Sedimentary rocks are the only rocks that can potentially containa.fossils.c.fractures.b.minerals.d.faults. Please select the be
2 answers:
You might be interested in
Answer:
A=50mΩ
B≅50mΩ
Explanation:
A) To answer this question we have to use the Current Divider Rule. that rule says:
(1)
Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.
We now have a set of two resistor in parallel, so:
(2)
where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).
substituting and rearranging (2)
(3)
Now substituting (3) in (1).
solving this, The value of R1 is: 50mΩ
This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.
Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ
If you substitude this values on (2) Req. will be 49.97 mΩ.
Answer:
v’= 279.66 m / s
Explanation:
We work this exercise using the conservation of the moment. For this we define the system formed by the two blocks, therefore the forces during the collision are internal of the action and reaction type.
Initial instant. Before the crash
p₀ = m v₀ + 0
Final moment. After the crash
p_f = m v + M v ’
how the tidal wave is preserved
p₀ = p_f
m v₀ = m v + M v ’
v =
let's calculate
v ’=
v ’=
v ’= 279.66 m / s
Answer:
4.86 seconds
Explanation:
Velocity of projection, u = 14 m/s
angle of projection, θ = 20°
Formula for the time of flight
For earth
Te = (2 x 14 x Sin 20) / 9.8
Te = 0.98 s
For moon
g' = g/6 = 1.64 m/s^2
Tm = ( 2 x 14 x Sin 20) / 1.64
Tm = 5.84 seconds
Tm - Te = 5.84 - 0.98 = 4.86 s
So, it takes 4.86 s more time of flight on moon than the earth.