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Helga [31]
2 years ago
9

A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only i

f the sharks are separated by 3.60 meters, one being that much farther away than the other. (a) If the ultrasound has a frequency of 105 kHz, show this ability is not limited by its wavelength m < 3.50 m (b) If this ability is due to the dolphin's ability to detect the arrival times of echoes, what is the minimum time difference (in ms) the dolphin can perceive? ms
Physics
1 answer:
Vlad1618 [11]2 years ago
6 0

Answer:

a) Wavelength of the ultrasound wave = 0.0143 m <<< 3.5m, hence its ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = Period of oscillation = 0.00952 ms

Explanation:

The frequency of the ultrasound wave = 105 KHz = 105000 Hz. The speed of ultrasound waves in water ≈ 1500 m/s. Wavelength = ?

v = fλ

λ = v/f = 1500/105000 = 0.0143 m <<< 3.5m

This value, 0.0143m is way less than the 3.5m presented in the question, hence, this ability is not limited by the ultrasound's wavelength.

b) Minimum time difference between the oscillations = The period of oscillation = 1/f = 1/105000 = 0.00000952s = 0.00952 ms

Hope this helps!

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A raft is made of a plastic block with a density of 650 kg/m 3 , and its dimensions are 2.00 m à 3.00 m à 5.00 m. 1. what is the
cupoosta [38]
1) The volume of the raft is the product between the lenghts of its three dimensions:
V = (2.00 m)(3.00m)(5.00m)=30 m^3

2) The mass of the raft is the product between its density, d, and its volume, V:
m=dV=(650 kg/m^3)(30 m^3)=19500 kg

3) The weight of the raft is the product between its mass m and the gravitational acceleration, g=9.81 m/s^2:
W=mg=(19500 kg)(9.81 m/s^2)=1.91 \cdot 10^5 N

4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
W_a = W- \rho_W V_{disp} g
where \rho _W = 1000 kg/m^3 is the water density and V_{disp} is the volume of displaced fluid.
The density of the raft (650 kg/m^3) is smaller than the water density (1000 kg/m^3), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
W=B=\rho _W V_{disp} g
and therefore, the apparent weight will be zero:
W_a = W-B=W-W=0

5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
B=\rho_W V_{disp} g
When the raft is completely immersed in the water, the volume of fluid displaced V_{disp} is equal to the volume of the raft, V_{disp}=V. Therefore the buoyancy in this situation is
B= \rho_W V g = (1000 kg/m^3)(30 m^3)(9.81 m/s^2)=2.94 \cdot 10^5 N
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
B=W=1.91 \cdot 10^5 N

6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
mg = m_{disp} g
where m_{disp}= \rho_W V_{disp} is the mass of the displaced water. From the previous equation, we obtain that m_{disp}=m=19500 kg.

7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship m=dV we can rewrite m=m_{disp} as:
d V =d_W V_{disp}
and so
V_{disp}= \frac{d V}{d_W}= \frac{(650 kg/m^3)(30m^3)}{1000kg/m^3}= 19.5 m^3

8) The volume of water displaced is (point 7) 19.5 m^3. This volume is now "filled" with part of the volume of the raft, therefore 19.5 m^3 is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, 30 m^3:
\frac{19.5 m^3}{30 m^3}\cdot 100= 65 \%
Viceversa, the volume of raft above the water level is 30 m^3-19.5 m^3 = 10.5 m^3. Therefore, the fraction of volume of the raft above water level is
\frac{10.5 m^3}{30 m^3}\cdot 100 = 35 \%

9) Let's repeat steps 5-8 replacing \rho _W, the water density, with \rho_E=806 kg/m^3, the ethanol density.

9-5) The buoyant force is given by:
B=\rho _E V_{disp} g = (806 kg/m^3)(30 m^3)(9.81 m/s^2)=2.37 \cdot 10^5 N
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.

9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft: 
m_E = m = 19500 kg

9-7) Using the relationship d= \frac{m}{V}, we can find the volume of displaced ethanol:
V_E =  \frac{m}{d_E} = \frac{19500 kg}{806 kg/m^3}=24.2 m^3

9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced: 24.2 m^3. Therefore, the fraction of raft's volume below the ethanol level is 
\frac{24.2 m^3}{30 m^3}\cdot 100 = 81 \%
Consequently, the raft's volume above the ethanol level is 
30 m^3 - 24.2 m^3 = 5.8 m^3
and the fraction of volume above the ethanol level is
\frac{5.8 m^3}{30 m^3}\cdot 100 = 19 \%
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2 years ago
A car covers 72 kilometers in the first hour of its journey. In the next hour, it covers 90 kilometers. What is the amount of wo
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The amount of work done can be solved using the formula:

Work = Force x Distance = Change in kinetic energy

Kinetic energy can be solved using the formula: KE = (1/2)*m*v^2
So, change in kinetic energy = (1/2)*m*(Vf)^2 - (1/2)*m*(Vo)^2

Where:

Vf = final velocity = 90 kph = 25 m/s
Vo = initial velocity = 72 kph = 20 m/s

substituting the given values:

Work = (1/2)*2500*(25^2) - (1/2)*2500*(20^2) = 281250 J, which can also be expressed as 2.8 x 10^5 Joules.

Among the choices, the correct answer is A.
4 0
2 years ago
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When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What
ser-zykov [4K]

Answer:

Electric field, E=4\times 10^5\ N/C

Explanation:

It is given that,

Magnitude of charge, q_o=2\ nC=2\times 10^{-9}\ C

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We need to find the electric field at the origin. It is given by :

F=q_o\times E

E=\dfrac{F}{q_o}

E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}

E=4\times 10^5\ N/C

So, the electric field at the origin is 4\times 10^5\ N/C. Hence, this is the required solution.

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2 years ago
(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the
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If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he
shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

F = \frac{GMm}{R^2} = mg

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

(mg)'= \frac{GMm}{(R+h)^2}

we need to find "h"

taking the ratio of two:

\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

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