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Elenna [48]
3 years ago
10

A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.

The observed frequency of the hum will be 152 Hz as the bumblebee flies away from you.
Physics
2 answers:
Veronika [31]3 years ago
8 0

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.

lyudmila [28]3 years ago
3 0

Answer:

<em>Less Than</em>

Explanation:

A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s. The observed frequency of the hum will be Less Than 152 Hz as the bumblebee flies away from you.

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A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire
Luda [366]

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

F=ILB sin \theta

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

\theta is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

\theta=90^{\circ} since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T

3 0
3 years ago
We need to find the launch velocity of our new marble launcher. we know that it will launch a 25g marble to a distance of 73 cm,
White raven [17]

The launch velocity of the marble launcher is 34.65 m/s

Given that the launch velocity of marble launcher, launches a 25g marble to a distance of 73 cm (0.73 m) and the marble roll up to 6.2 meters before stopping. The launch height is 20 cm (0.2 m).

The time for landing can be calculated by the second equation of motion formula:

h = ut + \frac{1}{2}gt^{2}

Let u = 0

0.2 = 0×t + \frac{1}{2} × 9.8 × t^{2}

t^{2} = \frac{0.2}{4.9}

t^{2} = 0.04

t = 0.2s

Now, the launch velocity of the marble launcher can be calculated by:

Speed = Distance / Time

Speed = \frac{0.73 + 6.3}{0.2}

Speed = \frac{6.93}{0.2}

Speed = 34.65 m/s

Therefore, the launch velocity of the marble launcher is 34.65 m/s

Know more about Launch velocity: -brainly.com/question/18883779

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3 0
1 year ago
Five moles of an ideal monatomic gas with an initial temperature of 121 ∘C expand and, in the process, absorb an amount of heat
liq [111]

Answer:

  T₂ ≈ 107.85∘C

Explanation:

The question didn't state if the volume is constant or not as such, we can apply the first law of thermodynamic

From the first law of thermodynamic,

ΔU =  Q - W

where ΔU = Internal Energy, Q = Quantity of heat absorbed, W = Amount of work done.

Q = 1200 J and W = 2020 J

∴ ΔU = 1200 -2020 = -820 J.

Using the ideal gas equation,

ΔU = 3/2nRΔT...................................equation 1

where n = number of moles, R = Molar gas constant, ΔT = Change in temperature = (T₂ - T₁).

Modifying equation 1,

ΔU = 3/2nR(T₂ -T₁)...............................equation 2.

making T₂ the subject of the relation in equation 2,

T₂ =  {2/3(ΔU)/nR}+T₁........................ equation 3

where T₁=121∘C, R= 8.314 J / mol, n=5 moles, ΔU=-820 J

Substituting these values into equation 3,

∴ T₂ ={ 2/3(-820)/(5×8.314)}+121

   T₂ = {2×(-820)/ (3×5×8.314)}+121  

   T₂={-1640/124.71}+ 121

   T₂ = {-13.151} + 121

 ∴T₂  = 121 - 13.151 = 107. 849∘C

    T₂ ≈ 107.85∘C

3 0
3 years ago
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