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3 years ago

What is the magnitude of the current in the R= 6 Ω resistor? kirchhoff

Physics

1 answer:

cupoosta [38]3 years ago

3 0

Answer:

Here's an explanation but not the answer

Explanation:

When a resistor is traversed in the same direction as the current, the ... Traversing the internal resistance r1 from c to d gives −I2r1. ... I1 = I2 + I3 = (6−2I1) + (22.5− 3I1) = 28.5 − 5I1.

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What do the length of the cord and gravity determine for a pendulum?

rusak2 [61]

Answer: C

Period/ Period of the pendulum.

Content:

Simple pendulum is a small diameter bob which is suspended from light cord or string. The string is strong enough to stretch.

Pendulums are quiet common in use such as clocks, swings etc.,

From the simple pendulum we can find conditions under which it performs simple harmonic motion and we can also derive the expressions for Period of pendulum, frequency etc.

Period of a pendulum/Time period is given by the following expression

 T =2π.√(L/g) seconds 

 T = period of pendulum in seconds

L = Length of the string/cord in meters

g = gravitational force in m/s² ( g = 9.8 m/s² )

Period of pendulum is independent on mass of the bob.

So, The relation between length of the cord and gravity is used to determine the period of pendulum

4 0

3 years ago

A clarinetist, setting out for a performance, grabs his 3.230 kg3.230 kg clarinet case (including the clarinet) from the top of

SpyIntel [72]

Answer:

- 0.5 m/s²

Explanation:

m = mass of the clarinet case = 3.230 kg

W = weight of the clarinet case in downward direction

a = vertical acceleration of the case

Weight of the clarinet case is given as

W = mg

W = 3.230 x 9.8

W = 31.654 N

F = Upward force applied = 30.10 N

Force equation for the motion of the case is given as

F - W = ma

30.10 - 31.654 = 3.230 a

a = - 0.5 m/s²

8 0

3 years ago

If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?

Arisa [49]

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm$

Magnification,

$m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25$

Hence, the magnification of the image is 0.25.

7 0

3 years ago

A sports car traveling along a straight line increases its speed from 19.8 mi/h to 59.9 mi/h.a. What is the ratio of the final t

Natasha2012 [34]

Answer

given,

initial speed of the car (v₁)= 19.8 mi/h

final speed of the car (v₂)= 59.9 mi/h

a) initial momentum = m v₁

P₁ = 19.8 m

final final momentum = m v₂

P₂ = 59.9 m

ratio = $\dfrac{P_2}{P_1}$

= $\dfrac{59.9 m}{19.8 m}$

ratio of momentum= $\dfrac{P_2}{P_1}=3.025$

b) initial kinetic energy= 1/2 m v₁²

K₁ = 196.02 m

final kinetic energy= 1/2 m v₂²

K₂ = 1794.005 m

ratio = $\dfrac{K_2}{K_1}$

= $\dfrac{1794.005 m}{196.02 m}$

ratio of Kinetic energy= $\dfrac{P_2}{P_1}=9.15$

8 0

3 years ago

Two 1 MHz radio antennas emitting in-phase are separated by600

algol [13]

Answer:

1 km

Explanation:

$d_0$ = Gap between antennas = 600 m

$\nu$ = Frequency = 1 MHz

$z_1$ = Distance to receiver = 2 km

c = Speed of light = $3\times 10^8\ m/s$

Wavelength is given by

$\lambda=\dfrac{c}{\nu}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1\times 10^6}\\\Rightarrow \lambda=300\ m$

Distance to be moved is given by

$D=\dfrac{z_1\lambda_0}{d_0}\\\Rightarrow D=\dfrac{2\times 300}{600}\\\Rightarrow D=1\ km$

The distance to be moved is 1 km north.

6 0

3 years ago