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3 years ago

What is the uppermost part of the tree referred to as?

Engineering

2 answers:

tangare [24]3 years ago

7 0

Answer:

canopy I believe it is called

elixir [45]3 years ago

5 0

Answer:

Canopy

Explanation:

The canopy of a tree is sometimes also called its “crown.” This part of the tree is the uppermost part, made up of branches, stems, and leaves for a deciduous tree. ... For evergreen trees, the canopy would include the branches, stems, and needles.

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n the Spring of 2015, three utility companies in the Ukraine received email purporting to come from Ukraine's parliament, the Ra

Tresset [83]

Answer:

Trojan horse

Explanation:

A trojan horse attack is a type of malware that misleads users, as it appears unsuspicious at first, but actually presents a threat to the user. A common example is that of an email that contains a malicious attachment. Another common example is that of a fake advertisement. The name comes from the Greek story of the Trojan horse that led to the fall of the city of Troy.

5 0

3 years ago

Which of the following should you avoid when pulling over to the curb?

Misha Larkins [42]

Answer:

Explanation:

When preparing to move to a curb or side of the road you should always accelerate quickly to move ahead of traffic.

7 0

3 years ago

Read 2 more answers

Find the number of Btu conducted through a wall in 8 hours. The wall is 8 feet high by 24 feet long and has a total R-value of 1

dedylja [7]

Answer:

ΔQ = 4930.37 BTu

Explanation:

given data

height h = 8ft

Δt = 8 hours

length L = 24 feet

R value = 16.2 hr⋅°F⋅ft² /Btu

inside temperature t1 = 68°F

outside temperature t2 = 16°F

to find out

number of Btu conducted

solution

we get here number of Btu conducted by this expression that s

$\frac{\Delta Q}{\Delta t} =\frac{-A}{R} (t2 -t1)$ ......................1

here A is area that is = h × L = 8 × 24 = 1492 ft²

put here value we get

$\frac{\Delta Q}{8} =\frac{-192}{16.2} (16-68)$

solve it we get

ΔQ = 4930.37 BTu

7 0

3 years ago

A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °

SOVA2 [1]

Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)= $\frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3} (Btu/s)/(ft.F)$

Thermal Conductivity is SI units:

$k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K$

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

$A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J$

In Btu:

$A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2$

Heat Q is:

$Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu$

PArt B:

At midpoint Length=L/2=0.1 m

$Q=\frac{k*A*(T_1-T_2)}{L}$

On rearranging:

$T_2=T_1-\frac{Q*L}{KA}$

$T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C$

4 0

3 years ago

*3–32. The rubber block is subjected to an elongation of 0.03 in. along the x axis, and its vertical faces are given a tilt so t

riadik2000 [5.3K]

Rubber block is not shown. I have attached an image of it.

Answer:

A) ε_x = 0.0075

B) ε_y = 0.00375

C) γ_xy = 0.0122 rad

Explanation:

We are given;

δ = 0.03 in

L = 4 in

ν_r = 0.5

θ = 89.3° = 89.3π/180 rad

Let's calculate ε_x in the direction of axis x

Thus, ε_x = δ/L = 0.03/4 = 0.0075

Let's calculate ε_y in the direction of axis y;

ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375

Now, shear strain is angle between π/2 rad surfaces at that point.

Thus,

γ_xy = π/2 - θ = π/2 - 89.3π/180

γ_xy = π(0.003889) = 0.0122 rad

3 0

3 years ago