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ololo11 [35]
3 years ago
14

The case whereby sound waves are heard at either a higher or lower pitch due to an object's motion is called:

Physics
2 answers:
omeli [17]3 years ago
5 0
Doppler effect!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ! ! ! ! ! ! ! ! ! ! ! !
e-lub [12.9K]3 years ago
5 0

Answer:

the Doppler effect

Explanation:

The doppler effect is what we called the effect that happens when the wavelength or frequency of a sound changes because the observer is moving relatevily form the object, could be moving closer or moving further away, but the wavelenght and the frequency changes alongside with the movement.

You might be interested in
What is keeping all variables the same except for the one being tested an example of?
stepan [7]

Answer:

A fair test.

Explanation:

Hi, a fair test is used to do scientifically valuable experiments, is a controlled investigation to answer a scientific question.

In a fair test two or more things are compared.

It consists in changing only one factor (the one bieng tested) and keeping all the other conditions the same during an experiment.

The factor is called a variable.

8 0
3 years ago
Read 2 more answers
NEED HELP ASAP
Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0
2 years ago
What are the top two gasses in the earth’s atmosphere ?
ch4aika [34]

Answer:

Nitrogen and oxygen are by far the most common; dry air is composed of about 78% nitrogen (N2) and about 21% oxygen (O2). Argon, carbon dioxide (CO2), and many other gases are also present in much lower amounts; each makes up less than 1% of the atmosphere's mixture of gases.

7 0
3 years ago
Leandra is baking cookies , and they are ready to come out of the oven . Before she takes them out of the oven , she puts her mi
ch4aika [34]

Answer: Leandra puts on her mittens because if you do not you will burn your self, due to extremely high temperatures.

Explanation:

8 0
3 years ago
uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station
Alex787 [66]

Answer:

The magnitude of the tangential velocity is v= 0.868 m/s

The magnitude of the resultant acceleration at that point is  a = 4.057 m/s^2

Explanation:

From the question we are told that

      The mass of the uniform disk is m_d = 40.0kg

       The radius of the uniform disk is R_d = 0.200m

       The force applied on the disk is F_d = 30.0N

Generally the angular speed i mathematically represented as

             w = \sqrt{2 \alpha  \theta}

Where \theta is the angular displacement given from the question as

           \theta  = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}

                 =1.257\  rad

   \alpha is the angular acceleration which is mathematically represented as

                    \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

    The moment of inertial is mathematically represented as

                     I = \frac{1}{2} m_dR^2_d

Substituting values

                    I = 0.5 * 40 * 0.200^2

                        = 0.8kg \cdot m^2

Considering the equation for angular acceleration

               \alpha = \frac{torque }{moment \ of  \ inertia}  = \frac{F_d * R_d}{I}

Substituting values

               \alph\alpha = \frac{(30.0)(0.200)}{0.8}

                   = 7.5 rad/s^2

Considering the equation for angular velocity

    w = \sqrt{2 \alpha  \theta}

Substituting values

     w =\sqrt{2 * (7.5) * 1.257}

         = 4.34 \ rad/s

The tangential velocity of a given point on the rim is mathematically represented as

                 v = R_d w

Substituting values

                    = (0.200)(4.34)

                     v= 0.868 m/s

The radial acceleration at hat point  is mathematically represented as

            \alpha_r = \frac{v^2}{R}

                  = \frac{0.868^2}{0.200^2}

                 = 3.7699 \ m/s^2

The tangential acceleration at that point is mathematically represented as

               \alpha _t = R \alpha

Substituting values

           \alpha _t = (0.200) (7.5)

                 = 1.5 m/s^2

The magnitude of resultant acceleration at that point is

                 a = \sqrt{\alpha_r ^2+ \alpha_t^2 }

Substituting values

                a = \sqrt{(3.7699)^2 + (1.5)^2}

                   a = 4.057 m/s^2

         

7 0
3 years ago
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