What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.
2 answers:
Answer:
<em>The distance between the compact car and pickup truck is 0.96048 m</em>
Explanation:
The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them. This is shown in equation 1;
............ 1
Where F is the gravitational force = 0.0001 N
G is the gravitational constant = 6.673 x
is the mass of the compact car = 900kg
is the mass of the pickup truck = 1600kg
d is the distance and its unknown ?
Let us make d the subject formula in equation 1
.... 2
Substituting into equation 2 we have
d = 0.96048m
Therefore the distance between the compact car and pickup truck is 0.96048 m
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required:
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
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b) C= 17.7 nF
c) Q = 159. 3 nC
Explanation:
a)
- By definition, the electric field is the electrostatic force per unit charge, and since the potential difference between plates is just the work done by the field, divided by the charge, assuming a uniform electric field, if V is the potential difference between plates, and d is the separation between plates, the electric field can be expressed as follows:
b)
- For a parallel-plate capacitor, applying the definition of capacitance as the quotient between the charge on one of the plates and the potential difference between them, and assuming a uniform surface charge density σ, we get:
From (1), we know that V = E*d, but at the same time, applying Gauss'
Law at a closed surface half within the plate, half outside it , it can be
showed than E= σ/ε₀, so finally we get:
c)
- From (3) we can solve for Q as follows:
Answer:
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Explanation:
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