A 1500-kg car drives at 30 m/s around a flat circular track 300 m in diameter. What are the magnitude and direction of the net f
1 answer:
Answer:
F=9000N
Direction is towards the center
Explanation:
Car is moving in a circular path so the force acting on the car is the centrifugal force which is always towards the center. In order to find that force we will use the following formula:
Where:
F is the centrifugal force
m is the mass of object
v is the velocity of object
r is the radius of circular path
Since diameter is given then r=diameter/2
r=300/2
r=150m
F=9000N
Direction is towards the center
So the magnitude of the force is 9000N and the direction is towards the center of circular track as car is moving in circular path.
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Answer:
Explanation:
Conceptual analysis
The electric field at a point P due to a point charge is calculated as follows:
E = k*q/r²
E: Electric field in N/C
q: charge in Newtons (N)
k: electric constant in N*m²/C²
r: distance from charge q to point P in meters (m)
The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Graphic attached
The attached graph shows the field due to the charges:
Ep₁: Total field at point P due to charge q₁. As the charge is positive ,the field leaves the charge.
Ep₂: Total field at point P due to charge q₂. As the charge is negative, the field enters the charge.
Known data
q₁ = 63 nC = 63×10⁻⁹ C
q₂ = -47 nC = -47×10⁻⁹ C
k = 8.99*10⁹ N×m²/C²
d₁ = 1.4cm = 1.4×10⁻² m
d₂ = 3.4cm = 3.4×10⁻² m
Calculation of r and β
Problem development
Ep: Total field at point P due to charges q₁ and q₂.
Ep₁ₓ = 0
Calculation of the electric field components at point P
