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Nadya [2.5K]
3 years ago
10

We have starin which is 0.4 and area is 0.05 what is sigma

Physics
1 answer:
Eddi Din [679]3 years ago
5 0
 <span>normal stress, shear stress, bulk stress, normal </span>strain<span>, shear </span>strain<span>, bulk </span>strain, ... The symbol for stress is often the Greek letter "sigma"<span> Stress </span>has<span> the units of force over </span>area<span> i.e. N/m</span>2<span> which is given the special name Pascal. ... 5.7. </span>0.4<span>. polystyrene. 3.8. 1.0. 4.7. </span>0.4<span>. polyethylene. 0.14. </span>0.05<span>. 0.35. </span>0.4<span> ...</span>
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What is the total wavelength if one-half of the wave is 3?
seraphim [82]

Answer:

6

Explanation:

Half the wave = 3

Wavelength = 3 x 2 = 6

3 0
2 years ago
Post WWII Germany was divided into four zones with the Americans occupying one zone and all of the following countries occupying
maxonik [38]

Answer:

b. Italy

Explanation:

Italy had sided with germany and lost in world war 2

After the Potsdam conference, Germany was divided into four occupied zones: Great Britain in the northwest, France in the southwest, the United States in the south and the Soviet Union in the east.

europeuncedu

3 0
2 years ago
A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They
soldier1979 [14.2K]

Answer:

a.\thta=71^{\circ}

b.v_f=3.78 m/s

Explanation:

We are given that

m_1=53 kg

v_1=2.9 m/s

m_2= 72 kg

v_2=6.2 m/s

a.We have to find the angle

\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}

\theta=71^{\circ}

b. We have to find the speed v_f

According to law of conservation of momentum

m_1v_1=(m_1+m_2)v_fcos\theta

53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f

v_f=\frac{153.7}{40.7}=3.78 m/s

3 0
3 years ago
In as small room fan of rating 50 watt is used for 10hrs 2bulb of rating 10v are used for 8hrs daily.calculate monthly. electric
worty [1.4K]

Answer:

The electric bill for June is Rs198000

Explanation:

Convert volt to watt, but in order to do so I need to know the amps and since it is not provided I converted if the amps was 1.

I multiple 50 with 10 then  with 30 so I know how much watt the fan takes at June.

Since there are 2 light bulb I multiple 10 with 2 than with 8 than with 30.

15000 watts for the fan,

4800 watts for light bulb,

add them and then times it by 10.

Rs198000

4 0
2 years ago
Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i
s2008m [1.1K]

Answer:

The final speed for the mass 2m is v_{2y}=-1,51\ v_{i} and the final speed for the mass 9m is v_{1f} =0,85\ v_{i}.

The angle at which the particle 9m is scattered is \theta = -66,68^{o} with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

\vec{p}=m\vec{v}

p_{xi} =p_{xf}

In the x axis before the collision we have

p_{xi}=9m\ v_{i} - 2m\ v_{i}

and after the collision we have that

p_{xf} =9m\ v_{1x}

In the y axis before the collision p_{yi} =0

after the collision we have that

p_{yf} =9m\ v_{1y} - 2m\ v_{2y}

so

p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}

then

p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}

<u>Conservation of kinetic energy:</u>

\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}

so

\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}

Putting in one side of the equation each speed we get

\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the  speed of the 9m particle:

v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}

the magnitude of the final speed of the particle 9m is

v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }

v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}

The tangent that the speed of the particle 9m makes with the -y axis is

tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}

As a vector the speed of the particle 9m is:

\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}

As a vector the speed of the particle 2m is:

\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}

8 0
3 years ago
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