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natita
1 day ago
7

How many times could Haley fly bewteen the two flowers in 1 minute (60 seconds)​

Physics
2 answers:
Sav [2.2K]1 day ago
8 0
The number of times Haley could complete the flight between the two flowers hinges on two fundamental parameters. First, the distance separating the two flowers, and second, Haley’s speed during this flight. The provided time frame is 1 minute, so the outcome is entirely dependent on the positioning of these two flowers and Haley’s flying speed. The shorter the distance between them, the more flights she can achieve; conversely, if they are further apart, Haley will take longer to travel between them.
Yuliya22 [2.4K]1 day ago
7 0
The answer is 30 seconds.
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Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,
Sav [2226]

Answer:

All three pendulums will have the same angular frequencies.

Explanation:

For a simple pendulum, the time period using the approximation sin(\theta )\approx \thetais expressed as:

T=2\pi\sqrt{\frac{l}{g}}

The angular frequency \omega is defined as

\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}

Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.

6 0
4 days ago
Two very small 8.55-g spheres, 15.0 cm apart from center to center, are charged by adding equal numbers of electrons to each of
ValentinkaMS [2425]

Answer:

1.43 x 10¹⁷.

They will move away from each other.

Explanation:

The force acting on each charged sphere is determined as F = mass x acceleration

= 8.55 x 10⁻³ x 25 x 9.8

= 2.095 N

Assuming Q is the charge on each sphere

F = \frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}

Using the values, 2.095 = \frac{9\times10^9\times Q^2}{(15\times10^{-2})^2}

We find that Q² = \frac{2.095\times(15)^2\times10^{-4}}{9\times10^9}

Thus, Q = 2.289 X 10⁻⁶

The quantity of electrons = Charge / charge of a single electron

= \frac{2.289\times10^{-6}}{1.6\times10^{-19}}

=1.43 x 10¹³.

They will accelerate away from each other.

4 0
14 days ago
Read 2 more answers
The amount of electric energy consumed by a 60.0-watt lightbulb for 1.00 minute could lift a
Sav [2226]

Answer:

Explanation:

For a 60W light bulb used for 1 minute:

P = 60 W

t = 1 minute = 60 seconds

This energy is capable of lifting an object weighing 10N.

W = 10N

This indicates conversion of electrical energy into potential energy.

Let's calculate the electrical energy:

Power describes the rate of work done.

Power = Work / time

Thus, work = power × time

Work = 60 × 60

Work = 3600 J

Potential energy calculation:

P.E = mgh

Where the weight is given by:

W = mg

Therefore, P.E = W·h

P.E = 10·h

Thus, we equate:

Potential energy = Electrical energy

P.E = Work

10·h = 3600

Dividing both sides by 10 gives:

h = 3600 / 10

h = 360m

The object can be lifted to a height of 360m.

6 0
27 days ago
An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. in
ValentinkaMS [2425]
Since the absolute values of the charges are identical, the changes in potential energy remain equivalent. Consequently, the changes in kinetic energy will also match. We have:

1 = Ke/Kp = m_e * v_e^2 / m_p * v_p^2, which simplifies to:

v_e/v_p = sqrt(m_p/m_e),

indicating that the velocity of the electron is sqrt(m_p/m_e) times greater than that of the proton.
4 0
16 days ago
Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45
Sav [2226]

Answer:

Please refer to the explanation

Explanation:

Race distance is 5km

Top speed = 45 yards

Converting yards to kilometers:

1km equals 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x indicates the maximum distance he can sustain his highest speed in kilometers.

Thus, from the data available, we can determine that Lamar will not be able to maintain his maximum speed for the full 5km race, as he can only sustain it for 0.0411 kilometers.

5 0
11 days ago
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