A periodic wave of frequency f and speed v travel what distance in one period?

John runs around a 126.5 m circular track 3.5 times in 4.17 minutes. What is his average speed?

sdas [7]

Average speed = distance traveled / time

average speed = (126.5 m * 3.5 laps) / (4.17 min)

= 106.2 m/min

6 0

3 years ago

Which wave has a disturbance that is parallel to the wave motion?

Leto [7]

The answer is D.<span>longitudinal</span>

6 0

3 years ago

Where is the potential energy equal to zero?

s2008m [1.1K]

Answer:

im sure your already past this but it's E.

Explanation:

This is because in this case potential energy is linear to height, which means that the higher the more potential energy.

4 0

3 years ago

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c

Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is $v_A= 4m/s$

Explanation:

From the question we are told that

The length of the string is $L = 1m$

The initial speed of block A is $u_A$

The final speed of block A is $v_A = \frac{1}{2}u_A$

The initial speed of block B is $u_B = 0$

The mass of block A is $m_A = 7kg$ gh

The mass of block B is $m_B = 2 kg$

According to the principle of conservation of momentum

$m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

$m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

$m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

$m_A \frac{u_A}{2} = m_Bv_B$

making $v_B$ the subject of the formula

$v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

$v_B =\frac{7 u_A}{4}$

This $v__B$ is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of $v__B'$ at the top of the vertical circle

The angular centripetal acceleration would be mathematically represented

$a= \frac{v^2_{B}'}{L}$

Note that this acceleration would be toward the center of the circle

Now the forces acting at the top of the circle can be represented mathematically as

$T + mg = m \frac{v^2_{B}'}{L}$

Where T is the tension on the string

According to the law of energy conservation

The energy at bottom of the vertical circle = The energy at the top of

the vertical circle

This can be mathematically represented as

$\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above

$(T + mg) L = m v^2_{B}'$

Substitute this into above equation

$\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = T + 5mgL$

The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero

This is mathematically represented as

$\frac{49 mv_A^2}{16} = 5mgL$

making $v_A$ the subject

$v_A = \sqrt{\frac{80mgL}{49m} }$

substituting values

$v_A = \sqrt{\frac{80* 9.8 *1}{49} }$

$v_A= 4m/s$

6 0

3 years ago

Question 9 of 27

Ksju [112]

The power of man performing 500 J of work in 8 seconds is 62.5 J/s.

Power can be defined as the pace at which work is completed in a given amount of time.

Horsepower is sometimes used to describe the power of motor vehicles and other machinery.

The pace at which work is done on an item is defined as its power. Power is a temporal quantity.

Which is connected to how quickly a project is completed.

The power formula is shown below.

Power = Energy / Time

Power = E / T

Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.

Here we have given Energy as 500 J and Time as 8 second.

Power = Energy / Time

Power = 500 / 8 Joule / sec

Power = 250 / 4 Joule / sec

Power = 125 / 2 Joule / sec

Power = 62.5 Joule / sec or 62.5 watt

Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.

So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.

Learn more about Power here:

brainly.com/question/1634438

#SPJ10

6 0

2 years ago