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Zanzabum
2 years ago
9

Rana writes a summary about a mass on a spring in simple harmonic motion as it moves upward from the equilibrium position toward

the maximum positive displacement. As the mass moves upward, the velocity is positive and decreasing while the acceleration is negative and decreasing. What error did Rana make?
Physics
2 answers:
Oksanka [162]2 years ago
5 0
The error is in saying that the acceleration is decreasing.


The statements says that the mass is moving from the equilibrium position toward the maximum positive displacement. That means that the mass is moving away from the equilibrium position.


As per the law of Hooke the force exerted by the spring is proportional to the displacement from the equilibrium, which means that the further is the spring from the position of equilibrim the greaer the force.


So, when the mass is moving away from the equilibrium position the force exerted by the spring on the mass is increasing.


Now you should see the mistake, because F = m*a when F (force) increases, a (acceleration) also increases. So it is a mistake saying that the acceleration is decreasing.  
Luba_88 [7]2 years ago
3 0

the answer D the acceleration should be increasing not decreasing

i took the got it right  

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A 4kg bird has 8 joules of kinetic energy, how fast is it flying?
Igoryamba
I believe the answer is 2m/s
7 0
2 years ago
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For a Physics course containing 10 students, the maximum point total for the quarter was 200. The point totals for the 10 studen
Talja [164]

Answer:

130.5

Explanation:

According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.

The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:

16 | 8 5 6

This means that the point for the students are 168, 165 and 166. Three students, three points.

If you watch the stemplot, the points for the students are:

116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.

The median can be calculated using the mean between the two values in the middle of the sequence.

In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:

Median = 128 + 133 / 2 = 130.5

5 0
2 years ago
When listening to tuning forks of frequency 256 Hz and 260 Hz, one hears the following number of beats per second. (A) 0 (B) 2 (
Degger [83]

Answer:

(C) 4 beats per second.

Explanation:

As we know that the no of beats can be calculated as.

No. of beats is equal to difference in the tuning forks frequencies.

So,

n= \nu _{1}- \nu _{2}.

Substitute the values of frequencies of 2 tuning forks in the above equation.

n=(260 Hz-256 Hz)\\n=4

Therefore the number of beats per second will be hear by the observer is 4 beats per second.

3 0
2 years ago
A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope
pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

L=\dfrac{\lambda}{4}

\lambda=4\times L

Put the value into the formula

\lambda=4\times31.2\times10^{-2}

\lambda=1.248\ m

We need to calculate the frequency of sound in pipe

Using formula of frequency

f=\dfrac{v}{\lambda}

Put the value into the formula

f=\dfrac{340}{1.248}

f=272.4\ Hz

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

L=\dfrac{l}{3}

L = \dfrac{129}{3}

L=43\ cm

We need to calculate the wavelength of the wire

Using formula of length

L=\dfrac{\lambda}{2}

\lambda=L\times 2

Put the value into the formula

\lambda=43\times2

\lambda=86\ cm

We need to calculate the speed of the wave

Using formula of frequency

f=\dfrac{v}{\lambda}

v=f\times\lambda

Put the value into the formula

v=272.4\times86\times10^{-2}

v=234.26\ m/s

Hence, The speed of transverse waves in the wire is 234.26 m/s.

4 0
2 years ago
A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t
ipn [44]

Answer:

ball clears the net

Explanation:

v_{o} = initial speed of launch of the ball = 20 ms^{-1}

\theta = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

v_{ox} = initial velocity = v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}

t = time of travel

X = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}       Eq-1

Consider the motion of the ball along the vertical direction

v_{oy} = initial velocity = v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}

t = time of travel

Y_{o} = Initial position of the ball at the time of launch = 2 m

Y = Final position of the ball at time "t"

a_{y} = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\

Since Y > 1 m

hence the ball clears the net

7 0
2 years ago
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