Which of the following types of electromagnetic radiation has the highest energy?

Can someone help me with the 10th one which is circled

PtichkaEL [24]

Force can be exerted into an object with out it moving, but if you were to move the object due to force it would be considered work. (yes, you can exert force without having the object moving)

6 0

3 years ago

Read 2 more answers

If a statement is true, select true. if it's false, select false. <br>

Elena L [17]

3 is false 2 is true and the rest true

7 0

3 years ago

Read 2 more answers

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min. Work out the difference be

Serggg [28]

Explanation:

We have,

Ajoba and Prav drive to work. Ajoba drives 45 miles in 2.5 hours. Prav drives 74 km in 1 hour 15 min.

1 mile = 1.6 km

45 miles = 72.42 km

74 miles = 119.0 km

1 hour 15 min means 1.25 hours

Average speed of Ajoba is :

$v_1=\dfrac{72.42 }{2.5}=28.96\ km/h$

Average speed of Prav,

$v_2=\dfrac{119}{1.25}=95.2\ km/h$

Difference in average speed of Ajoba and Prav is :

$v=v_2-v_1\\\\v=v_2-v_1\\\\v=95.2-28.96\\\\v=66.24\ km/h$

So, the difference in average speed of Ajoba and Prav is 66.24 km/h.

7 0

3 years ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

3 years ago

A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motio

docker41 [41]

Answer:

Angular velocity is same as frequency of oscillation in this case.

ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$

Explanation:

- write the equation F(r) = -K $r^{4}$ with angular momentum <em>L</em>

- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.

- Write the energy of the orbit in relative to r = 0, and solve for "E".

- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.

- Solve for effective potential

- ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$

3 0

3 years ago