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3 years ago

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Physics

1 answer:

lyudmila [28]3 years ago

7 0

Answer:

4,800?

First I subtracted 8 from 24 because there are 24 hours in a day ad she spends 8 of those hours sleeping. I then took the answer which was 16, and multiplied that by 60 to find out how many minutes were in 16 hours and I got 960. I then multiplied 960 by 5 because she winks 5 times each minute. This left me with the answer 4,800.

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Explain what cause air pressure in your own words. Why do aircraft need pressurized cabins?

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Why are some of the rocks on the moon older than the oldest known rocks on Earth?

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How many joules of heat must be transferred to a 410-g aluminum pizza pan to raise its temperature from 32oC to 232oC? The speci

xxTIMURxx [149]

Answer:

recall that heat absorbed released is given by

Q = mc*(T2 - T1)

where

m = mass (in g)

c = specific heat capacity (in J/g-k)

T = temperature (in C or K)

*note: Q is (+) when heat is absorbed and (-) when heat is released.

substituting,

Q = (480)*(0.97)*(234 - 22)

Q = 98707 J = 98.7 kJ

Explanation:

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3 years ago

Three electromagnetic waves travel through a certain point P along an x axis. They are polarized parallel to a y axis, with the

chubhunter [2.5K]

Answer:

$4.846*10^{-5} sin(5*10^{14}t )$

Explanation:

Frequency of each electromagnetic wave is same thus we can interfere them along X axis in Phase diagram.

In phase diagram indicate each wave as vector with amplitude as length and phase difference as angle. (shown in attachment)

Now take component of E2 and E3 amplitude along X direction.

Net amplitude of all E in X direction = $4*10^{-5}$ + cos(45)[ $6*10^{-6}$ + $6*10^{-6}$ ]

⇒ $4.846*10^{-5}$

Final resultant have no change in frequency

So , resultant is $4.846*10^{-5} sin(5*10^{14}t )$

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4 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

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

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

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>

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

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

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