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Alexxx [7]
3 years ago
15

A sound wave traveling through dry air has a frequency of 16 Hz, a wavelength of 22 m, and a speed of 350 m/s. When the sound wa

ve passes through a cloud of nitrous oxide, its wavelength changes to 16 m, while its frequency remains the same. What is its new speed? (The equation for the speed of a wave is v= f x 1.)
A. 350 m/s
B. 16 m/s
C. 5,600 m/s
D. 256 m/s​

It's D
Physics
2 answers:
IRISSAK [1]3 years ago
8 0

Answer:

D. 256 m/s​

Explanation:

d. d as in dog

nataly862011 [7]3 years ago
5 0

Answer:

D. 256 m/s

Explanation:

f = Frequency of the wave = 16\ \text{Hz}

\lambda = Wavelength = 16\ \text{m}

Speed of a wave is given by

v=f\lambda

\Rightarrow v=16\times 16

\Rightarrow v=256\ \text{m/s}

The speed of the wave as it passes through the cloud of nitrous oxide is 256\ \text{m/s}.

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An object located near the surface of Earth has a weight of a 245 N
marusya05 [52]

Answer:

The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.

Explanation:

Weight of the object on the surface of Earth, W = 245 N

On the surface of Earth, acceleration due to gravity, g = 10 m/s²

Weight of an object is given by :

W = mg

m is mass

m=\dfrac{W}{g}\\\\m=\dfrac{245\ N}{10\ m/s^2}\\\\=24.5\ kg

So, the mass of the object is 24.5 kg

Acceleration due to gravity on Mars, g' = 3.72 m/s²

Weight of the object on Mars,

W' =mg'

W' = 24.5 kg × 3.72 m/s²

= 91.14 N

So, the weight of the object on Mars is 91.14 N.

4 0
3 years ago
A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height
AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

E_i=\frac{1}{2}m_1v_1^2

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

E_f = (m_1 +m_2)gh_{max}

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m

4 0
3 years ago
You are on roller blades on top of a small hill. Your potential energy is equal to 1,000.0 joules. The last time
irinina [24]
It is -59 i think or just ask someone esle for help
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Which of the following has mechanical energy?
KATRIN_1 [288]
Mechanical energy can have mechanical systems. The only mechanical system in the list is the compressed spring. A car battery and a glowing incandescent lightbulb have electrical energy, a nucleus of atom has potential (internal) energy.

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Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n
jenyasd209 [6]

The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.

2. This is immediate from (2.2.7).

3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =

3 0
3 years ago
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