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2 years ago

Which phenomenon supports the particle model of light?

Physics

2 answers:

Korolek [52]2 years ago

7 0

Answer:

The photoelectric effect

kondaur [170]2 years ago

3 0

Answer:

Explanation:

your answer is c Destructive interference

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What is the momentum of an object with mass 8 kg and velocity 5 m/s

Slav-nsk [51]

Momentum = (mass) x (speed)

Momentum = (8 kg) x (5 m/s)

Momentum = 40 kg-m/s

6 0

3 years ago

Light traveling in water (n = 1.33) into an unknown medium.If rhe angles if incidence and refraction are 40 degrees and 25 degre

Bess [88]

40 degreased and 25 dregrees=65 degrees‍♀️

8 0

3 years ago

Need help due in 4 minutes

sergey [27]

Answer:

speed

Explanation:

3 0

3 years ago

Read 2 more answers

Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont

SVEN [57.7K]

Answer:

(a) 2.34 s

(b) 6.71 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

$T = \frac{2 u Sin\theta }{g}$

$T = \frac{2 \times 20 \times Sin35 }{9.8}$

T = 2.34 second

(b) The formula for the maximum height is given by

$H = \frac{u^{2} \times Sin^{2}\theta }{2g}$

$H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}$

H = 6.71 m

$R = \frac{u^{2} \times Sin 2\theta }{g}$

$R = \frac{20^{2} \times Sin 2 \times 35}{9.8}$

R = 38.35 m

(d) It hits with the same speed at the initial speed.

8 0

3 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago