What's an easy way to create an interference pattern of waves?

Lien uses a spring scale to pull a block toward the right across the lab table. The scale reads 8 N. Which force should Lien con

Alja [10]

Answer

The force on the left across the lab table.

Explanation

The Newton's third law of motion states that; <em>action and reaction are equal but a ct in opposite direction. </em>

When the block of is pulled on the right with a force of X Newtons then there is a force -X Newtons applying and equal force on the left. For every action there must be a reaction with is equal and applying in the opposite direction.

So, if the block is pulled on the right by a force of 8 N there is another equal force applying on the left.

5 0

3 years ago

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You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe

steposvetlana [31]

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, $L=800\ mH=800\times 10^{-3}\ H=0.8\ H$

Capacitance, $C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F$

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

$Z=\sqrt{R^2+(X_C-X_L)^2}$ ...........(1)

Where

$X_C$ is the capacitive reactance, $X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega$

$X_L$ is the inductive reactance, $X_L={2\pi fL}$

$X_L={2\pi \times 60\times 0.8}=301.59\ \Omega$

So, equation (1) becomes :

$Z=\sqrt{(100)^2+(265.65-301.59)^2}$

Z = 106.26 ohms

or

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.

8 0

3 years ago

Please i really need help..

stiv31 [10]

C.) chemical to electrical to light

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3 years ago

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lasie4. A 4 kg object is displaced to the right by a distance of 12 m underthe influence of the following forces: a 17 Nforce pu

Oksanka [162]

The work done by a constant force in a rectilinear motion is given by:

$W=Fd\cos\theta$

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.

In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is: