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3 years ago

What does the "coefficient of friction" tell you?

Physics

1 answer:

Arte-miy333 [17]3 years ago

8 0

a coefficient of fraction is a value that shows the relationship between two objects and the normal reaction between the objects that are involved.

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Calculate the wavelength (in nm) of the blue light emitted by a mercury lamp with a frequency of 6.88 × 1014 Hz. The speed of li

Nezavi [6.7K]

Answer: 430 nm.

Explanation:

The relation of wavelength and frequency is:

Formula used : $\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency = $6.88\times 10^{14}Hz$

$\lambda$ = wavelength = ?

c = speed of light = $3.00\times 10^{8}m/s$

Now put all the given values in this formula, we get

$6.88\times 10^{14}=\frac{3.00\times 10^{8}m/s}{\lambda}$

$\lambda=\frac{3.00\times 10^{8}m/s}{6.88\times 10^{14}}=0.43\times 10^{-6}m=430m$ $(1nm=10^{-9}m)$

Thus the wavelength (in nm) of the blue light emitted by a mercury lamp is 430 nm.

6 0

3 years ago

Which of the following is not a major function of the roots?

sashaice [31]

it is c. it is photosynthesis because is under the ground and away from the sun

7 0

3 years ago

The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).

sammy [17]

Answer:

(a) $emf_L=-LI_{max}\omega cos(\omega t)$

(b) neither increasing or decreasing

Explanation:

The current through an inductor of inductance L is given by:

$I(t)=I_{max}sin(\omega t)$ (1)

(a) The induced emf is given by the following formula

$emf_L=-L\frac{dI}{dt}$ (2)

You derivative the expression (1) in the expression (2):

$emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)$

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

$emf_L=-\omega LI_{max}$

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

6 0

2 years ago

An object of mass 0.400.40 kg, hanging from a spring with a spring constant of 8.08.0 N/m, is set into an up-and-down simple har

sineoko [7]

Answer:

Acceleration will be equal to $2m/sec^2$

Explanation:

We have given mass of the object m = 0.4 kg

Spring constant k = 8 N/m

Maximum displacement of the spring is given x = 0.1 m

From newton's law force is equal to $F=ma$ .....eqn 1

By hook's law spring force is equal to $F=kx$ .....eqn 2

From equation 1 and equation 2

$ma=kx$

$0.4\times a=8\times 0.1$

$a=2m/sec^2$

So acceleration will be equal to $2m/sec^2$

8 0

3 years ago

A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 47 N. How f

Novosadov [1.4K]

Answer:

7.53 m

Explanation:

Force, F = 47 N

initial velocity, u = 0

Final kinetic energy, Kf = 354 J

Let the distance traveled by the student is s.

According to the work energy theorem,

Work done by all forces = Change in kinetic energy

Force x distance = final kinetic energy - initial kinetic energy

F x s = kf - ki

47 x s = 354

s = 7.53 m

4 0

3 years ago