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stira [4]
3 years ago
13

Dolphins rely on echolocation to be able to survive in the ocean. In a 20 °C ocean, a dolphin produces an ultrasonic sound with

a frequency of 125 kHz. What is the wavelength of this sound, in meters?While remaining stationary, the dolphin emits a sound pulse and receives an echo after 0.220 s. How far away, in meters, is the reflecting object from the dolphin?
Physics
1 answer:
Firdavs [7]3 years ago
8 0

Answer:

wavelength =  0.01 m

distance = 162.8 m

Explanation:

Given that;

Speed of sound in water = 1,480 meters per second

Frequency of ultrasound = 125KHZ

From=

v=λf

v= speed of sound

λ= wavelength of sound

f= frequency of sound

λ= 1,480 ms-1/125 * 10^3 Hz

λ= 0.01 m

From

v = 2x/t

where;

v= velocity of sound in water

x= distance traveled

t = time taken

x = vt/2

x = 1,480 ms-1 *  0.220 s/2

x= 162.8 m

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A hypothesis is _______.
Arada [10]

Answer:

A hypothesis is a basically a theory proposed to a subject or refrence to an act with limited evidences.

6 0
3 years ago
Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes
pantera1 [17]

Answer:

The  magnitude of F1 is

|F1|=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.  

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.  

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

\displaystyle -2F\ sin21^o+F\ cos\alpha =0

The sum of the vertical components of F1 and F2 must equal the total force Ft

\displaystyle -2F\ cos21^o+F\ sin\alpha =460

Solving for \alpha in the first equation

\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o

\displaystyle cos\alpha =0.717=>\alpha =44.214^o

\displaystyle F(2cos21^o+sin\alpha)=460

\displaystyle F=\frac{460}{2cos21^o+sin\alpha}

\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}

\displaystyle F=179.37\ N

The  magnitude of F1 is

|F1|=2*F=358.74 \ N

The magnitude of F2 is

|F2|=179.37\ N

And the direction of F2 is

\alpha = 44.214^o

4 0
3 years ago
S Problem Set<br> 2.) 6.4 x 109 nm to cm
anyanavicka [17]

Answer:

6.4\cdot 10^2 cm

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

1 nm = 10^{-9} m

So we have:

6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m

Now we can convert from metres to centimetres, keeping in mind that

1 m = 10^2 cm

So, we find:

6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm

8 0
3 years ago
a 3000 kg and a 7000 kg Mass attract each other with a force of 0.0015 N. What distance separates the two objects (Radius) (Plea
frutty [35]

Answer:

<h3> 3.057m</h3>

Explanation:

According to law of gravitation;

F = GMm/d²

G is the universal gravitation

M and m are the masses

d is the distance between the masses

d² = GMm/F

d² = 6.67408 × 10-11 *3000*7000/0.0015

d² = 140.15568*10^-5/0.0015

d² = 1.4016*10^-3/0.0015

d² = 1.4016*10^-3/1.5*10^-3

d²  = 0.9344*10

d² = 9.344

d = √9.344

d = 3.057m

Hence the distance between the two objects is  3.057m

3 0
3 years ago
To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i
ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

m_1(-v_1)+m_2v_2=(m1+m2)(-v2)

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s

hence, the time is t=0.42 s

4 0
3 years ago
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