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Contact [7]
3 years ago
11

NU U

Physics
2 answers:
stich3 [128]3 years ago
7 0

Answer:

Explanation:

I cannot help either is there a diagram for this??

OlgaM077 [116]3 years ago
5 0
I can’t help you with your problem because you’re not showing a picture of any sort of diagram
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PLEASEEEEEE HELP
emmainna [20.7K]

Answer:

8.89 m/s² west

Explanation:

Assume east is +x.  Given:

v₀ = 120 m/s

v = 0 m/s

t = 13.5 s

Find: a

v = at + v₀

0 m/s = a (13.5 s) + 120 m/s

a = -8.89 m/s²

a = 8.89 m/s² west

8 0
3 years ago
Which of the following is a potential result of preparing appropriately before starting an experiment in a lab?
oksano4ka [1.4K]

Answer:

Your project goes well.

Explanation:

Because that's how it works.

6 0
3 years ago
Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll
Andrew [12]

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

Q = \dfrac{4mME }{(m+M)^2}

However; from the total stopping power & power loss of the electron;

\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

∴

radiational energy loss = collisional energy loss  =\dfrac{82 \times 1.9}{800}

= 0.19475

b)

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.

4 0
3 years ago
All of the following would be questions that could be scientifically investigated except:
ra1l [238]
All of the following would be questions that could be scientifically investigated except A.
That is an opinion and cannot become a fact.
6 0
4 years ago
Read 2 more answers
A stone whirled at the end of the a rope 30cm long, makes 10 complete resolution in 2 seconds Find: A. The angular velocity in r
lord [1]

Answers:

A: Angular velocity \omega=31.40 \frac{r a d}{s}

B: Linear velocity v=9.42 \frac{m}{s}

C: Linear Distance d=47.1 \mathrm{m}

Given:

Radius of the rope r=30cm=0.3m

Angular distance\Delta \theta=10 revolutions

Time taken t=2seconds

To find:

A: Angular velocity in radians

B: Linear speed

C: Distance covered in 5 seconds

<u>Step by Step Explanations:</u>

Solution:

A: Angular velocity in radians;

According to the formula, Angular velocity can be calculated as

Angular Velocity = angular distance/ time

\omega=\Delta \theta / \Delta t

Where \omega=Angular velocity

\Delta \theta=Angular distance=10 revolutions

Changing revolutions to radians multiply with 2 \pi, so that we get

=10 \times 2 \pi

=10 \times 2(3.14)  

=62.80 rad/rev

\Delta t =Change in time

Substitute the known values in the above equation we get

\omega=62.80 / 2  

\omega=31.40 \frac{r a d}{s}

B. Linear speed of the rope;

As per the formula

Linear speed = angular speed × radius

v=\omega \times r  

Where \omega=Angular velocity

v=Linear speed of the rope

r=Radius of the rope

Substitute the known values in the above equation we get

v=31.40 \times 0.30

v=9.42 \frac{m}{s}

C. Dsitance covered in 5 seconds;

Linear distance = linear speed × time

d=v \times t

Where d= Linear distance of the rope

v=Linear speed of the rope

t=Time taken

Substitute the known values in the above equation we get

d=9.42 \times 5

d=47.1 \mathrm{m}

Result:

Thus A: Angular velocity of the rope \omega=31.40 \frac{r a d}{s}

B Linear speed of the rope v=9.42 \frac{m}{s}

C: Distance covered in 5 seconds d=47.1 \mathrm{m}

6 0
3 years ago
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