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Ivanshal [37]
3 years ago
11

Based on the number on the fuse, what do you think is the maximum current allowed in this circuit?

Physics
1 answer:
Inessa [10]3 years ago
3 0

Answer:

0.3 A

Explanation:

Build the circuit attached. Make sure the switch is turned off as you build the circuit. Set the Selected battery voltage to 10 volts.

A.      Based on the number on the fuse, what do you think is the maximum current allowed in this circuit?

Solution:

A fuse is a safety device which is used in circuits for preventing very high currents which can cause overloading or start a fire. A fuse is made up of a low resistance thin piece of metal such that the flow of high current can cause the metal to melt leading to an open circuit. A fuse lets current flow up to a maximum value before it melts and the circuit will be broken.

From the image attached, since the maximum rating of the fuse is 0.3 A, hence the maximum current that can flow through the circuit is 0.3 A.

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6 0
3 years ago
The work function of a metal surface is 4.80 × 10-19 J. The maximum speed of the electrons emitted from the surface is vA = 7.7
poizon [28]

Answer:

\lambda_A=2.65177\times 10^{-7}\ m

\lambda_B=3.32344\times 10^{-7}\ m

Explanation:

h = Planck's constant = 6.63\times 10^{-34}\ m^2kg/s

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

m = Mass of electron = 9.11\times 10^{-31}\ kg

W_0 = Work function = 4.8\times 10^{-19}\ J

v_A = Velocity of A particle = 7.7\times 10^5\ m/s

v_B = Velocity of B particle = 5.1\times 10^5\ m/s

The wavelength is given by

\lambda=\frac{hc}{\frac{1}{2}mv^2+W_0}

\lambda_A=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(7.7\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_A=2.65177\times 10^{-7}\ m

The wavelength \lambda_A=2.65177\times 10^{-7}\ m

\lambda_B=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(5.1\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_B=3.32344\times 10^{-7}\ m

The wavelength \lambda_B=3.32344\times 10^{-7}\ m

5 0
4 years ago
A spring with spring constant 15.5 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.
Aloiza [94]

Answer:

A) 138.8g

B)73.97 cm/s

Explanation:

K = 15.5 Kn/m

A = 7 cm

N = 37 oscillations

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ω² = k/m and m = k/ω²

Also, angular frequency (ω) = 2π/T

Now, T is the time it takes to complete one oscillation.

So from the question, we can calculate T as;

T = 22/37.

Thus ;

ω = 2π/(22/37) = 10.5672

So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g

B) In simple harmonic motion, velocity is given as;

v(t) = vmax Sin (ωt + Φ)

It is from the derivative of;

v(t) = -Aω Sin (ωt + Φ)

So comparing the two equations of v(t), we can see that ;

vmax = Aω

Vmax = 7 x 10.5672 = 73.97 cm/s

6 0
3 years ago
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