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Alex_Xolod [135]
3 years ago
10

When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the

period of oscillation, T , neglecting the mass of the spring itself.
Physics
1 answer:
topjm [15]3 years ago
4 0

Answer:

The period of oscillation is 1.33 sec.

Explanation:

Given that,

Mass = 275.0 g

Suppose value of spring constant is 6.2 N/m.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=\sqrt{\dfrac{k}{m}}

Where, m = mass

k = spring constant

Put the value into the formula

\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}

\omega=4.74\ rad/s

We need to calculate the period of oscillation,

Using formula of time period

T=\dfrac{2\pi}{\omega}

Put the value into the formula

T=\dfrac{2\pi}{4.74}

T=1.33\ sec

Hence, The period of oscillation is 1.33 sec.

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A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go ?
AnnZ [28]

So, this is a problem where the accleration is not provided, since it is implied.  The only acceleration is acceleration due to gravity (9.8 m/s)


The equation we will use for this problem is V^2 =V_{0}^2 + 2a (X-X_0)

V is the final velocity, V₀ is the initial velocity, a is the acceleration, X is the final height, and X₀ is the starting height.


We can assume that the ball starts on the ground since no height is given, so now we plug our numbers in.

We will use 0 as the final velocity, since the ball will stop moving upwards when it is the highest.  We will use -9.8 since that is the acceleration due to gravity and we will use 22m/s as V₀ since that is the starting velocity.

V^2 =V_{0}^2 + 2a (X-X_0)\\0^2 = 22^2 + 2\times-9.8(X-0)\\0=484-19.6x\\-484=-19.6x\\24.69387755 = x\\x\approx24.69


So, the ball will go 24.69 meters up


4 0
3 years ago
How old is a bone that has 12.5% of the original amount radioactive carbon 14 remaining?
nlexa [21]
Nuclear decay formula is N(t)=N₀*2^-(t/T), where N(t) is the amount of nuclear material in some moment t, N₀ is the original amount of nuclear material, t is time and T is the half life of the material, in this case carbon 14. In our case N(t)=12.5% of N₀ or N(t)=0.125*N₀, T=5730 years and we need to solve for t:

0.125*N₀=N₀*2^-(t/T), N₀ cancels out and we get:

0.125=2^-(t/T), 

ln(0.125)=ln(2^-(t/T))

ln(0.125)=-(t/T)*ln(2), we divide by ln(2),

ln(0.125)/ln(2)=-t/T, multiply by T,

{ln(0.125)/ln(2)}*T=-t, divide by (-1) and plug in T=5730 years,

{ln(0.125)/[-ln(2)]}*5730=t

t=3*5730=17190 years.

The bone is t= 17190 years old. 

4 0
3 years ago
A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If
vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

  • The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
  • The traslational part can be written as follows:

       K_{trans} = \frac{1}{2}* M* v_{cm} ^{2}  (1)

  • The rotational part can be expressed as follows:

       K_{rot} = \frac{1}{2}* I* \omega ^{2}  (2)

  • where I = moment of Inertia regarding the axis of rotation.
  • ω = angular speed of the rotating object.
  • If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

       v = \omega * R (3)

  • For a solid cylinder, I = M*R²/2 (4)
  • Replacing (3) and (4)  in (2), we get:

       K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2}  (5)

  • Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

       K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2}  +\frac{1}{4}* M* v_{cmc}^{2}  =  \frac{3}{4}* M* v_{cmc} ^{2} (6)

  • Repeating the same steps for the spherical shell:

        I_{sph} = \frac{2}{3} * M* R^{2} (7)  

       K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2}  (8)

      K_{sph} = \frac{1}{2}* M* v_{cms} ^{2}  +\frac{1}{3}* M* v_{cms}^{2}  =  \frac{5}{6}* M* v_{cms} ^{2} (9)

  • Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
  • And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
  • Rearranging, and taking square roots on both sides, we get:

       \frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)

  • This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
3 0
3 years ago
The potential difference across a and b is 15 v. determine the electrical charge on the 3 μf capacitor?
Slav-nsk [51]

The potential difference across a and b is 15 v. determine the electrical charge on the 3 μf capacitor will be 45 *  10^{-6} C

Capacitance, property of an electric conductor, or set of conductors, that is measured by the amount of separated electric charge that can be stored on it per unit change in electrical potential. Capacitance also implies an associated storage of electrical energy.

Charge (Q) stored in a capacitor is the product of its capacitance (C) and the voltage (V) applied to it. The capacitance of a capacitor should always be a constant, known value. So we can adjust voltage to increase or decrease the cap's charge. More voltage means more charge, less voltage... less charge.

charge = capacitance * voltage

Q = CV

   =  3  * 10^{-6} * 15 v

   = 45 *  10^{-6} C

To learn more about capacitance here

brainly.com/question/14746225

#SPJ4

8 0
2 years ago
Please help with this!!!!!
34kurt
(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K)  I hoped that helped
4 0
3 years ago
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