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laiz [17]
3 years ago
13

A mass of 1.5 kg is attached to a spring and placed on a horizontal surface. The spring has a spring constant of 120 N/m, and th

e spring is compressed 0.25 m past its natural length. If the mass is released from this compressed position, what is the speed of the mass as it passes the natural length of the spring? A. 3.4 m/s B. 1.5 m/s C. 0.99 m/s D. 2.2 m/s
Physics
2 answers:
dusya [7]3 years ago
8 0
I even need this ! Help me tooooooooo Thanks  HAve a good day
IrinaK [193]3 years ago
4 0

Answer:

Speed of the mass is 2.2 m/s

Explanation:

It is given that,

Mass of the object, m = 1.5 kg

Spring constant of the spring, k = 120 N/m

The spring is compressed by a distance of, x = 0.25 m

The mass is released from this compressed position, then the initial kinetic energy of block is equal to the spring potential energy i.e.

\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2

v is the speed of the mass.

v=\sqrt{\dfrac{kx^2}{m}}

v=\sqrt{\dfrac{120\times (0.25)^2}{1.5}}

v = 2.23 m/s

The speed of the mass as it passes the natural length of the spring is 2.2 m/s. Hence, the correct option is (d).            

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rjkz [21]

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umm the lower the frequency the higher the pitch

Explanation:

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2 years ago
A sound wave traveling at 340 m/s is generated by a 480 Hz tuning fork.
Shtirlitz [24]

Answer:

Wavelength = 0.7083 meters

Explanation:

Given the following data;

Speed of wave = 340 m/s

Frequency = 480 Hz

To find how long is the sound wave, we would determine its wavelength;

Mathematically, the wavelength of a waveform is given by the formula;

Wavelength = velocity/frequency

Wavelength = 340/480

Wavelength = 0.7083 meters

8 0
2 years ago
A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

F = \frac{kq_1q_2}{d^2}

Here

k = Coulomb's Constant

q_{1,2} = Charge of each object

d = Distance

Our values are given as,

q_1 = 1 \mu C

q_2 = 6 \mu C

d = 1 m

k =  9*10^9 Nm^2/C^2

a) The electric force on charge q_2 is

F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}

F_{12} = 54 mN

Force is positive i.e. repulsive

b) As the force exerted on q_2 will be equal to that act on q_1,

F_{21} = F_{12}

F_{21} = 54 mN

Force is positive i.e. repulsive

c) If q_2 = -6 \mu C, a negative sign will be introduced into the expression above i.e.

F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}

F_{12} = F_{21} = -54 mN

Force is negative i.e. attractive

6 0
3 years ago
Thomas and John are carrying a 43kg cylinder head on a 510cm X 510mm board. The cylinder head with dimensions of 43cm X 250mm li
tatyana61 [14]

John carry the heaviest load.

<h3>How to find out who is carrying the heavy load?</h3>

Write down given data from questions:

Board=510cm X 510mm.

Cylinder head with dimensions=43cm X 250mm.

Cylinder lies across the board 210cm from john.

Find out: Who is carry the heaviest load?​

Calculation:

We assume that mass of cylinder head = x kg

Then weight=x x 9*81

                 W=9.81x  Newton.

Weight per unit length= Weight/Total leanth

Weight per unit length= 9.81x/43

(w/l)=0.23x N/cm

From equation contition: (F_{J} +F_{T} =9.81x n)

F_{T} (510)=9.81x  (210+21.5)

F_{T} (510)=9.81 x (231.5)

F_{T} =4.452x N.

F_{J} =9.81x-4.452x

F_{J} =5.358 xN

Therefore F_{J} \geq F_{T}

To learn more about mass per unit length, refer to:

brainly.com/question/24180692

#SPJ9

4 0
1 year ago
The length and mass of the arm are Larm = X1 = 50 cm and Marm = 0.3 kg, X2 = 15 cm, and the mass of the object is MObject = 0.25
max2010maxim [7]

Answer: 0.5N

Explanation: if the system is at equilibrium, sum of the torque will be equal to zero.

But if they are not in equilibrium.

U will find the difference in the two torque

find the attached file for solution

3 0
2 years ago
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