Total Current = 2 Amps
Req total = 60 ohms
Current on EF mesh = 1/3 Amp
Current on 24 ohms resistor = 1/6 Amp
Voltage on the 40ohms resistor is lefting.
Answer:
q = 224 mm, h ’= - 98 mm, real imagen
Explanation:
For this exercise let's use the constructor equation
where f is the focal length, p and q are the distance to the object and the image respectively.
In a mirror the focal length is
f = R / 2
indicate us radius of curvature is equal to the diameter of the eye
R = 3,50 10² mm
f = 3.50 10² /2 = 1.75 10² mm
they also say that the distance to the object is p = 0.800 10³ mm
1 / q = 1 / f - 1 / p
1 / q = 1 / 175 - 1 /800
1 / q = 0.004464
q = 224 mm
to calculate the size let's use the magnification ratio
m =
h '=
h ’= - 224 350 / 800
h ’= - 98 mm
in concave mirrors the image is real.
Answer:
161.86 N
Explanation:
mass of box m= 55.0 kg
weight of the box, mg= 55×9.81
g here is acceleration due to gravity =9.81 m/sec^2
coefficient of friction between the box and the surface μ= 0.3
the friction force F_s= μmg= 0.3×55×9.81
=161.86 N
to move the ball horizontal force required is 161.86 N
Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
I think its a tbh bc it seems to be the best answer out of a b c and d