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Anna71 [15]
3 years ago
9

A loudspeaker located in air generates sound waves of frequency 1,000 Hz. Some of these sound waves enter a pool of water, where

the speed of sound is 1,400 m/s. What is the wavelength of the sound waves in water?
Physics
1 answer:
prisoha [69]3 years ago
8 0

Answer:

1.4 m

Explanation:

v = Speed of sound in water = 1400 m/s

f = Frequency of sound = 1000 Hz

\lambda = Wavelength

When we multiply the frequency and the wavelength of a wave we get the velocity of sound in that medium

v=f\lambda\\\Rightarrow \lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{1400}{1000}\\\Rightarrow \lambda=1.4\ m

The wavelength of the sound waves in water is 1.4 m

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One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa
Ludmilka [50]

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

<u>ΔP.E = 6.48 x 10⁸ J</u>

7 0
3 years ago
jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh
Greeley [361]

Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

  • velocity of jet, v_j=500\ mph
  • direction of velocity of jet, east relative to the ground
  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

8 0
3 years ago
The device that measures current in a wire by using the deflections of an electromagnet in an external magnetic field is
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A galvanometer hope this helps

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6 0
3 years ago
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A car takes 15 minutes to travel along a road that is 20 km long.
iogann1982 [59]

Answer:

C

Explanation:

15. 20

x4. x4

60. 80

80km/h

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A trunk is dragged 3.0 meters across an attic floor by a force of 2.0N how much positive work is done on the trunk
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The formula for work is

F*d

Therefore work=2.0N*3.0=6N*m

8 0
3 years ago
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