Answer: The volume of an irregularly shaped object is 0.50 ml
Explanation:
To calculate the volume, we use the equation:
Density of object =
mass of object = 3.0 g
Volume of object = ?
Putting in the values we get:
Thus the volume of an irregularly shaped object is 0.50 ml
Complete Question
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.
How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?
For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.
Give your answer in centimeters, to two significant figures.
Answer:
The radius of the dish is
Explanation:
From the question we are told that
The radius of the orbit is =
The power output of the power is
The electric vector amplitude is given as
The area of thereciever is
Generally the intensity of the dish is mathematically represented as
Where A is the area orbit which is a sphere so this is obtained as
Then substituting into the equation for intensity
Now the intensity received by the dish can be mathematically evaluated as
Where c is thesped of light with a constant value
is the permitivity of free space with a value
is the electric filed on the dish
So since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish
Now making the eletric field intensity the subject of the formula
substituting values
The incident power on the dish is what is been reflected to the receiver
Where is the power incident on the dish which is mathematically represented as
And is the power incident on the dish which is mathematically represented as
Now equating the two
Making R the subject we have
Substituting values
Answer:
4 (please see the attached file)
Explanation:
While the angular speed (counterclockwise) remained constant, the angular acceleration was just zero.
So, the only force acting on the bug (parallel to the surface) was the centripetal force, producing a centripetal acceleration directed towards the center of the disk.
When the turntable started to spin faster and faster, this caused a change in the angular speed, represented by the appearance of an angular acceleration α.
This acceleration is related with the tangential acceleration, by this expression:
at = α*r
This acceleration, tangent to the disk (aiming in the same direction of the movement, which is counterclockwise, as showed in the pictures) adds vectorially with the centripetal force, giving a resultant like the one showed in the sketch Nº 4.
Answer:
E ≈ 1.70 10⁵ N/C
Explanation:
The electric field is a vector quantity, so we can calculate the field of each plate and then add them. To calculate the field of a plate let's use Gauss's law
Φ = ∫ E. dA = / ε₀
To apply this law we must create a Gaussian surface that takes advantage of the symmetry of the problem. The electric field lines on the surface are perpendicular, so the Gaussian surface that will be a cylinder with the base parallel to the plate.
On this surface the normal to the base (A) is parallel to the field lines whereby the scalar product is reduced to the ordinary product. The normal on the sides of the cylinder is perpendicular to the field, therefore, the product scale is zero.
∫I E dA = /ε₀
Let's look for the load under the cylinder, let's use the concept of load density
σ = / A
= σ A
Let's write Gauss's law for this case
E A = /ε₀
E A = σ A / ε₀
E = σ / ε₀
As the field is emitted for each side of the plate the value to only one side is
E = G / 2ε₀
This expression is the same for each plate, now let's add the electric field at the requested point
R = (0.50, 0.00, 0.00) cm
We see that this point is on the X axis, between the plates that are at the points y = -1.0 cm and y = 1.0 cm, as the plates are very large the test point is between them
The negative plate has an incoming field and the positive plate has an outgoing field, the test load is always positive. The field due to the negative plate goes to the left, the field through the positive plate goes to the left at this point whereby two are added
E = E_ + E +
E = σ1 / 2ε₀ + σ2 / 2ε₀
E = 1 / 2o (σ1 + σ2)
Let's calculate the value
E = 1/2 8.85 10⁻¹² (1.00 10⁻⁶ + 2.00 10⁻⁶)
E = 3 10⁻⁶ / 17.7 10⁻¹²
E = 1,695 10⁵ N / C
E ≈ 1.70 10⁵ N/C