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2 years ago

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

Physics

2 answers:

Gnom [1K]2 years ago

8 0

Answer:

Explanation:

Number of times Haley could fly between two flowers depends on the two basic parameters.

1- Distance between two flowers.

2- Haley's speed during this.

As the time interval is given which is of 1 minute, it completely depends on where these two flowers are located and how speedily can Haley fly.

If the distance between flowers is less, number of turns will be more and if the distance between flowers is more, Haley will probably take time to travel between the two.

dmitriy555 [2]2 years ago

7 0

30 seconds is the answer

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2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

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2 years ago

Determine the maximum weight of the bucket that the wire system can support so that no single wire develops a tension exceeding

stellarik [79]

Let there be N number of wires.

Maximum tension a wire can withstand = 100 lb

so, Total tension N wires can withstand = 100 N

now, total tension in N wires = Maximum weight of bucket

100 N = W

so, W = 100N

W is the weight of bucket and 100N is its maximum value.

8 0

2 years ago

There is an electric field in the region between the two plates. The magnitude of this electric field is ed. This imposes anothe

krok68 [10]

Answer:

it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

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The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.

To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

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5 0

2 years ago

The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet

Naily [24]

Answer:

Speed of comet before collision is

$v_{2_{i}}=-2.5\times10^{3}\quad km/h$

Explanation:

Correction: (As stated after collision comet moves away from moon so velocity of moon and moon and comet must be opposite in direction. as spped of moon after collision is −4.40 × 10^2km/h so that comet's must be 5.740 × 10^3km/h instead of -5.740 × 10^3km/h)

Solution:

$mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h$

Case is considered as partially inelastic collision, by conservation of momentum

$m_{1}v_{1_{i}}+m_{2}v_{2_{i}}=m_{1}v_{1_{f}}+m_{2}v_{2_{f}}\\\\m_{1}v_{1_{i}}+0.5m_{1}v_{2_{i}}=m_{1}v_{1_{f}}+0.5m_{1}v_{2_{f}}\\\\v_{1_{i}}+0.5v_{2_{i}}=v_{1_{f}}+0.5v_{2_{f}}\\\\v_{2_{i}}=2(v_{1_{f}}+0.5v_{2_{f}}-v_{1_{i}})\\\\v_{2_{i}}=2(-4.40 \times 10^2+0.5(5.740 \times 10^3)-3.680 \times 10^3 )\\\\v_{2_{i}}=-2.5\times10^{3}\quad km/h$

8 0

2 years ago

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

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