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3 years ago

The portion of a sound wave with higher-than-normal pressure is called a(n)________

Physics

1 answer:

Leokris [45]3 years ago

4 0

Answer:

Condensation

Explanation:

Increased pressure is called condensation or compressions because they are regions of high air pressure whilst rarefractions are low air pressure.

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An object's distance from a converging lens is 3.78 times the focal length. a) Determine the location of the image. Express the

Romashka-Z-Leto [24]

Answer:

a) v=1.36f

b)m=.36

c) real

d) Inverted

Explanation:

In this question we have given,

object distance from converging lens,u=-3.78f

focal length of converging lens,=f

we have to find location of image,v=?

and magnification,m=?

we know that u, v and f are related by following formula

$\frac{1}{f} =\frac{1}{v}- \frac{1}{u}$ .............(1)

put values of f u in equation (1)

we got,

$\frac{1}{f} =\frac{1}{v}- \frac{1}{-3.78f}$

$\frac{1}{f}-\frac{1}{3.78f} =\frac{1}{v}$

$\frac{2.78}{3.78f} =\frac{1}{v}$

v=1.36f

b) Magnification, $m=\frac{v}{u} \\m=\frac{1.36f}{3.78f}\\m=.36$

Here the object is located further away from the lens than the focal point therefore image is real image and inverted.

d) image is inverted

7 0

3 years ago

When do you produce more pressure on the ground...standing or laying down?

ycow [4]

When you are Standing you produce more pressure on the ground

7 0

3 years ago

A soccer player icks a rock horizontally off a 40m high cliff into a pool f water if the player hears the sound of the splash s

Semenov [28]

Answer:

v = 9.936 m/s

Explanation:

given,

height of cliff = 40 m

speed of sound = 343 m/s

assuming that time to reach the sound to the player = 3 s

now,

time taken to fall of ball

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 40}{9.8}}$

t = 2.857 s

distance

d = v x t

d = v x 2.875

time traveled by the sound before reaching the player

$t_0 = t - t_{fall}$

$t_0 = 3 - 2.875$

$t_0 = 0.143 s$

distance traveled by the wave in this time'

r = 0.143 x 343

r= 49.05 m

now,

we know.

d² + h² = r²

d² + 40² = 49.05²

d =28.387 m

v x 2.875=28.387 m

v = 9.936 m/s

7 0

3 years ago

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

4 0

3 years ago

Dan is gliding on his skateboard at 4.00m/s . He suddenly jumps backward off the skateboard, kicking the skateboard forward at 6

frozen [14]

To solve this problem we will apply the concepts related to the conservation of the Momentum. For this purpose we will define the momentum as the product between mass and velocity, and by conservation the initial momentum will be equal to the final momentum. Mathematically this is,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$