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LuckyWell [14K]
2 years ago
6

A book sitting on a shelf is an example of ____________.

Engineering
2 answers:
lutik1710 [3]2 years ago
8 0

Answer:

A) inertia

Explanation:

inertia: "A object at rest stay at rest A object in motion will stay in motion unless a outside force meet." - Sir Issac Newton

lyudmila [28]2 years ago
5 0
Answer is A.) Inertia
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What engineers call moment, scientists call
Svet_ta [14]

Answer:

yes

Explanation:

7 0
3 years ago
Out
olchik [2.2K]

ave you ever seen a Rube Goldberg machine in action? You probably have, even if you didn’t know what it was. A Rube Goldberg machine is a contraption that uses a chain reaction to carry out a simple task. It performs a very basic job in a complicated way.

7 0
3 years ago
I'll mark brainliest plz help
Citrus2011 [14]

Answer:

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

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
What is the De Broglie wavelength of an electron under 150 V acceleration?
yanalaym [24]

Answer:

0.1 nm

Explanation

Potential deference of the electron is given as V =150 V

Mass of electron m=9.1\times 10^{-31}

Let the velocity of electron = v

Charge on the electron =1.6\times 10^{-19}C

plank's constant h =6.67\times 10^{-34}

According to energy conservation eV =\frac{1}{2}mv^2

v=\sqrt{\frac{2eV}{M}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 150}{9.1\times 10^{-31}}}=7.2627\times 10^{-6}m/sec

Now we know that De Broglie wavelength \lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm

4 0
3 years ago
Read 2 more answers
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