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Len [333]
3 years ago
8

What is the frequency of a wave that has a wavelength of 0.39 m and a speed

Physics
1 answer:
gogolik [260]3 years ago
7 0

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

f =  \frac{c}{ \lambda}  \\

where

c is the velocity

From the question

wavelength = 0.39 m

c = 86 m/s

We have

f =  \frac{86}{0.39}  \\  = 220.512820...

We have the final answer as

<h3>200 Hz</h3>

Hope this helps you

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At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction betwe
Salsk061 [2.6K]
Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.

5 0
2 years ago
NEED HELP TODAY
Free_Kalibri [48]

Answer:

I am pretty sure it is B My friend hope you are well

Explanation:

6 0
3 years ago
Read 2 more answers
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
Un objeto se suelta desde determinada altura y emplea un tiempo t en caer al suelo. Si se cuadruplica la altura desde la cual se
blondinia [14]

When an object falls from a h height, you should work with the uniformly accelerated linear movement equations:

y=½*a*t²+Vo*t+yo

You should consider:

a=-g=-10m/s²

yo=h

If it’s a freefall, it means it starts from rest, which means it has no initial velocity:

Vo=0

Replacing that information in the equation:

y=½*(-10m/s²)*t²+0*t+h=-5m/s²*t²+0+h=-5m/s²*t²+h

So this is the

Besides, if you want to find out how long it takes for it to get to the floor, you should put the height of the floor as final height, which would be 0 (assuming the initial height has been measured from there):

y=0

0=-5m/s²*t²+h

5m/s²*t²=h

t²=h/(5m/s²)

t=√(h/(5m/s²))

t=√(hs²/(5m))

t=(√(h/(5m)))s

<span>If we <span>quadruple </span>h:</span>

t2=(√(h2/(5m)))s=(√(4*h1/(5m)))s=(√4)*(√h1/(5m)))s=2*(√h1/(5m)))s=2*t1

This 4 goes inside the square root, so then it converts to 2. So the new time is twice as much the previous time.

Concerning velocity, you have to use the other equation:

v=at+vo

As I said before, a is gravity and vo is zero.

v=-10m/s²*t+0=-10m/s²*t

Final velocity is directly related to time, so if time is doubled, so is velocity.

v2=-10m/s²*t2=-10m/s²*(2*t1)=2*(-10m/s²*t1)=2*v1

<span>So the correct answer is A, and the other ones are false.</span>

8 0
3 years ago
The freezer compartment in a conventional refrigerator can be modeled as a rectangular cavity 0.3 m high and 0.25 m wide with a
Mkey [24]

Answer:

Thickness of Styrofoam insulation is 0.02741 m.

Explanation:

Given that,

Height = 0.25 m

Depth = 0.5 m

Power = 400 W

Temperature = 33°C

We need to calculate the area of Styrofoam

Using formula of area

A=2(lb+bh+hl)

Put the value into the formula

A=2(0.3\times0.5+0.25\times0.5+0.5\times0.3)

A=0.85\ m^2

Inner surface temperature of freezer

T_{i}=-10°C=263\ K

Outer surface temperature of freezer

T_{o}=33+273=306\ K

We need to calculate the thickness of Styrofoam insulation

Using Fourier law,

q=\dfrac{kA}{L}(T_{o}-T_{i})

L=\dfrac{kA}{q}(T_{o}-T_{i})

Put the value into the formula

L=\dfrac{0.30\times0.85}{400}(306-263)

L=0.02741\ m

Hence, Thickness of Styrofoam insulation is 0.02741 m.

3 0
3 years ago
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