Which statement is TRUE?

A mountain climber at the peak ha ___________ energy.

Levart [38]

Its Kinetic, hope this helps you

7 0

3 years ago

A honey bee flaps its wings 200 times per second. How much time is required for one wingbeat? Give your answer in milliseconds.

kakasveta [241]

A bees wings move so rapidly that studying them, even seeing them, has proved difficult.
The honeybees have a rapid wing beat honeybee flaps its wings 230 times every second.

Therefore, a bee flaps its wings over 200 times in about 1ms

3 0

3 years ago

Read 2 more answers

This term means that we are interacting with an item at a high-level, with lower-level internal details hidden from the user.

katovenus [111]

Option d) abstraction is the correct answer.

Abstraction is the process of taking away or removing characteristics from something in order to reduce it to a set of essential characteristics. In object-oriented programming, abstraction is one of three central principles Through this process, a programmer hides all but the relevant data about an object in order to reduce complexity and increase efficiency.

Learn more about Abstraction here:

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7 0

2 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

3) connect two lamps to a power supply in series and current drawn from the power supply is is. connect the same two lamps in pa

notka56 [123]

The current drawn by the series circuit is(R₁ + R₂)/R₁R₂ times the current drawn by the parallel circuit.

Let the resistance of the two lamps are R₁ and R₂.

Then the equivalent resistance in series combination is: R = R₁ + R₂.

And, the equivalent resistance in parallel combination is:

r = R₁R₂/(R₁ + R₂).

So, if the supply voltage is V,

Then, current drown in series combination; $i_s$ = V/R = V/(R₁ + R₂)

And, current drown in parallel combination; $i_p$ = V/r = V(R₁ + R₂)/R₁R₂

So , $\frac{i_s}{i_p}$ = [ V/(R₁ + R₂)] /[V(R₁ + R₂)/R₁R₂]

= (R₁ + R₂)/R₁R₂

Hence, the ratio of current drawn in series and current drown in parallel is (R₁ + R₂)/R₁R₂. So, he current drawn by the series circuit is(R₁ + R₂)/R₁R₂ times the current drawn by the parallel circuit.

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4 0

1 year ago