Answer:
<em>Second option</em>
Explanation:
<u>Linear Momentum</u>
The linear momentum of an object of mass m and speed v is
P=mv
If two or more objects are interacting in the same axis, the total momentum is
Where the speeds must be signed according to a fixed reference
The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right
The second cart of mass m goes to the right at a speed v
The total momentum before the impact is
The total momentum after the collision is negative, both carts will join and go to the left side
The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven
The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either
The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".
The second option is the correct one because the mass 
has a negative momentum and then the sum of both masses keeps being negative
Answer:
E = (-3.61^i+1.02^j) N/C
magnitude E = 3.75N/C
Explanation:
In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:
![\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%3D-k%5Cfrac%7Bq%7D%7Br%5E2%7Dcos%5Ctheta%5C%20%5Chat%7Bi%7D%2Bk%5Cfrac%7Bq%7D%7Br%5E2%7Dsin%5Ctheta%5C%20%5Chat%7Bj%7D%5C%5C%5C%5C%5Cvec%7BE%7D%3Dk%5Cfrac%7Bq%5E2%7D%7Br%7D%5B-cos%5Ctheta%5C%20%5Chat%7Bi%7D%2Bsin%5Ctheta%5C%20%5Chat%7Bj%7D%5D)
(1)
Where the minus sign means that the electric field point to the charge.
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q = -4.28 pC = -4.28*10^-12C
r: distance to the charge from the point P
The point P is at the point (0,9.83mm)
θ: angle between the electric field vector and the x-axis
The angle is calculated as follow:
The distance r is:
You replace the values of all parameters in the equation (1):
![\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5Cfrac%7B4.28%2A10%5E%7B-12%7DC%7D%7B%2810.21%2A10%5E%7B-3%7Dm%29%7D%5B-cos%2815.84%5C%C2%B0%29%5Chat%7Bi%7D%2Bsin%2815.84%5C%C2%B0%29%5Chat%7Bj%7D%5D%5C%5C%5C%5C%5Cvec%7BE%7D%3D%28-3.61%5Chat%7Bi%7D%2B1.02%5Chat%7Bj%7D%29%5Cfrac%7BN%7D%7BC%7D%5C%5C%5C%5C%7C%5Cvec%7BE%7D%7C%3D%5Csqrt%7B%283.61%29%5E2%2B%281.02%29%5E2%7D%5Cfrac%7BN%7D%7BC%7D%3D3.75%5Cfrac%7BN%7D%7BC%7D)
The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C
Explanation:
The difference between a seismograph and a Seismogram is that:
a Seismograph is referred to a tool or instrument used in measuring the magnitude or effect of an earth quake
While
Seismogram is a type of output report or data that is generated out of a seismograph.
So in summary, a Seismograph is an instrument or tool while Seismogram is an output data generated from the usage of a seismograph.
Answer:
15 m/s to the right
Explanation:
Let's say right is positive and left is negative.
Momentum is conserved:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(6.0 kg) (25.0 m/s) + (15 kg) (0 m/s) = (6.0 kg) (-12.5 m/s) + (15 kg) v₂
v₂ = 15 m/s
Answer: The time of contact will be 0.006 seconds.
Explanation:
An impulse of a force is the product of average force and the time interval when the force acts.
Mathematically,
Where,
J = impulse = 3.6 Ns
F = Force = 600N
= Time = ?s
Putting the values in above equation, we get:
t = 0.006 seconds
Hence, the time of contact will be 0.006 seconds.
