Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated
The upward force exerted on the board by the support is mathematically given as
Fu= 764.8 N
<h3>What is the upward force exerted on the board by the support?</h3>
Generally, the equation for is mathematically given as
Considering that the Net Force on the system is null
The weight of the children plus the weight of the board equals the upward force imposed on the support.
The upward force
Fu= 440 + 272 + 52.8 N
Fu= 764.8 N
In conclusion, he upward force exerted on the board by the support
Fu= 764.8 N
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<span>Bit level for a CCD (Charged coupled device) with a greatest possible pixel value of 4095:The relationship between the bit level and pixel value is given as:pixel value = 2^bit level.Most charged coupled devices (CCDs) have 8-bit, 16-bit, 32-bit levels.Using simple mathematics, we can see that 2^12 = 4096.Since the maximum number of pixels is 4095, the bit level is 12., i.e. the CCD has 12-bit level.</span>
here we have to calculate the net positive charge present on he surface of the conducting sphere.
as the sphere is a conducting one the,the charge will be stored on its surface.
though the charge is present at the surface,still it will behave just like the total charge is concentrated at the centre of the sphere.
the electric field at outside of the sphere is E=
as E=
so V=
here V=27 volt,radius of sphere is 0.800 m and the point at which the potential is considered is at 1.20 m from th centre of the sphere.hence the point is at the outside of the sphere .
hence we have 27=
q=27× coulomb
=3.6× coulomb [ans]
Answer:
T = 27.92 N
Explanation:
For this exercise let's use Newton's second law
T - W = m a
The weight
W = mg
The acceleration can be found by derivatives
a = dv / dt
v = 2 t + 0.6 t²
a = 2 + 0.6 t
We replace
T - mg = m (2 + 0.6t)
T = m (g + 2 + 0.6 t) (1)
Let's look for the time for the speed of 15 m / s
15 = 2 t + 0.6 t²
0.6 t² + 2 t - 15 = 0
We solve the second degree equation
t = [-2 ±√(4 - 4 0.6 (-15))] / 2 0.6
t = [-2 ±√40] / 1.3 = [-2 ± 6.325] / 1.2
We take the positive time
t = 3.6 s
Let's calculate from equation 1
T = 2.00 (9.8 + 2 + 0. 6 3.6)
T = 27.92 N