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alexandr1967 [171]
4 years ago
9

A 9.0-cm-long spring is attached to the ceiling. when a 1.8 kg mass is hung from it, the spring stretches to a length of 18 cm .

you may want to review ( pages 219 - 221) . part a what is the spring constant k?
Physics
1 answer:
Alex Ar [27]4 years ago
5 0
The spring constant is computed by:
F = kx

Where: F is the force applied in newtons (N)

k is the spring constant measured in newtons per meter (N/m); and

x is the distance the spring is stretched (m)
and

F = mg

Where: F is the force pulling objects in the direction of the Earth.

m is the mass of the object.

g is the acceleration due to gravity; 
So plugging our values in the formula:

F = mg

 = (1.8) (9.81) = 17.658N 

k =

F/x = 17.658 /0.09 = 196.2 N/meter
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7 0
4 years ago
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The block accelerates down the 30 degree incline at 3m/s2. What is the coefficient of friction of these surfaces
dolphi86 [110]
Fnet = Fg sin 30 - Ff
ma = mg sin 30 - mew Fg cos 30
ma = mg sin 30 - mew mg cos 30
a = g sin 30 - mew gcos30
a - g sin 30 = - mew g cos 30
mew = -(a - g sin30)/(g cos 30)
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8 0
4 years ago
A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera
AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel r=0.3\ m

Initial angular velocity \omega _0=0

It rotates 3 revolution in 5 s therefore

\omega =2\pi 3=\6\pi =18.85\ rad/s

using \omega =\omega _0+\alpha t

where \alpha =angular\ acceleration

\omega =Final\ angular\ velocity

t=time

\alpha =\frac{18.85}{5}=3.77 rad/s^2

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

\omega at t=1

\omega =0+3.77\times 1=3.77 rad/s

a_c=\omega ^2\cdot r

a_c=(3.77)^2\cdot 0.3=4.26 m/s^2

Tangential acceleration a_t=\alpha \times r

a_t=3.77\times 0.3=1.13 m/s^2

a_{net}=\sqrt{a_t^2+a_c^2}

a_{net}=\sqrt{(1.13)^2+(4.26)^2}

a_{net}=4.41 m/s^2

                       

7 0
4 years ago
A curved, transparent object used to refract light
zlopas [31]
It's called a concave lens.
6 0
3 years ago
A glider is initially moving at a constant height of 3.72 m. It is suddenly subject to a wind such that its velocity at a later
My name is Ann [436]

Answer:

17.89\ \text{m/s}

Explanation:

Velocity is given by

v(t) = 16.02\hat{i}-7.96(1 + t)\hat{j} + 0.76t^3\hat{k}

Initial velocity is asked so t = 0

v(0)=16.02\hat{i}-7.96(1+t)\hat{j}+0.76\times 0\hat{k}\\\Rightarrow v(0)=16.02\hat{i}-7.96\hat{j}

Magnitude is given by

|v|=\sqrt{16.02^2+(-7.96)^2}\\\Rightarrow |v|=17.89\ \text{m/s}

The initial velocity of the glider was 17.89\ \text{m/s}.

5 0
3 years ago
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