A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w
2 answers:
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Answer:
7500 m
Explanation:
The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.
We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so
d = 2L
In a time of
Electromagnetic waves travel in a vacuum at the speed of light, which is equal to
Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:
(1)
where
, where L is the distance between the radar and the object
Re-arranging eq(1) and substituting, we find L:
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
If an object is projected with vertical speed given as
now the time of flight of the object that time in which it comes back on ground
now here we will have
now the range of projectile is given as
now here we know that
now the range is given as
now in order to have maximum range we can say
so we will have
so now we can say
so both speed must be same to have maximum horizontal range