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Talja [164]
3 years ago
11

The theory that light exists in quantized "chunks" was proposed in response

Physics
2 answers:
Dafna1 [17]3 years ago
8 0
Its D because of the emission spectra
qwelly [4]3 years ago
5 0

Answer: C.

Explanation:

The ultraviolet catastrophe.

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Which is TRUE about static electricity?
Fed [463]

Answer:

the first one stationary charge

3 0
2 years ago
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A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a force of 250 n at an angle of 32.0° ab
Phoenix [80]
According to Newton's second law, the resultant of the forces acting on the box is equal to the product between its mass and its acceleration:
\sum F = ma (1)

we are only concerned about the horizontal direction, so there are only two forces acting on the box in this direction:
- the horizontal component of the force exerted by the rope, which is equal to
F_x = F cos \theta = (250N)(\cos 32.0^{\circ} )=212.0 N
- the frictional force, acting in the opposite direction, which is equal to
F_f = \mu mg=(0.350)(50.0 kg)(9.81 m/s^2)=171.7 N

By applying Newton's law (1), we can calculate the acceleration of the box:
F_x - F_f = ma
a= \frac{F_x - F_f}{m}= \frac{212.0 N-171.7 N}{50.0 kg} =0.81 m/s^2

6 0
3 years ago
A car travels 10 meters east in 4 seconds. What is the car's velocity? *
MrMuchimi

Answer:

your velocity is 2.5 m/sec

6 0
2 years ago
Read 2 more answers
If the mass of a material is 50 grams and the volume of the material is 5 cm^3, what would the density of the material be?
kotykmax [81]
D = M / V

D = 50 / 5

D = 10 g/cm^3
5 0
3 years ago
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Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

5 0
3 years ago
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