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2 years ago

Monday Unit test: Ms Nayr's Class:

2 answers:

6 0

1. is A 7
2. reticulated means divide or mark something in such a way as to resemble a net or network.
3. A dhole is an Asian dog that has a Sandy coat and black, bushy tail and lives in packs.

elena-14-01-66 [18.8K]2 years ago

5 0

1.) 7, humans have the same amount of neck bones. Think of it that way

2.) It has a pattern diamond like or web like.

3.) Indian wild dog

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Which explains why more energy is released in nuclear reactions than in chemical reactions? Chemical reactions are always endoth

Natali [406]

More energy is released in nuclear reactions than in chemical reactions; this is because in nuclear reactions, mass is converted to energy. Nuclear energy released in nuclear fission and fusion is several 100 million times as large as an ordinary chemical reaction like the combustion process. The reason why nuclear energy release so much energy is because tremendous amounts of energy is released at one time. The nuclei in a nuclear reaction undergo a chain reaction, causing the neutrons to move extremely fast and release high amounts of energy.

3 0

3 years ago

Read 2 more answers

Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg objec

yKpoI14uk [10]

Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

The frequency of vibration, f = 3.84 Hz

When the object is suspended, the gravitational force is balanced by the spring force as :

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{3.74\times 9.8}{0.0161}$

k = 2276.52 N/m

The frequency of vibration is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}$

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

8 0

2 years ago

To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv

Lesechka [4]

Answer:

E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

λ = Q / L

If we derive from the length we have

λ = dq/dx ⇒ dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k $\int\limits^{d+L}_d {\lambda/x^{2}} \, dx$

We take out the constant magnitudes and perform the integral

E = k λ (-1/x) ${(-1/x)}^{d+L} _{d}$

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]

3 0

2 years ago

A centrifuge in a medical research laboratory rotates at an angular speed of 3,650 rev/min. When switched off, it rotates 46.0 t

erma4kov [3.2K]

Answer:

The constant angular acceleration of the centrifuge = -252.84 rad/s²

Explanation:

We will be using the equations of motion for this calculation.

Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.

First of, we write the given parameters.

w₀ = initial angular velocity = 2πf₀

f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s

w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s

θ = 46 revs = 46 × 2π = 289.14 rad

w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)

α = ?

Just like v² = u² + 2ay

w² = w₀² + 2αθ

0 = 382.38² + [2α × (289.14)]

578.29α = -146,214.4644

α = (-146,214.4644/578.29)

α = - 252.84 rad/s²

Hope this Helps!!!

6 0

3 years ago

Read 2 more answers

Strontium has density of 2.64 g/cm3 and crystallizes with the face-centered cubic unit cell. Calculate the radius of a strontium

pantera1 [17]

Answer:

The radius of strontium atom is

2.14 E-8cm

Explanation:

Go through the attached file for a comprehensive detailed explanation.

6 0

2 years ago