag%2Asin%5Ctheta)
Here, g = gravity = 9.8 m/s²
![F=(0.2+0.2)*9.8*sin37=2.3591N](https://tex.z-dn.net/?f=F%3D%280.2%2B0.2%29%2A9.8%2Asin37%3D2.3591N)
![v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3D%5Cfrac%7BF%2Adelta%28t%29%7D%7B2m%7D%20%3D%5Cfrac%7B2.3591%2A0.5%7D%7B2%2A0.2%7D%20%3D2.9489m%2Fs)
Answer:
approximately 1.625 m/s , about 16.6% that on Earth's surface or 0.166 ɡ. Over the entire surface, the variation in gravitational acceleration is about 0.0253 m/s.
The initial velocity of the car is 11 m/s.
Explanation:
- Here, we can use the equations of motions.
- The first of equation of motion is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
- The given parameters are,
Final velocity, v = 20 m/s
acceleration of the care, a = 2 m/![s^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D)
time, t = 3 s
Initial velocity, u = ?
- Hence, the equation can be written as u = v - at
once you substitute all these vales, you will get u = 11 m/s
Answer:
The magnetic field strength is 2.11 T.
Explanation:
It is given that,
Length of the wire, L = 37 cm = 0.37 m
Current, I = 3.2 A
Force, F = 2.5 N
Magnetic force is given by :
![F=qvB\ sin\theta](https://tex.z-dn.net/?f=F%3DqvB%5C%20sin%5Ctheta)
Here, ![\theta=90,sin\ \theta=1](https://tex.z-dn.net/?f=%5Ctheta%3D90%2Csin%5C%20%5Ctheta%3D1)
![F=qvB](https://tex.z-dn.net/?f=F%3DqvB)
We know that,
Velocity, ![v=\dfrac{L}{t}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7BL%7D%7Bt%7D)
![F=\dfrac{qLB}{t}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7BqLB%7D%7Bt%7D)
Since, ![\dfrac{q}{t}=I](https://tex.z-dn.net/?f=%5Cdfrac%7Bq%7D%7Bt%7D%3DI)
![F=ILB](https://tex.z-dn.net/?f=F%3DILB)
![B=\dfrac{F}{IL}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7BF%7D%7BIL%7D)
![B=\dfrac{2.5\ N}{3.2\ A\times 0.37\ m}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B2.5%5C%20N%7D%7B3.2%5C%20A%5Ctimes%200.37%5C%20m%7D)
B = 2.11 T
So, the magnetic field strength is 2.11 T. Hence, this is the required solution.