Answer:
The magnitude of the horizontal net force is 13244 N.
Explanation:
Given that,
Mass of car = 1400 kg
Speed = 17.7 m/s
Distance = 33.1 m
We need to calculate the acceleration
Using equation of motion
Where, u = initial velocity
v = final velocity
s = distance
Put the value in the equation
Negative sign shows the deceleration.
We need to calculate the net force
Using newton's formula
Negative sign shows the force is opposite the direction of the motion.
The magnitude of the force is
Hence, The magnitude of the horizontal net force is 13244 N.
Answer:
to make calculation more easy to get
Explanation:
if you are using chart or calculate Thermodynamic problems you will not never solve this problem with out using data table for thermodynamic
Answer:
<h2>12 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
f is the force
m is the mass
From the question
f = 6000 N
m = 500 kg
We have
We have the final answer as
<h3>12 m/s²</h3>
Hope this helps you
His role as a field research is that of a: Complete participant!
(Option D.)
~Good luck!
Answer:
Somewhere between the two wires, but closer to the wire carrying λ₂
Explanation:
Electric Field for a point at distance x from an electric charge Q is Ef = K*Q/x².
Electric Fied due to an electric charge is a vector and its direction is such that if we place a positive charge in the point it will be rejected ( equal sign charge repulse each other and different attract each other)
According to that previous explanation, it is no possible two have Ef=0 out of the two wires region, since above the upper wire and below the lower wire we have to add the two electric fields (both have the same direction). Therefore we only have possibilities of Ef = 0 inside the two wires, where the repulsion produced over a positive charge due to the two wires are opposite
In the particular case in which λ₁ and λ₂ are equals then all the points exactly in the middle of d (distance between the two wires ) will have Ef =0.
As we can see at the beginning of the step by step explanation Electric field is proportional to the electric charge, or for a bigger charge, bigger Ef (keeping constant distance). In our case λ₁ >λ₂ then E₁ (Electric field produced by a wire carrying λ₁ will be bigger than (Electric field produced by wire carrying λ₂ at the middle way between the wires.
But for points closer to wire with λ₂ ( where E₂ is bigger than E₁ ) we will surely find an appropriate distance to get equals E and then Ef = 0
