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3 years ago

As elevation increases temperature decreases. At which elevation will sound travel fastest

Physics

2 answers:

Y_Kistochka [10]3 years ago

8 0

Answer:

higher evaluations = faster sound travel

mestny [16]3 years ago

5 0

Answer:

The higher the better. Inside the atmosphere since sound cant travel without a atmosphere

Explanation:

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A spring with spring constant 31 N/m is attached to the ceiling, and a 4.5-cm-diameter, 1.5 kg metal cylinder is attached to its

aalyn [17]

Answer:

0.315758099469 m

Explanation:

m = Mass of cylinder = 1.5 kg

$\rho$ = Density of water = 1000 kg/m³

V = Volume = Ah

A = Area = $\pi r^2$

k = Spring constant = 31 N/m

x = Displacement

g = Acceleration due to gravity = 9.81 m/s²

Here the forces are conserved

Weight of cylinder = Buoyant force + Spring force

$mg=\rho Vg+kx\\\Rightarrow mg=\rho Ahg+kx\\\Rightarrow 1.5\times 9.81=1000\times \pi(2.25\times 10^{-2})^2x\times 9.81+31x\\\Rightarrow 1.5\times 9.81=46.60213x\\\Rightarrow x=\dfrac{1.5\times 9.81}{46.60213}\\\Rightarrow x=0.315758099469\ m$

The length of the submerged cylinder is 0.315758099469 m

4 0

4 years ago

Please help i don’t get this

antoniya [11.8K]

Answer:

i think it is with friction because when we move we go against friction so like when were walking friction is trying to decrease our speed

Explanation:

3 0

3 years ago

If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object

777dan777 [17]

Answer:

+9.8m/s^2

Explanation:

The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

The constant value for acceleration due to gravity is 9.8m.s^2

Since the object is falling, hence the acceleration due to gravity is positive.

Rate of gravity working on the object will be +9.8m/s^2

4 0

3 years ago

an object that is 8 cm tall is located 2.5 m in front of a plane mirror. The image formed by the mirror appears to be. Find di a

UkoKoshka [18]

The image is at 2.5 m behind the mirror (virtual image)

Explanation:

We can solve the problem by using the mirror equation:

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

where

f is the focal length of the mirror

$d_o$ is the distance between the object and the mirror

$d_i$ is the distance between the image and the mirror

For a plane mirror, the focal length is infinite:

$f=\infty$

And in this problem, the distance of the object from the mirror is

$d_o = 2.5 m$

Substituting and solving for $d_i$ , we find the location of the image:

$0=\frac{1}{d_o}+\frac{1}{d_i}\\d_i = -d_o = -2.5 m$

And the negative sign means that the image is located behind the mirror, so it is virtual.

Learn more about reflection and refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

6 0

3 years ago

A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is

Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
$P = i^2R$

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

$P = (2.0^2)(50) = \boxed{\text{ B. }200 W}$

7 0

2 years ago