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Ket [755]
3 years ago
9

g suppose a spring with spring constant of 50 N/m is hanging from the ceiling. You hang 2.0 kg mass from the spring. How far is

the spring stretched
Physics
1 answer:
adelina 88 [10]3 years ago
8 0

Answer:

The extension of the spring is 0.392 m.

Explanation:

Given;

spring constant, k = 50 N/m

mass attached to the spring, m = 2.0 kg

let the extension of the spring = x

The extension of the spring is calculated by applying Hook's law;

F = kx

mg = kx

x = \frac{mg}{k} \\\\x = \frac{2 \times 9.8}{50} \\\\x = 0.392 \ m

Therefore, the extension of the spring is 0.392 m.

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A ball bounces changing velocity from vi=15m/s[D] to vf=15m/s[U] in t=0.01s. The balls acceleration is *
fomenos

Answer:

Option (A) is correct

Explanation:

a= (vf-vi)/ t

put the values

hence,

a= ( 15-15)/0.01

a=0

5 0
2 years ago
Read 2 more answers
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 31.5°c.
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The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant =  8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec
8 0
3 years ago
An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
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v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
3 years ago
If a train travels 500 kilometers from Stockholm to
o-na [289]

Hello,

Average speed is total distance divided by total time. From the problem, our total distance is given as 500 kilometers and given time is 5 hours. Therefore, the average speed is:

\displaystyle{v_\text{average}=\sum_{i=1}^n \dfrac{s_i}{t_i}}\\\\\displaystyle{v=\dfrac{500\ \text{km}}{5 \ \text{h}}}\\\\\displaystyle{v=100 \ \text{km/h}}

Therefore, the average speed is 100 km/h. Please let me know if you have any questions!

4 0
1 year ago
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