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3 years ago

What is tan θ for the given triangle?

Physics

2 answers:

yulyashka [42]3 years ago

7 0

Tan = opposite/adjacent
= 20/15
=4/3

Nuetrik [128]3 years ago

4 0

Answer:

it's C

Explanation: 4/3=1.33

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CO + 7H2 - C3Hg + 3H20

Assoli18 [71]

Answer:

Espanol 7h2 that's c coupon

6 0

3 years ago

Leila is building an aluminum-roofed shed in her backyard to store her garden tools.The flat roof will measure 2.0 x 3.0m in are

IrinaK [193]

Answer: 0.0138 m^2 = 138 cm^2

Explanation:

The thermal expansion is the term use for the physical phenomena of dilation of the objects when they are exposed to changes in temperature.

The objects dilate when they are heated and contract when they are cooled.

The dilation is proportional to the change in temperatur.

For linear dilation, the proportionality constant is called linear dilation coefficient of the materials, it is named α and is measured in °C ^-1.

ΔL = α * Lo * ΔT, which means that the dilation (or contraction) is proportional to the product of the original length (Lo) and the change of temperature (ΔT).

There is also superficial dilation, for which the dilation is:

ΔA = β * Ao * ΔT, which means that the superficial dilation (or contraction) is proportional to the product of the original area (Ao) and the change of temperature (ΔT).

It is very interesting and important to solve problems that β = 2α, because regularly you will find the values of α for different materials and so, you just to multiply it times 2 to use β.

For this problem:

- Original area, Ao = area of the flat roof at - 10°C = 2.0m * 3.0m = 6.0 m^2.
- α for aluminum = 24 * 10^ -6 °C^-1.
- ΔT = 38°C - (-10°C) = 48°C

So, ΔA = 6.0m^2 * (2 * 24*10^-6 °C&-1) * 48°C = 0.0138 m^2

And that is the area that should stick out in summer to fit the structure during cold winter nights.

You can pass that number to cm^2 to grasp better the idea of this size:

0.0138 m^2 * (100 cm)^2 / m^2 = 138 cm^2

3 0

3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

3 years ago

Pls pls help me out AHH, what is not true about MEIOSIS?

agasfer [191]

Answer:

Explanation:

This is not true as the number of chromosomes in the daughter cells are half the number in the parent's cells

4 0

3 years ago

How far away is a cliff if an echo is heard 0.486 s after the original sound? Assume that sound traveled at 343 m/s on that day.

goldfiish [28.3K]

143m/s if you just perhaps by what you know you'll figure it out

4 0

3 years ago