Answer:
The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.
Explanation:
Let u is the initial speed of the launch. Using first equation of motion as :
a=-g
The velocity of the shell at launch and 5.4 s after the launch is given by :
So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.
Answer:
The acceleration is 14.28 km/h^2
Explanation:
Step one:
Given data
initial speed u= 0 km/h
final speed v= 140km/h
time t= 9.8 seconds
Required
The acceleration of the car
Step two:
From a= v-u/t
substitute
a= 140-0/9.8
a=140/9.8
a=14.28 km/h^2
<u>Answer:</u>
The acceleration of the car is 
<u>Explanation:</u>
In the question it is given that car initially heads north with a velocity 
. It then accelerates for 
and in the end its velocity is 
.
initial velocity 
time 
final velocity 
The equation of acceleration is
The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.
A. scatter plot (100% positive, btw quizzes is a good tool for answers)
Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5
Explanation: In order to solve this problem we have to use the gaussian law in the mentioned regions.
Region 1; 0<r<2
∫E.ds=Qinside the gaussian surface/ε0
inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.
Region 2; 2<r<4;
E.4*π*r^2=8,84/ε0
E=8,84/(4*π*ε0*r^2)
Region 3; 4<r<5
E=0 because is inside the conductor.
Finally
Region 4; r>5
E.4*π*r^2=(8,84-2.02)/ε0
