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antoniya [11.8K]
3 years ago
6

Find the current passing through each of the 3 resistors connected parallel to each other as shown in the figure (i1, i2, I3). S

how your actions clearly.

Physics
1 answer:
Thepotemich [5.8K]3 years ago
3 0

Answer:

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

Explanation:

From the image, we see that the resistors are connected in parallel. This means that the voltage passing through them is the same.

Now, formula for current is; I = V/R

In this case, V which is voltage is denoted by ε.

Thus;

I1 = ε/R1

I2 = ε/R2

I3 = ε/R3

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What injuries could occur due to incorrect movement and handling of equipment with any sport or activity? (Select ALL that apply
NeTakaya
Sprains/Strains, tears to muscles, tendons ligaments, not hypothermia bc thats when you get really cold and not dehydrated
4 0
4 years ago
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
3 years ago
A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s
Ede4ka [16]

When acceleration is constant, the average velocity is given by

\bar v=\dfrac{v+v_0}2

where v and v_0 are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}

where x,x_0 are the final/initial displacements, and t,t_0 are the final/initial times, respectively.

Take the car's starting position to be at t_0=0\,\mathrm s. Then

\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t

So we have

x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m

You also could have first found the acceleration using the equation

v=v_0+at

then solve for x via

x=x_0+v_0t+\dfrac12at^2

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of a anyway.

7 0
3 years ago
Suppose light from a 632.8 nm helium-neon laser shines through a diffraction grating ruled at 520 lines/mm. How many bright line
Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

y=\frac{m\lambda D}{d}

where

m is the order of the maximum

\lambda is the wavelength of the light

D is the distance of the screen

d is the separation between two adjacent slit

Here we have:

\lambda=632.8 nm = 632.8\cdot 10^{-9} m is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

n=520 lines/mm is the number of lines per mm, so the spacing between two lines is

d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m

Therefore, substituting m = 1, we find:

y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m

So, on the distant screen, there is 1 bright fringe every 33 cm.

6 0
3 years ago
Please i need your help w/h/w you make the following measurements if an object:42kg, and 22m​
Rufina [12.5K]

Answer:

1.9 kg/m^3

Explanation:

The density of an object is given by

d=\frac{m}{V}

where

m is the mass of the object

V is its volume

In this problem,

m = 42 kg

V = 22 m^3

Substituting into the equation, we find the object's density:

d=\frac{42 kg}{22 m^3}=1.9 kg/m^3

8 0
3 years ago
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