Explanation:
Given that,
(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :
(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :
Therefore, this is the required solution.
Answer:
the claim is not valid or reasonable.
Explanation:
In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:
η(max) = 1 - T₁/T₂
where,
η(max) = maximum efficiency = ?
T₁ = Sink Temperature = 300 K
T₂ = Source Temperature = 400 K
Therefore,
η(max) = 1 - 300 K/400 K
η(max) = 0.25 = 25%
Now, we calculate the actual frequency of the engine:
η = W/Q
where,
W = Net Work = 250 KJ
Q = Heat Received = 750 KJ
Therefore,
η = 250 KJ/750 KJ
η = 0.333 = 33.3 %
η > η(max)
The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.
<u>Therefore, the claim is not valid or reasonable.</u>
Answer:
v = 1.32 10² m
Explanation:
In this case we are going to use the universal gravitation equation and Newton's second law
F = G m M / r²
F = m a
In this case the acceleration is centripetal
a = v² / r
The force is given by the gravitational force
G m M / r² = m v² / r
G M/r = v²
Let's calculate the mass of the planet
M = v² r / G
M = (1.75 10⁴)² 5.00 10⁶ / 6.67 10⁻¹¹
M = 2.30 10²¹ kg
With this die we clear the equation to find the orbit of the second satellite
v = √ G M / r
v = √ (6.67 10⁻¹¹ 2.30 10²¹ / 8.75 10⁶)
v = 1.32 10² m
Answer:
20 meters.
Explanation:
In the graph, the x-axis (the horizontal axis) represents the time, while the y-axis (the vertical axis) represents the distance.
If we want to find the distance covered in the first T seconds, you need to find the value T in the horizontal axis.
Once you find it, we draw a vertical line, in the point where this vertical line touches the graph, we now draw a horizontal line. This horizontal line will intersect the y-axis in a given value. That value is the total distance travelled by the time T.
In this case, we want to find the total distance that David ran in the first 4 seconds.
Then we need to find the value 4 seconds in the horizontal axis. Now we perform the above steps, and we will find that the correspondent y-value is 20.
This means that in the first 4 seconds, David ran a distance of 20 meters.
