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3 years ago

1. Transverse waves _____.

Physics

2 answers:

poizon [28]3 years ago

8 0

D. have particle moment perpendicular to the direction of the energy.

kumpel [21]3 years ago

7 0

The answer is D to you question

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You work at a retail store. Recently, paychecks have had errors and there have been many employee complaints. You have been assi

Nata [24]

Answer:

C,B,A

Explanation:

5 0

3 years ago

Read 2 more answers

Skater 1 has a mass of 45 kg and is at rest. Skater 2 has a mass of 50 kg and is moving slowly at a constant velocity of 3.2 m/s

algol13

Answer:

None

Explanation:

Force, F is given by ma where m is the mass of an object and a is acceleration

Acceleration is the rate of change in velocity per unit time. Since skaters with mass of 75 kg and 50 kg are moving at a constant speed, there is no acceleration hence F=50*0=0 and F=75*0=0

For skater of 45 kg, he is at rest to mean the initial and final velocitu of the skater is zero hence no acceleration, the force will be 45*0=0

Therefore, none of the skaters will experience a greater net force.

3 0

3 years ago

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R. The centra

Dmitriy789 [7]

Answer:

(a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

We need to calculate the electric field due to the rod

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

$E=0$

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}$

$E\propto\dfrac{1}{z^2}$

(c). In terms of R,

We need to calculate the positive distance

If $E\rightarrow E_{max}$

Then, $\dfrac{dE}{dz}=0$

$\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0$

Taking only positive distance

$z=\dfrac{R}{\sqrt{2}}$

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

$E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

$E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}$

$E_{max}=15418.7\ N/C$

$E_{max}=1.54\times10^{4}\ N/C$

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

6 0

2 years ago

Radiation with a wavelength of 238 nm shines on a metal surface and ejects electrons that have a maximum speedof 3.75 X 105 m/s.

Ulleksa [173]

The Answer is e) - Gold.

4 0

3 years ago

Calculate the length of a simple pendulum that oscillates with a frequency of 0.4Hz g=10m/s2 , ^=3.142

earnstyle [38]

Answer:

Explanation:

For simple pendulum the formula is

$T=2\pi\sqrt{\frac{l}{g} }$

Where T is time period , l is length and g is acceleration due to gravity .

$\frac{1}{n} =2\pi\sqrt{\frac{l}{g} }$

n is frequency

Putting the values

$\frac{1}{.4} =2\pi\sqrt{\frac{l}{10} }$

$\frac{l}{10} = .1584$

l = 1.584 m

4 0

3 years ago