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3 years ago

On March 21, a stick casts the following shadow. What is the most likely time of day?

Physics

2 answers:

Tpy6a [65]3 years ago

8 0

Answer: Yep it is 9:00

Explanation:

jenyasd209 [6]3 years ago

5 0

Answer:

9:00 AM

Explanation:

I took the test and that was the answer

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Based on the free-body diagram, the net force acting<br> on this firework is

Strike441 [17]

0N. The net force acting on this firework is 0.

The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.

Based on the free-body diagram, we have:

The force of gases is Fgases = 9,452N

The force of the rocket Frocket = -9452

Then, the net force acting is:

Fnet = Fgases + Frocket

Fnet = 9,452N - 9,452N = 0N

4 0

3 years ago

Read 2 more answers

Which event is an example of condensation?

ale4655 [162]

Answer: D

If the fog disappears when the Sun comes out, then this is an example of condensation because:

the Sun actually dries up the fog, and it makes it into higher clouds.

Hope this helps you!

3 0

3 years ago

A book is to be produced with pages of thickness 0.125mm.If the book is to be exactly 1cm thick what is the maximum number of pa

miskamm [114]

Answer:

1cm=10mm

10cm/0.125mm

80pages

5 0

2 years ago

An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result

mylen [45]

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

$s=ut+\frac{1}{2}gt^{2}$

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

$s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}$

So the distance covered by the acorn is 3.136 m.

8 0

3 years ago

The conducting path between the right hand and the left hand can be modeled as a 9.0-cm-diameter, 140-cm-long cylinder. The aver

Crank

Answer:

The potential difference is 121.069 V

Solution:

As per the question:

Diameter of the cylinder, d = 9.0 cm = 0.09 m

Length of the cylinder, l = 40 cm = 1.4 m

Average Resistivity, $\rho = 5.5\ \Omega-m$

Current, I = 100 mA = 0.1 A

Now,

To calculate the potential difference between the hands:

Cross- sectional Area of the Cylinder, A = $\pi (\frac{d}{2})^{2} = 6.36\times 10^{- 3}\ m^{2}$

Resistivity is given by:

$\rho = R\frac{A}{l}$

$R = \rho \frac{l}{A}$

$R = 5.5\times \frac{1.4}{6.36\times 10^{- 3}} = 1210.69\ \Omega$

Now, using Ohm's Law:

V = IR

$V = 0.1\times 1210.69 = 121.069\ V$

4 0

3 years ago