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4 years ago

Figure 1 shows a wave movement during one second. What is the frequency of the wave

Physics

2 answers:

AURORKA [14]4 years ago

8 0

2 hertz, as explained previously- I'm afraid m/s is a completely different measurement

Andreyy894 years ago

4 0

This is 2 hertz. You can mark out 2 full wavelengths in the second of time.

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Genes are organized on matching pairs of _______.

natta225 [31]

Answer:

homologous chromosomes

Explanation:

6 0

3 years ago

8. A meter reader determines that a business has used 5000 kW.h of energy in 4 months. If

ladessa [460]

Answer:

ENERGY AND COST. One kllowatt hour is 1,000 watts of power for one hour of time. ... Determine power: P = V XI ... Calculate the total kilowatt hours used. ... If the electric costs are 150 per kWh, how much does it cost to run the refrigerator in ... 8. A room was lighted with three 100-watt bulbs for 5 hours per day. If the cost of.

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3 years ago

A force of 60 N is applied to a skier to pull him along a horizontal surface so that his speed remains constant. If the coeffici

matrenka [14]

Answer:

$1200\ \text{N}$

Explanation:

F = Force on the skier = 60 N

$\mu$ = Coefficient of friction = 0.05

w = Weight of skier

Force is given by

$F=\mu w$

$\Rightarrow w=\dfrac{F}{\mu}$

$\Rightarrow w=\dfrac{60}{0.05}=\dfrac{6000}{5}$

$\Rightarrow w=1200\ \text{N}$

Weight of the skier on which the force is being applied is $1200\ \text{N}$ .

7 0

3 years ago

A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N forc

Setler79 [48]

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = $\frac{1}{2}ke^{2}$

==> $\frac{1}{2}* 550* 0.02^{2}$ = 0.11 J

3 0

3 years ago

A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro

Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia = 1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0

3 years ago