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skad [1K]
3 years ago
6

Is 4x a term? or does it have to remove the 4 to be a term

Physics
1 answer:
Gnesinka [82]3 years ago
5 0

Answer:

4x is a term oonly

No need to be romoved

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In the context of depth perception, which of the following is a monocular cue?
olga55 [171]
Hello there.

<span>In the context of depth perception, which of the following is a monocular cue?

</span><span>(C) Convergence
</span>
8 0
3 years ago
The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting
White raven [17]

Answer:

\frac{\dot Q}{A} =20.129\ W.m^{-2}

T_1=27.58\ ^{\circ}C & T_2=2.41875\ ^{\circ}C

Explanation:

Given:

  • interior temperature of box, T_i=30^{\circ}C
  • height of the walls of box, h=3\ m
  • thickness of each layer of bi-layered plywood, x_p=1.25\ cm=0.0125\ m
  • thermal conductivity of plywood, k_p=0.104\ W.m^{-1}.K^{-1}
  • thickness of sandwiched Styrofoam, x_s=5\ cm=0.05\ m
  • thermal conductivity of Styrofoam, k_s=0.04\ W.m^{-1}.K^{-1}
  • exterior temperature, T_o=0^{\circ}C

<u>From the Fourier's law of conduction:</u>

\dot Q=\frac{dT}{(\frac{x}{kA}) }

\dot Q=\frac{dT}{R_{th} } ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

R_{th}=R_p+R_s+R_p

R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}

R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})

R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})

R_{th}=\frac{1.4904}{A} .....................(2)

Putting the value from (2) into (1):

\dot Q=\frac{30-0}{\frac{1.4904}{A} }

\dot Q=\frac{30\ A}{1.4904}

\frac{\dot Q}{A} =20.129\ W.m^{-2} is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}

20.129=0.104\times \frac{30-T_1}{0.0125}

T_1=27.58\ ^{\circ}C

&<u>For Styrofoam-plywood interface from inside:</u>

\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}

20.129=0.04\times \frac{27.58-T_2}{0.05}

T_2=2.41875\ ^{\circ}C

4 0
3 years ago
Which of the following in NOT an ionic compound?<br><br> NaBr<br> MgCl2<br> O SCI2<br> K (NO3)
Nonamiya [84]
It’s could be O SCI2
4 0
2 years ago
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0
2 years ago
if your reading this never take life for granted i had corona 19 i found out what life really is and god is the only cure
Shtirlitz [24]

Answer:

omg i'm so sorry, i hope you get better <3!

Explanation:

     

8 0
3 years ago
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