What is the smallest organism?

Why are circuit breakers and fuses important?

Lesechka [4]

Answer:

A fuse and circuit breaker both serve to protect an overloaded electrical circuit by interrupting the continuity, or the flow of electricity. ... Fuses tend to be quicker to interrupt the flow of power, but must be replaced after they melt, while circuit breakers can usually simply be reset.

6 0

3 years ago

A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc

german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = $\frac{1}{2} kx^2$ and the kinetic energy gained by the box is given by KE = $\frac{1}{2} mv^2$

Since work done by the spring = kinetic energy gained by the box we have

$\frac{1}{2} mv^2$ = $\frac{1}{2} kx^2$ therefore we have v = $\sqrt{\frac{kx^2}{m} }$ = $x\sqrt{\frac{k}{m} }$ = $0.15\sqrt{\frac{300}{0.2} }$ = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = $\frac{v^2}{2g}$ = $\frac{5.81^{2} }{2*9.81}$ = 1.72 m

6 0

3 years ago

Q1. Which statement is correct

marishachu [46]

Answer:

C

D

Explanation:

7 0

3 years ago

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

<span>The current I is any motion of charge from one region to another, so this is given by:

</span> $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$
<span>
Finally, for the drift velocity magnitude vd, we find:

</span> $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

3 years ago

When we apply the energy conversation principle to a cylinder rolling down an incline without sliding, we exclude the work done

NikAS [45]

Answer:

D. the linear velocity of the point of contact (relative to the inclined surface) is zero

Explanation:

The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction becomes zero . Hence work done by friction becomes zero.

5 0

3 years ago