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Nesterboy [21]
3 years ago
7

A cube and a square pyramid were joined to form the composite solid. A cube with side lengths of 12 inches. A square pyramid wit

h triangular sides with a height of 9 inches. What is the total surface area of the composite solid? 792 square inches 936 square inches 1,080 square inches 1,152 square inches Mark this and return
Physics
2 answers:
oee [108]3 years ago
7 0

Answer:

The answer is 936 inches squared on Edg 2020 <33

Explanation:

Vesna [10]3 years ago
6 0

Answer:

i think it is the first option.

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Consider the data collected in science class. Different masses were thrown with varied amounts of
lesya [120]

Answer:

A: In all cases, the acceleration was the same.

Explanation:

I know this because its a clear obvious answer not only that it was one of my USA TESTPREP questions and it was right.

All you mainly have to do is the math - F=ma , In each case , the acceleration is 5 m/s squared

4 0
3 years ago
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<img src="https://tex.z-dn.net/?f=2.25%20%5Ctimes%2030" id="TexFormula1" title="2.25 \times 30" alt="2.25 \times 30" align="absm
Paraphin [41]
67.5 is the answer i believe!!
7 0
2 years ago
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Which one of the following statements is false?
emmasim [6.3K]

Answer:

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

Explanation:

Affirmations

a) true. The orbits are accurate in the Bohr model and probabilistic in quantum mechanics

b) True. If both give the same results and use the same quantum number (n)

c) True. If in angular momentum it is quantized, in the Bohr model too but it does not justify it

d) False. If the angular momentum is zero, it implies in electro without turning, which would create a collapse towards the nucleus, so in both models the moment must be different from zero

7 0
3 years ago
1) An object travels 15 m in 3 s. What is its' speed?​
Sav [38]

Answer:

5m/s

Explanation:

Distance/Time= Speed

D/T=S

15/3=S

5=S

8 0
2 years ago
Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the
Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

F = G \frac{m_1m_2}{r^2}

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

F = m_2a

Therefore, we can see that

a(r) = G \frac{m_1}{r^2}

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}

a(R) is actually the constant acceleration at sea level

and

a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}

Therefore:

a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}

6 0
3 years ago
Read 2 more answers
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