Answer:
<h2>10,000 J</h2>
Explanation:
The kinetic energy of an object can be found by using the formula

m is the mass
v is the velocity
From the question we have

We have the final answer as
<h3>10,000 J</h3>
Hope this helps you
Answer:
I think, (remember think) it might be 2.0 m/s
Explanation:
If it's wrong I'm truly sorry.
Answer:
i. 0.34
ii. 0.4
iii. 1700 w/m²
iv. 2211.36 w/m²
Explanation:
Given that
Irradiation of the plate, G = 2500 w/m²
Reflected rays, p = 500 w/m²
Emissive power, E = 1200 w/m²
See attachment for calculations
If the box is moving at constant velocity, net force must be zero, so:
F + fr = 0
fr = -F
<u>fr = -40 N</u>
Answer:
each resistor is 540 Ω
Explanation:
Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance
defined by the formula:

Therefore, R/3 is the equivalent resistance of the initial circuit.
In the second circuit, two of the resistors are in parallel, so they are equivalent to:

and when this is combined with the third resistor in series, the equivalent resistance (
) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

The problem states that the difference between the equivalent resistances in both circuits is given by:

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:
