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Nastasia [14]
3 years ago
8

If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis

tic) formula here.)
Physics
1 answer:
arlik [135]3 years ago
3 0
The kinetic energy of the electron is
K= \frac{1}{2}mv^2
where m=9.1 \cdot 10^{-31} kg is the mass of the electron and v its speed. Since we know the value of the kinetic energy, K=4.1 \cdot 10^{-18} J, we can find the value of the speed v:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} }  = 3\cdot 10^6 m/s
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4. A car, initially traveling east with a speed of
PilotLPTM [1.2K]

Answer:

150 m

Explanation:

Given,

u=5m/s

a=2m/s2

t=10s

v=?

s=?

Now,

v=u+at

=5+2×10

=5+20

=25m/s

So,

s=u+v/2×t

=5+25/2×10

=30/2×10

=15×10

=150m

5 0
3 years ago
Joe and Max shake hands and say goodbye. Joe walks east 0.40 km to a coffee shop, and Max flags a cab and rides north 3.65 km to
katen-ka-za [31]

Answer:

3.67 km

Explanation:

Joe distance towards coffee shop is,

OB=0.40 km

And the Max distance towards bookstore is,

OA=3.65 km

Now the distance between the Joy and Max will be,

By applying pythagorus theorem,

AB=\sqrt{OB^{2}+OA^{2}}

Substitute 0.40 km for OB and 3.65 km for OA in the above equation.

AB=\sqrt{0.40^{2}+3.65^{2}}\\AB=\sqrt{13.4825} \\AB=3.67 km

Therefore the distance between there destination is 3.67 km.

6 0
3 years ago
Naming covalent compounds <br> P4S5
Andrew [12]

Answer:

what the heck is sakurfa

Explanation:

3 0
3 years ago
A 10,000-watt radio station transmits at 535 kHz. Determine the number of joules transmitted per second. 10,000 J/s 10 J/s 535 J
REY [17]
1 nanowatt  =  1 nanojoule/sec
1 watt  =  1 joule/sec
10 watts  =  10 joules/sec
100 watts  =  100 joules/sec
742.914 watts  =  742.914 joules/sec
1,000 watts  =  1,000 joules/sec
10,000 watts  =  10,000 joules/sec
100,000 watts  =  100,000 joules/sec
1 megawatt  =  1 megajoule/sec
1 gigawatt  =  1 gigajoule/sec
1 petawatt  =  1 petajoule/sec

We don't care what frequency the transmission is using,
or who their morning DJ is.

5 0
3 years ago
A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m + t/(0.20 s))], where x is in m and t in s.
Len [333]

Corrected and Formatted Question:

A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))], where x is in m and t in s.

(a) In what direction is this wave traveling?

(b) What are the wave speed, frequency, and wavelength?

(c) At t = 0.50 , what is the displacement of the string at x = 0.20 m?

Answer:

The wave is travelling in the negative x direction

The wave speed = 12.0m/s

The frequency = 5Hz

The wavelength = 2.4m

The displacement at t = 0.50s and x = 0.20m is -0.029m

Explanation:

The general wave equation is given by;

y(x, t) = y cos (2\pi(x/λ) - 2\pift)    --------------------------------(i)

Where;

y(x, t) is the displacement of the wave at position x and a given time t

y = amplitude of the wave

f = frequency of the wave

λ = wavelength of the wave

Given;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]   ------------------(ii)

Which can be re-written as;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m)) + 2π(t/(0.20 s))]  -------------(iii)

Comparing equations (i) and (iii) we have that;

=> 2π(x/(2.4 m) = 2π(x/λ)

=> λ = 2.4m

Therefore the wavelength of the wave is 2.4m

Also, still comparing the two equations;

=> 2π(t/(0.20 s) = 2πft

=> f = 1 / 0.20

=> f = 5Hz

Therefore the frequency of the wave is 5Hz

To get the wave speed (v), it is given by;

v = f x λ

Where f = 5Hz and λ = 2.4m

=> v = 5 x 2.4

=> v = 12.0m/s

Therefore, the speed of the wave is 12.0m/s

At t = 0.50s and x = 0.20m;

The displacement, y(x,t) of the string wave is given by

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]

<em>Convert the amplitude of 3.0cm to m</em>

=> 3.0cm = 0.03m

<em>Substitute this back into the equation</em>

=> y(x, t) = (0.03m) × cos[2π(x/(2.4 m) + t/(0.20 s))]

<em>Substitute the values of t and x into the equation above;</em>

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m) + 0.50/(0.20 s))]

<em>Carefully solve the equation</em>

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m)) + 2π(0.50/(0.20 s))]

=> y(x, t) = (0.03m) × cos[0.08π + 5π]

=> y(x, t) = (0.03m) × cos[5.08π]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × cos[15.96]

=> y(x, t) = (0.03m) × -0.9684

=> y(x, t) = 0.029m

Therefore the displacement at those points is -0.029m

Also, the sign of the displacement shows that the direction of the wave is in the negative x direction.

8 0
3 years ago
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