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trasher [3.6K]
2 years ago
14

This question is about sound waves and light waves.

Physics
2 answers:
kirill [66]2 years ago
5 0

Answer: your correct it’s c

kenny6666 [7]2 years ago
4 0

Answer:

you had it right

Explanation:

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A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.1 m/s. How long does h
balandron [24]
We need to use the kinematic equation
S=ut+(1/2)at^2
where
S=displacement (+=up, in metres)
u=initial velocity (m/s)
t=time (seconds)
a=acceleration (+=up, in m/s^2)

Substitute values
S=displacement = 1.96-2.27 = -0.31 m (so that shot does not hit his head)
u=11.1
a=-9.81 (acceleration due to gravity)

-0.31=11.1t+(1/2)(-9.81)t^2
Rearrange and solve for t
-4.905t^2+11.1t-0.31=0
t=-0.02756 or t=2.291 seconds 
Reject the negative root to give
t=2.29 seconds (to 3 significant figures)

3 0
3 years ago
What kind of light would be the best to use to look inside a cold dark cloud and see the warm stars forming inside?
lys-0071 [83]
<h2>Answer: Infrared light</h2>

A dark nebula is a cloud of dust and cold gas, which does not emit visible light and hides the stars it contains.

These types of nebulae are composed mainly of the hydrogen they obtain from nearby stars, which is their fuel.

It is using infrared light that we can "observe" and analyze in detail what happens in the inner parts of these nebulae.

7 0
2 years ago
. A weather balloon is inflated to a volume of 2.2 x 103 L with 37.4 g of helium. What is the density of helium in grams per lit
Margarita [4]
That is the mst best eway to find its solution.
37.4/2.2*10^3 =  0.017 gm/liter or  1.7*10^-2
so we conclude that option b is sorrect
6 0
3 years ago
A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Lelechka [254]

Answer:

a. 16 s b. -1.866 kJ

Explanation:

a. Since the initial rotational speed ω₀= 3313 rev/min = 3313/60 × 2π rad/s = 346.94 rad/s. Its rotational speed becomes ω₁ = 0.75ω₀ in time t = 4 s.

We find it rotational acceleration using α = (ω₁ - ω₀)/t = (0.75ω₀ - ω₀)/t = ω₀(0.75 - 1)/t = -0.25ω₀/t = (-0.25 × 346.94 rad/s)/4 s = -21.68 rad/s².

Since the turntable stops at ω = 0, the time it takes to stop is gotten from

ω = ω₀ + αt and t = (ω - ω₀)/α = (0 - 346.94 rad/s)/-21.68 rad/s² = (-346.94/-21.68) s = 16 s.

So it takes the turntable 16 s to stop.

b. The workdone by the turntable to stop W equals its rotational kinetic energy change.

So, W = 1/2Iω² - 1/2Iω₀² = 1/2 × 0.031 kgm² × 0² - 1/2 × 0.031 kgm² × (346.94 rad/s)² = 0 - 1865.7 J = -1865.7 J = -1.8657 kJ ≅ -1.866 kJ

3 0
2 years ago
¿Qué distancia recorre un móvil que lleva una aceleración de 5m/s durante 10seg?
coldgirl [10]

Answer:

250 m

Explanation:

The car in this problem is moving of uniform accelerated motion, so we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

Assuming the car starts from rest,

u = 0

Also we know that

a = 5 m/s^2 (acceleration of the car)

t = 10 s

Substituting, we find the distance covered:

s=0+\frac{1}{2}(5)(10)^2=250 m

3 0
2 years ago
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