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daser333 [38]
3 years ago
6

What are 2 types of force fields

Physics
1 answer:
Serjik [45]3 years ago
6 0
Gravitational force and normal force
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The Hubble space telescope has been maintained by which of these?
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6 0
3 years ago
Two resistors have resistances R(smaller) and R(larger), where R(smaller) &lt; R(larger). When the resistors are connected in se
just olya [345]

Answer:

1.61ohms and 4.39ohms

Explanation:

According to ohm's law which States that the current (I) passing through a metallic conductor at constant temperature is directly proportional to the potential difference (V) across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the current

Rt is the effective resistance

Let the resistances be R and r

When the resistors are connected in series to a 12.0-V battery and the current from the battery is 2.00 A, the equation becomes;

12 = 2(R+r)

Rt = R+r (connection in series)

6 = R+r ...(1)

If the resistors are connected in parallel to the battery and the total current from the battery is 10.2 A, the equation will become;

12 = 10.2(1/R+1/r)

Since 1/Rt = 1/R+1/r (parallel connection)

Rt = R×r/R+r

12 = 10.2(Rr/R+r)

12(R+r) = 10.2Rr ... (2)

Solving equation 1 and 2 simultaneously to get the resistances. From (1), R = 6-r...(3)

Substituting equation 3 into 2 we have;

12{(6-r)+r} = 10.2(6-r)r

12(6-r+r) = 10.2(6r-r²)

72 = 10.2(6r-r²)

36 = 5.1(6r-r²)

36 = 30.6r-5.1r²

5.1r²-30.6r +36 =

r = 30.6±√30.6²-4(5.1)(36)/2(5.1)

r = 30.6±√936.36-734.4/10.2

r = 30.6±√201.96/10.2

r = 30.6±14.2/10.2

r = 44.8/10.2 and r = 16.4/10.2

r = 4.39 and 1.61ohms

Since R+r = 6

R+1.61 = 6

R = 6-1.61

R = 4.39ohms

Therefore the resistances are 1.61ohms and 4.39ohms

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2 years ago
There are five basic health-related components that one must have in order to be physically fit.
kirza4 [7]

Answer: the answer is true :-)

Explanation:

7 0
3 years ago
NaCl solid is an example of a/an<br> A. Insulator<br> B. Conductor<br> OC. Nonmetal<br> D. Metalloid
julia-pushkina [17]
It is an Insulator I’m pretty sure
6 0
1 year ago
Problem 1: Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructi
Anton [14]

1) Destructive interference

The condition for constructive interference to occur is:

\delta = m\lambda (1)

where

\delta =|d_1 -d_2| is the path difference, with

d_1 is the distance of the point from the first source

d_2 is the distance of the point from the second source

m is an integer number

\lambda is the wavelength

In this problem, we have

d_1 = 78.0 m\\d_2 = 143 m\\\lambda=26.0 m

So let's use eq.(1) to see if the resulting m is an integer

\delta =|78.0 m-143 m|=65 m\\m=\frac{\delta }{\lambda}=\frac{65 m}{26.0 m}=2.5

It is not an integer so constructive interference does not occur.

Let's now analyze the condition for destructive interference:

\delta = (m+\frac{1}{2})\lambda (2)

If we apply the same procedure to eq.(2), we find

m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{65.0 m}{26.0 m}-0.5=2

which is an integer: so, this point is a point of destructive interference.

2) Constructive interference

In this case we have

d_1 = 91.0 m\\d_2 =221.0 m

So the path difference is

\delta =|91.0 m-221.0 m|=130.0 m

Using the condition for constructive interference:

m=\frac{\delta }{\lambda}=\frac{130.0 m}{26.0 m}=5

Which is an integer, so this is a point of constructive interference.

3) Destructive interference

In this case we have

d_1 = 44.0 m\\d_2 =135.0 m

So the path difference is

\delta =|44.0 m-135.0 m|=91.0 m

Using the condition for constructive interference:

m=\frac{\delta }{\lambda}=\frac{91.0 m}{26.0 m}=3.5

This is not an integer, so this is not a point of constructive interference.

So let's use now the condition for destructive interference:

m=\frac{\delta}{\lambda}-\frac{1}{2}=\frac{91.0 m}{26.0 m}-0.5=3

which is an integer: so, this point is a point of destructive interference.

3 0
3 years ago
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