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2 years ago

The amplitude of the wave in the picture below is... Select one: a. 1 cm b. 1.5 cm c. 2 cm d. 3 cm

Physics

1 answer:

antiseptic1488 [7]2 years ago

8 0

Answer:

1cm

Explanation:

The amplitude of a wave is the distance from the center line (in this case the 1 cm marker on the vertical ruler) to the highest or lowest point. For this image the highest point is 2 cm, 2cm-1cm=1cm. The lowest point is 0cm, 1cm-0cm=1cm.

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How much work is done by 0.070 m3 of gas, when the volume remains constant with pressure of 63 x 105 Pa?

ICE Princess25 [194]

Answer:

W = 0 J

Explanation:

The amount of work done by gas at constant pressure is given by the following formula:

$W = P\Delta V$

where,

W = Work done by the gas

P = Pressure of the gas

ΔV = Change in the volume of the gas

Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:

$W = P(0\ m^3)\\$

5 0

3 years ago

According to Coulomb’s Law, what happens to the force when the distance increase between 2 particles?

ohaa [14]

Answer:

The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, <u>the attraction or repulsion becomes weaker</u>, decreasing to one-fourth of the original value.

Explanation:

Coulomb’s law, mathematical description of the electric force between charged objects. Formulated by the 18th-century French physicist Charles-Augustin de Coulomb, it is analogous to Isaac Newton’s law of gravity.

Both gravitational and electric forces decrease with the square of the distance between the objects, and both forces act along a line between them. In Coulomb’s law, however, the magnitude and sign of the electric force are determined by the electric charge, rather than the mass, of an object. Thus, charge determines how electromagnetism influences the motion of charged objects. Charge is a basic property of matter. Every constituent of matter has an electric charge with a value that can be positive, negative, or zero.

Coulomb's Law says that the force between 2 charges is proportional to the product of the quantities of charge on each and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is $F=k\frac{q_{1}q_{2} }{r^{2} }$ .

$F$ is the force.

$k$ is the Coulomb's constant ( $8.987*10^{9} \frac{Nm^{2} }{C^{2} }$ ).

$q_{1}$ is the electric charge of object 1.

$q_{2}$ is the electric charge of object 2.

$r$ is the distance between the two charges.

Electric force is inversely proportional to ( $r^{2}$ ) instead of ( $r$ ). As the distance between charges increases, the electric force decreases by a factor of $\frac{1}{r^{2} }$ .

8 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is: