A block with a weight of 20 N sits on a table. How much force does the table exert
1 answer:
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The spring constant is
Explanation:
For an object in a simple harmonic motion, the acceleration of the object is related to the displacement by
where
a is the acceleration
is the angular frequency
x is the displacement
The angular frequency is defined as
where
k is the spring constant
m is the mass
Substituting the second equation into the first one, we get
In this problem we have
m = 1 g = 0.001 kg
And at t=0,
x = 43.75 cm
a = -1.754 cm/s
Therefore, we can re-arrange the equation above to find the spring constant:
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Answer:
v₁ = 4 [m/s].
Explanation:
This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.
where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
where:
m₁ = mass of the tank = 500 [kg]
v₁ = velocity of the tank after firing the missile [m/s]
m₂ = mass of the missile = 20 [kg]
v₂ = velocity of the missile after firing = 100 [m/s]
Answer:
E_total = 5.8 10⁴ N /C
Explanation:
In this problem they ask to find the electric field at two points, the electric field is a vector magnitude, so we can find the field for each charged shoah and add them vectorally at the point of interest.
To find the electric field of a charged conductive sheet, we can use the Gauss law,
Ф = E. d S = / ε₀
Let us use as a Gaussian surface a small cylinder, with the base parallel to the sheet, the electric field between the sheet and the normal one next to the cylinder has 90º, so its scalar product is zero, the electric field between the sheet and the base has An Angle of 0º, therefore the scalar product is reduced to the algebraic product.
Let's look for the electric field for plate 1
The total flow is the same for each face, as there are two sides of the cylinder
2E A = q_{int} /ε₀
For the internal load we use the concept of surface density
σ = q_{int1} / A
q_{int1} = σ₁ A
Let's replace
2E A = σ₁ A /ε₀
E₁ = σ₁ / 2ε₀
For the other plate we have a field with a similar expression, but of negative sign
E₂ = -σ₂ / 2ε₀
The total field is,
E_total = σ₁ / 2ε₀ + σ₂ / 2ε₀
E_total = (σ₁ + σ₂) / 2ε₀
Let us apply this expression to our case, when placing a sheet without electric charge, a charge is induced for each sheet, the plate 1 that has a positive charge the electric field is protruding to the right and the plate 2 that has a negative charge creates a incoming field, to the right, as the two fields have the same address add
The conductive sheet in the middle pate undergoes an induced load that is created by the other two plates, but because the conductive plate the charges are mobile and are replaced.
E_total = (0.51 +0.52) 10⁻⁶ / 2 8.85 10⁻¹²
E_total = 5.8 10⁴ N /C
Note that the field is independent of the distance between the plates