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3 years ago

Question 1

Physics

2 answers:

irina [24]3 years ago

5 0

Dang that’s a lot wait i will answer in comments

artcher [175]3 years ago

5 0

Answer:

question 3 c,6d,7c,10a

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A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

2 years ago

1. What is the difference between a patent, latent, and plastic shoe Impression?

Irina-Kira [14]

Answer:

Patent fingerprints can be made by blood, grease, ink, or dirt. This type of fingerprint is easily visible to the human eye. Plastic fingerprints are three-dimensional impressions and can be made by pressing your fingers in fresh paint, wax, soap, or tar.

I Hope this helps!

Sincerely; Victoria<3

Explanation:

4 0

3 years ago

Read 2 more answers

Jane does 5 joules of work when she closes her bedroom window. She applies a force of 6 newtons to do the job. How far does she

insens350 [35]

Statement:

Jane does 5 joules of work when she closes her bedroom window. She applies a force of 6 newtons to do the job.

To find out:

The displacement of the window when she pulled it.

Solution:

Work done (W) = 5 J
Force (F) = 6 N
Let the displacement of the window be s.
We know, the formula of work done, i.e., W = Fs
Putting the values in the above formula, we get
5 J = 6 N × s
or, s = (5 ÷ 6) m
or, s = 0.83 m

Answer:

She pulls the window by 0.83 m.

Hope you could understand.

If you have any query, feel free to ask.

4 0

3 years ago

Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mas

ASHA 777 [7]

M*U + 0 = m*v'1 + 2m*v'2
the zero means deuteron has no velocity
where v'1 and v'2 are the post-collision velocities.
The equatio becomes
U = v'1 + 2v'2
U = v'2- v'1 
v'2 = U + v'1 

U = v'1 + 2(U + v'1) = 2U + 3v'1 
U = -3V 
V = -U / 3 
The speed ratio is 1/3 

B) Since KE is proportional to the square of the speed, if the speed is 1/3, then KE is 1/9 

C) (1/3)ⁿ = 1/729 
3ⁿ = 729 
n = 6

5 0

3 years ago

Read 2 more answers

A glass rod that has been charged to +12.0 nC touches a metal sphere. Afterward, the rod's charge is +8.0 nC. a) What kind of ch

SpyIntel [72]

Answer:

Explanation:

charge transferred to sphere = 12 - 8 = 4 nC

a )

The positive charge is decreased on glass rod , that means electrons must have entered the glass rod from the sphere which reduces its positive charge.

b )

charge transferred = 4 nC = 4 x 10⁻⁹ C

charge on one electron = 1.6 x 10⁻¹⁹ C

No of electrons transferred

= Total charge transferred / charge on one electron

= 4 x 10⁻⁹ / 1.6 x 10⁻¹⁹

= 2.5 x 10¹⁰ .

4 0

3 years ago