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Maksim231197 [3]
3 years ago
15

If the strength of the magnetic field at B is 20 units, the strength of the magnetic field at A is _____.

Physics
2 answers:
xz_007 [3.2K]3 years ago
8 0
The answer is A = 20 units
Fantom [35]3 years ago
3 0

Answer: The answer is actually 80 units

Explanation: I did the assignment and 80 units was the answer that was correct.

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Can animal cells be broken down further into a living unit
mezya [45]

Answer: No! Animal cells cannot be broken down further into living cells.

5 0
3 years ago
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What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
Nastasia [14]

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

8 0
3 years ago
A 5.00 L air sample at a temperature of -50 °C has a pressure of 107 kPa. What will be the new pressure if the temperature is ra
Serjik [45]
Its simple use formuila ,
PV=nRT
n,R is constant as the both have same moles.
so,
(p1v1)/T1 = (p2v2)/T2
so, 128.53338kpa
4 0
3 years ago
A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the
Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

m is an integer, m = 1, 2, 3...

We here know that as sin\theta as w are constant, then

\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

3 0
3 years ago
A charge of 90 C passes through a wire in 1 hour 15 minutes . what is the current in the wire
timurjin [86]
We calculate current from the formula:
I= \frac{q}{t} , where q is a electric charge transferred over time t 
Time should be converted to seconds:
1h 15 min= 75min= 4500s
I=\frac{90C}{4500s}=0,02A Result is in unit-Ampere
5 0
3 years ago
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