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1 year ago

We see a bolt of lightning and 4 s later we hear the thunderclap. If the speed of

Physics

1 answer:

Alex73 [517]1 year ago

6 0

Answer:

i guess 0.8 miles away

Explanation:

mathematically:-

speed is given in question

Time is given

and we have to find distance

simply by using speed formula (s) = d/t

we get answer

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11. Examine the table provided. Which of the objects is producing

ivann1987 [24]

Answer:

b is the answer tennis ball

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2 years ago

The site<br> of proteins is called what

ValentinkaMS [17]

Answer: Ribosomes are the site of protein synthesis, and they convert the coded information in messenger RNA into an actual protein molecule.

I hope this helped! OvO

8 0

2 years ago

Read 2 more answers

What is the angle of the 2nd order bright fringe produced by two slits that are 8.25x10-5m apart if the wavelength of the incide

dexar [7]

In order to calculate the angle, we can use the formula below for a constructive interference (the interference is constructive because the fringe is bright):

$d\sin\theta=m\lambda$

Where d is the distance between the slits, m is the order of the interference and lambda is the wavelength.

So, using d = 8.25 * 10^-5, m = 2 and lambda = 4.5 * 10^-7, we have:

$\begin{gathered} 8.25\cdot10^{-5}\cdot\sin\theta=2\cdot4.5\cdot10^{-7}\\ \\ \sin\theta=\frac{9\cdot10^{-7}}{8.25\cdot10^{-5}}\\ \\ \sin\theta=1.091\cdot10^{-2}\\ \\ \theta=0.625° \end{gathered}$

Therefore the correct option is the second one.

5 0

7 months ago

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

2 years ago

Consider an electron, of charge magnitude e = 1.602 10-19 C and mass me = 9.11 10-31 kg, moving in an electric field with an ele

Tems11 [23]

Answer: v= 7.509 x 10^6 m/s

B) the lower plate because the electron is negatively charged

Explanation:

From the question

Electronic charge (q) =1.602 x 10^-19c

Electric field intensity (E) = 8 x 10² = 800N/C

Mass of electron (m) = 9.11 x 10^-31 kg

Length of plate (L) = 50cm=0.5m

Distance between plates (D) = 20cm = 0.2m

Since the electron is entering a uniform electric field, the resulting motion will be of a constant acceleration and can be defined by the equations of motion with a constant acceleration.

From newton's law of motion.

F= ma

The force (F) is coming from the electric field which is

F=Eq.

Thus F = 800 x 1.602 x 10^-19

F = 1.2816 x 10^-16 N

Acceleration of electron (a) = F/m where m is the mass of electron (given above)

Hence

a = 1.2816 x 10^-16 / 9.11 x 10^-31

a= 1.41 x 10¹⁴ m/s².

Using Newton laws of motion to get velocity, we recall that v²= u² + 2ad

Where v is final velocity, u is initial velocity (zero in this case because the electron starts it motion from rest), a= acceleration, d= distance traveled (which in this case is distance between plates)

v² = 0² + 2(1.41 x 10¹⁴) x 0.2

v² = 5.64 x 10¹³

Thus v = √ 5.64 x 10¹³

v = 7.509 x 10^6 m/s

Since the electric field is upwards it denotes that the positive plate is downward and the negative is upward ( this is because electric flux from a positive charge has an outward flow and that from a negative charge has an inward flow. So from our questions, if the electric field is upward, it means it is starting from the bottom plate which will be positive) the electron (which is negatively charged) will be attracted to the positive plate which is downward for this question of ours.

So therefore, the electron is attracted to the downward plate because it (electron) is negative

Option B

4 0

2 years ago