Counter clockwise torque is 360Nm.
Clockwise torque is 240Nm.
40 * 9 = 360
80 * 3 = 240
No. Motion is the thing that when you're moving, you're in it.
But it IS possible for one person to say you're moving and another person to say you're not moving, both at the same time, and both of them are correct !
Answer:
450X
Explanation:
When a specimen is been viewed, both
objective and ocular lenses works together so that the object is magnified.
From the question,objective lenses are;
1)10x
2)45x
ocular lens= 10x
Highest magnification
= 10X ocular × 45X objective
=450X
This implies that the image that was viewed will appear 450 times the actual size.
Answer:
The resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.
Explanation:
We are given 10 voltage power source and we have two Resistors with resistance of 20 ohm and 5ohm.
We need to find the orientation in which these two resistors would be arranged so that the circuit could get a current of 2.5Ampere.
Using ohm's law we have
V = I*R
V= voltage
I= current
R= resistance
10 = 2.5*R
R = 10/2.5 = 4ohm
that means we need a total of 4ohm resistance from these two resistors.
since the net Resistance(4ohm) is lower than the smallest resistance(5ohm) available that means the orientation of the resistors will be in parallel.

R(net) =4ohm
Now the orientation of the resistors are in parallel so the current will be divided.
we know that the current will divide in opposite manner the arm which provides more resistance less current will flow from there and vice versa.
We know that the voltage in parallel remains same
In 20 ohm resistance
again using ohms law
V = i1*R1
10 = i1*20
i1 = 0.5A
in 5ohm resistor
V=i2*R2
10 = I2*5
i2 =2A
and i1+i2 = 0.5+2= 2.5A which means our calculation is correct.
Therefore the resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.
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).
Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.
Answer:
The electric field vector of the satellite broadcast as measured at the surface of the earth is 
Explanation:
From the question we are told that
The height of the satellite is 
The power output of the satellite is 
Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is mathematically represented as

substituting values


This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be mathematically represented as

Where
is the amplitude of the electric field vector of the satellite broadcast so

substituting values

