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ipn [44]
3 years ago
6

Can anyone help me with any parts in question number 1: parallel circuit?

Physics
1 answer:
Leno4ka [110]3 years ago
4 0
Question #1:

a).  The sketch is attached to this answer.

b).  The equivalent resistance of 30Ω and 50Ω in parallel is

1 / (1/30 + 1/50)  =  18.75 Ω

c).   I = V/R = (100/30) = (3 and 1/3) Amperes

d).  Follow the wires, and you see that the 50Ω resistor is
connected directly to the battery, and so is the voltmeter.
So the voltage across the 50Ω resistor, and the reading 
on the voltmeter, is 100 volts.

e).  I = V/R
Through the 30Ω resistor:  I = 3-1/3 A
Through the 50Ω resistor:  I = 2 A

f).  In the parallel circuit, both resistors are connected
directly to the battery.  So neither resistor even knows
that the other one is there.  Each resistor sees 100 volts,
and the current through each resistor is 100/R, just as if
it were the only resistor in the circuit.

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A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring balance C and D. Balance C is
vagabundo [1.1K]

A uniform metal rod with of length 80cm and a mass of 3.2kg is supported horizontally by two vertical spring balances C and D. Balance C is 20cm from one end while D is 30cm from the other end would show the reading of 1.06 Kg and 2.13 kg respectively

<h3>What is gravity?</h3>

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another, The gravity varies according to the mass and size of the body for example the force of gravity on the moon is the 1/6th times of the force of gravity on the earth.

As given in the problem, A uniform metal rod of the length of 80cm and mass of 3.2kg is supported horizontally by two vertical springs balance C and D. Balance C is 20cm from one end while D is 30cm from the other end

The weight of the rod acting downward is from the center of the rod at 40 cm

Let us suppose the reading on the spring balance C and D are P and Q respectively

By using the equilibrium for the vertical force

Fv=0

P + C = 3.2

By using the equilibrium for the moment around the left corner

20×P+ 50×Q= 40 ×3.2

By solving for both P and Q from the above two equations we would get

P =1.06 and Q = 2.13

Thus, the reading on the spring balance C and D would be 1.06 Kg and 2.13 kg respectively

Learn more about gravity from here

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5 0
11 months ago
9) For a horizontally launched projectile, decreasing the height of the
Volgvan

Answer:

Increasing the launch height increases the downward distance, giving the horizontal component of the velocity greater time to act upon the projectile and hence increasing the range.

Explanation:

4 0
2 years ago
An astronaut is on a 100-m lifeline outside a spaceship, circling the ship with an angular speed of
nataly862011 [7]

Answer:

D = 72.68 m

Explanation:

given,

R = 100 m

angular speed = 0.1 rad/s

distance she can be pulled before the centripetal acceleration reaches 5g = 49 m/s².

using conservation of Angular momentum

I_i\omega_i= I_f\omega_f

mr_i^2\omega_i=m r_f^2\omega_f

\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i

\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i

we know,

centripetal acceleration

a = \dfrac{v^2}{r}

v = r ω

a =\omega_f^2 r_f

a =(\dfrac{r_i^2}{r_f^2}\times \omega_i)^2 r_f

a =\dfrac{r_i^4\times \omega_i^2}{r_f^3}

r_f^3=\dfrac{100^4\times 0.1^2}{5\times 9.8}

r_f^3=20408.1632

r_f = 27.32\ m

distance she has reached inward is equal to

D = 100 - 27.23

D = 72.68 m

3 0
3 years ago
List in order of imcreasing​
valkas [14]

Radio waves , microwaves , infrared radiation , visible light , ultraviolet waves , x-rays , gamma rays

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8 0
2 years ago
A 0.4 kg ball is thrown straight up into the air with a velocity of 12.2 m/s. What maximum height will it reach?
AysviL [449]

Answer:

Maximum height is 7.59 m.

Explanation:

We have,

Mass of a ball is 0.4 kg

It is thrown straight up into the air with a velocity of 12.2 m/s.

It is required to find the maximum height reached by the ball.

Concept used : Law of conservation of energy.

Solution,

Here, the energy of the ball remains conserved. Let h is the maximum height reached by the ball such that,

\dfrac{1}{2}mv^2=mgh\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(12.2)^2}{2\times 9.8}\\\\h=7.59\ m

So, the maximum height reached by the ball is 7.59 m.

6 0
3 years ago
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