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Keith_Richards [23]
2 years ago
15

You hang an object with mass m = 0.380 kg from a vertical spring that has negligible mass and force constant k = 60.0 N/m. You p

ull the mass down to a distance 0.030 m from its equilibrium position and release it. You measure the period of the subsequent motion to be 0.500 s.(a) If you pull the mass down0.010 mfrom its equilibrium position and release it, the new period would be which of the following?less than 0.500 s0.500 s greater than 0.500 sThere is no way to tell.(b) If, instead, you change the spring to one with a force constant of56.0 N/m but keep everything else the same, the new period will be which of the following?less than 0.500 s0.500 s greater than 0.500 sThere is no way to tell.(c) If, instead, you change the hanging object to one with a mass of0.530 kg but keep everything else the same, the new period will be which of the following?less than 0.500 s0.500 s greater than 0.500 sThere is no way to tell.
Physics
1 answer:
joja [24]2 years ago
8 0

Answer:

a) 0.500 s

b) greater than 0.500 s

c) greater than 0.500 s

Explanation:

The time period of an oscillating spring-mass system is given by:

T=2\pi \sqrt{\frac{m}{k}}

where, m is the mass and k is the spring constant.

a) As the period of oscillation does not depend on the distance by which the mass is pulled, the period would remain same as 0.500 s for the given system.

b) As the period varies inversely with the square root of spring constant, so with the decrease in the spring constant, the period would increase. So, the new period would be greater than 0.500 s.

c) As the period varies directly with the square root of mass, so with the increase in mass, the period will also increase. The new period will be greater than 0.500 s.

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What is the mass of an object weighing 63 N on Earth?
avanturin [10]
Weight expressed in Newtons is expressed in the equation whereby Weight= the mass of an object * the force of gravity. The force of gravity on earth is a constant 9.8 meters per second squared. Therefore if weight (w) = 63 N and the force of gravity is 63 N then the mass must equal 6.43 kg. Because the equation for weight is w=mg so 63 N (w) = m * 9.8 m/s^2. 
3 0
2 years ago
answers Collision derivation problem. If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the in
Natasha2012 [34]

Answer:

4.8967m

Explanation:

Given the following data;

M = 0.2kg

∆p = 0.58kgm/s

S(i) = 2.25m

Ratio h/w = 12/75

Firstly, we use conservation of momentum to find the velocity

Therefore, ∆p = MV

0.58kgm/s = 0.2V

V = 0.58/2

V = 2.9m/s

Then, we can use the conservation of energy to solve for maximum height the car can go

E(i) = E(f)

1/2mV² = mgh

Mass cancels out

1/2V² = gh

h = 1/2V²/g = V²/2g

h = (2.9)²/2(9.8)

h = 8.41/19.6 = 0.429m

Since we have gotten the heigh, the next thing is to solve for actual slant of the ramp and initial displacement using similar triangles.

h/w = 0.429/x

X = 0.429×75/12

X = 2.6815

Therefore, by Pythagoreans rule

S(ramp) = √2.68125²+0.429²

S(ramp) = 2.64671

Finally, S(t) = S(ramp) + S(i)

= 2.64671+2.25

= 4.8967m

3 0
2 years ago
Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener
Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 \frac{m^{3} }{kg}

% Error =  20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 \frac{m^{3} }{kg}

% Error =  - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 \frac{J}{kg k}

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ \frac{V}{m} = \frac{R T}{P}

⇒ Here \frac{V}{m} = Specific volume = v

⇒ v =  \frac{R T}{P}

Put all the values in the above formula we get

⇒ v = \frac{323}{10^{6} } ×81.49

⇒ v = 0.02632 \frac{m^{3} }{kg}

This is the specific volume by ideal gas equation.

Actual value = 0.021796 \frac{m^{3} }{kg}

Error =  0.02632 - 0.021796 =   0.004524 \frac{m^{3} }{kg}

% Error =  \frac{0.004524}{0.021796} × 100

% Error =  20.75 %

(b). From the generalized compressibility chart the value of specific volume

 \frac{V}{m} = v = 0.0142 \frac{m^{3} }{kg}

The actual value = 0.021796 \frac{m^{3} }{kg}

Error = 0.0142 - 0.021796 =  \frac{m^{3} }{kg}

% Error = \frac{- 0.0076}{0.021796} × 100

% Error =  - 34.85 %

3 0
2 years ago
The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt
nevsk [136]

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

6 0
2 years ago
A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0
Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

F_{lift}=F_{up} - F_{dw}=0.5*p*V^2

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

F_{lift}=0.5*p*V^2=W

By replacing the known values it is possible to find the plate's weight:

F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W

W=72.6 N

When the plate kept to 30° from the vertical the moment equation balance is written as:

F_{lift}=0.5*p*V^2=W*sen(30\°)

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

V=\sqrt(W*sen(30\°)/0.5p)

V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})

V=\sqrt(60.5 \frac{N}{kg/m^3})

V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})

V=\sqrt(60.5 \frac{m^2}{s^2})

V= 7.78 m/s

7 0
2 years ago
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