Help me please..

Two forces have magnitudes in the ratio

rewona [7]

Answer:

F1=26N and F2=09N ..this is from the two simultaneously equations

8 0

2 years ago

Why does the light from the touch reach upto only certain distance

ad-work [718]

It is also likely (but not certain) that the photons will be absorbed by atoms. ... Light particles( or photons) never”run out” or loose their energy, so they can go an infinite distance, or until it reaches an object, that reflects the light or obsorbs it. Ie, a planet, or a mirror.

6 0

3 years ago

A heavy ball with a weight of 100 NN is hung from the ceiling of a lecture hall on a 4.4-mm-long rope. The ball is pulled to one

allochka39001 [22]

Answer:

175.3 N

Explanation:

The motion of the ball is a uniform circular motion, therefore the net force on it must be equal to the centripetal force.

There are two forces acting on the ball at the lowest point of motion:

- The tension in the string, T , upward

- The weight of the ball, $mg$ , downward

The net force (centripetal force) has the same direction as the tension (upward, towards the centre of the circular path), so we can write:

$T-mg=m\frac{v^2}{r}$

where the term on the right is the expression for the centripetal force, and where:

T is the tension in the string

$mg=100 N$ is the weight of the ball

$m=\frac{mg}{g}=\frac{100}{9.8}=10.2 kg$ is the mass of the ball

v = 5.7 m/s is the speed of the ball at the lowest point

r = 4.4 m is the length of the rope, so the radius of the circle

Solving for T, we find the tension in the string:

$T=mg+m\frac{v^2}{r}=(100)+(10.2)\frac{5.7^2}{4.4}=175.3 N$

7 0

3 years ago

An LC circuit consists of a 3.4-µF capacitor and a coil with a self-inductance 0.080 H and no appreciable resistance. At t = 0 t

alexira [117]

Answer

given,

capacitance = C = 3.4-µF

inductance = L = 0.08 H

frequency is expressed as

$f = \dfrac{1}{2\pi\sqrt{LC}}$

time period

$T = \dfrac{1}{f}=2\pi\sqrt{LC}$

after time T/4 current reach maximum

$t = \dfrac{T}{4}$

$t = \dfrac{2\pi\sqrt{LC}}{4}$

$t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}$

t = 8.2 x 10⁻⁴ s

t = 0.82 ms

b) using law of conservation

$\dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2$

$I^2 = \dfrac{CV^2}{L}$

$I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}$

$I =\sqrt{\dfrac{Q^2}{CL}}$

$I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}$

I = 0.010 A

I = 10 mA

4 0

3 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

gayaneshka [121]

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law, PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

$V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}$

Given;

$\frac{dP}{dt}$ (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) = 600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( $\frac{dv}{dt}$ );

(600 x 40) + (150 x $\frac{dv}{dt}$ ) = 0

$\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min$

Therefore, the volume is decreasing at 160 cm³/min

3 0

3 years ago