What are two ways thermal energy can be increased in a system?

HElP thank you pLeAsE

olga nikolaevna [1]

Answer:

Cell membrane

Explanation:

I've done this before

3 0

3 years ago

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If a projectile is fired with an initial velocity of v0 meters per second at an angle α above the horizontal and air resistance

lawyer [7]

Answer:

a) The bullet hits the ground 51.02 s after it was fired.

b) 22,092.3 units

c) 3188.8 units

Explanation:

a) Assuming that the level the bullet was fired from is the ground level.

The bullet hits the ground when y = 0

y = (v₀ sin(α))t − (1/2) gt²

v₀ = 500 m/s

α = 30°

y = 0

0 = (500 sin 30) t - 0.5(9.8)t²

4.9t² - 250t = 0

t(4.9t - 250) = 0

t = 0 s or (4.9t - 250) = 0

The t = 0 s indicates that the bullet was indeed fired from the ground level.

The time it eventually hits the ground back

4.9t = 250

t = 51.02 s

The bullet hits the ground 51.02 s after it was fired.

b) The distance from the firing point that the bullet lands.

x = (v₀ cos(α))t

At this horizontal distance, t = 51.02 s

Substituting the parameters

x = (500 cos 30°) × 51.02

x = 22,092.3 units

c) Maximum height attained by the bullet

Maximum height is given by

H = (u² sin² α)/2g

H = (500² sin² 30°)/(2×9.8)

H = 3188.8 units

Hope this Helps!!!

6 0

4 years ago

A hydraulic cylinder causes the distance between points A and O to decrease at a constant rate of 3 inches per second. a) Determ

Ierofanga [76]

Answer:

a) The speed of the slider is 4.28 in/s

b) The velocity vector is 2.33 in/s

Explanation:

a) According to the diagram 1 in the attached image:

$r_{C/A} =12*cos55i-12*sin55j\\r_{C/A}=6.883i-9.829j$

Also:

$v_{C} =v_{A}+w_{AC}*r_{C/A}\\v_{Ci}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC} \\6.883&-9.829&0\end{array}\right]\\v_{Ci}=-3j+(0+9.829w_{AC} i-(0-6.883w_{AC})j\\v_{Ci}=9.829w_{AC}i+(-3+6.883w_{AC})j$

If we comparing both sides of the expression:

$-3+6.883w_{AC}=0\\w_{AC}=0.435rad/s$

$v_{C}=9.829*0.435=4.28in/s$

b) According to the diagram 2 in the attached image:

$r_{C/A}=12cos50i-12sin50j=7.713i-9.192j\\r_{B/C}=-3.856i+4.596j$

$v_{C}=v_{A}+w_{AC}r_{C/A}\\v_{C}=-3j+\left[\begin{array}{ccc}i&j&k\\0&0&w_{AC}\\7.713&-9.192&0\end{array}\right] \\v_{Ci}=-3j+(9.192w_{AC})i+7.713w_{AC}j\\v_{Ci}=9.192w_{AC}i+(7.713w_{AC}-3)j$

Comparing both sides of the expression:

$7.713w_{AC}-3=0\\w_{AC}=0.388rad/s\\v_{C}=3.575i$

$v_{B}=v_{C}+w_{AC}r_{B/C}\\v_{B}=3.57i+\left[\begin{array}{ccc}i&j&k\\0&0&0.388\\-3.856&4.59&0\end{array}\right] \\v_{B}=3.57i+(0-1.78)i-(0+1.499)j\\v_{B}=1.787i-1.499j\\|v_{B}|=\sqrt{1.787^{2}+1.499^{2} } =2.33in/s$

6 0

3 years ago

The built-up beam is formed by welding together the thin plates of thickness 5 mm. determine the location of the shear center o.

atroni [7]

Answer:

The location of the shear center o is 0.033 or 33 m

Explanation:

Solution

Recall that,

The moment of inertia of the section is = I = 0.05 * 0.4 ^3 /12 + 0.005 * 0.2 ^3/12

= 30 * 10 ^ ⁻⁶ m⁴

Now,

The first moment of inertia is

Q =ῩA = [ (0.1 -x) + x/2] (0.005 * x)

= 0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²

Thus,

The shear flow is,

q = VQ/I

so,

P = (0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²)/ 30 * 10 ^⁻⁶

P = (16.67 x - 83. 33 x²)

The shear force resisted by the shorter web becomes

Vw,₂ = 2∫ = ₀.₁ and ₀ = P (16.67 x - 83. 33 x²) dx = 0.11x

Then,

We take the moment at a point A

∑Mₐ = 0

- ( p * e)- (Vw₂ * 0.3 ) = 0

e = 0.11 p * 0.3/p

which gives us 0.033 m

= 33 m

Therefore the location of the shear center o is 0.033 or 33 m

Note: Kindly find an attached diagram to the question given above as part of the explanation solved with it.

4 0

3 years ago

Find the pressure exerted by a 3000 N crate that has an area of 2m squared

ipn [44]

Pressure = Force/ Area = 3000/2 = 1500 pascal.

8 0

3 years ago