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olchik [2.2K]
3 years ago
8

Which statement best describes how light waves travel in a uniform medium. A. in straight lines. . B. in extending circles. . C.

in bent lines. . D. in shrinking ripples
Physics
2 answers:
Tanya [424]3 years ago
6 0
Light waves travel in straight lines when they are travelling in a uniform medium. This is because the waves are travelling at the same speed.
g100num [7]3 years ago
6 0

The answer is a. Straight lines

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Two capacitors have the same size of plates and the same distance 4 mm between the plates The potentials of the two plates in ca
OverLord2011 [107]

Answer:

    E = -4000 N / C

Explanation:

The potential and electric field are related

         V = - E s

          E = - V / s

we reduce the magnitudes to the SI system

          s = 4 mm (1 m / 1000 mm) = 0.004 m

we calculate

          E = - 16 /0.004

          E = -4000 N / C

8 0
3 years ago
A person jogs 4.0 km in 32 minutes, then 2.0 km in 22 minutes, and finally 1.0 km in 16 minutes. What is the jogger's average sp
kupik [55]

Answer:

22

Explanation:

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7 0
3 years ago
Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ =  4·m·v₃

\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0
2 years ago
If the velocity of gas molecules is doubled, then it’s kinetic energy will?( hint formula for kinetic energy is KE=.5 x mv2)
eduard

Answer:

D quadruple

Explanation:

E = \frac{1}{2}mv^2\\E'=\frac{1}{2}m(2v)^2=\frac{1}{2}m4 v^2 = 4(\frac{1}{2}mv^2)=4E

7 0
3 years ago
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a 85 Kg girl, running 1.5 m/s, runs for 2550 seconds. How much force is required to stop her in 5 seconds
adoni [48]

450x force to stop her 5 seconds

3 0
3 years ago
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