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drek231 [11]
3 years ago
9

A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav

els down the cord and reaches the other support in 0.88 ss .What is the tension in the cord? (Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Agata [3.3K]3 years ago
4 0

Answer:

The tension in the string is T = 1.49*10^{-6}N.

Explanation:

For a string with tension T and linear density \mu_d carrying a transverse wave at speed v it is true that

v = \sqrt{\dfrac{T}{\mu_d} }

solving for T we get:

T = \dfrac{v^2}{\mu_d}.

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s

And it's linear density (mass per unit length) is

\mu_d = \dfrac{0.35kg}{7.4*10^{-3}m} \\\\\mu_d = 47.3kg/m

Therefore, the tension in the cord is

T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.

\boxed{T = 1.5*10^{-6}N}

or in micro newtons

T =1.5\mu N

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ikadub [295]

(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

7 0
3 years ago
A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of
Citrus2011 [14]

Answer:

5.959 m/s

Explanation:

m = Mass of gymnast

u = Initial velocity

v = Final velocity

h_i = Initial height

h_f = Final height

From conservation of Energy

\frac{1}{2}mv^2+mgh_f=\frac{1}{2}mu^2+mgh_i\\\Rightarrow\frac{1}{2}mv^2+mg0=\frac{1}{2}m0^2+mgh_i\\\Rightarrow \frac{1}{2}mv^2=mgh_i\\\Rightarrow v=\sqrt{2gh_i}

h_i=2r

v=\sqrt{4gr}\\\Rightarrow v=\sqrt{4\times 9.81\times 0.905}\\\Rightarrow v=5.959\ m/s

Velocity of gymnast at bottom of swing is 5.959 m/s

5 0
3 years ago
If the 50-kg crate starts from rest and achieves a velocity of v = 4 m/s
Kay [80]

Answer:

P = 227 N

Explanation:

Assuming the crate is on horizontal ground and subject to a horizontal force.

F = ma

P - μmg = ma

P = m(a + μg)

P = m(v²/2s + μg)

P = 50(4²/(2(5))+ 0.3(9.8))

P = 227 N

6 0
3 years ago
An electric field is produced by the very long, uniformly charged rod drawn above. If the strength of the electric field is E1 a
Debora [2.8K]

Answer: hello the complete question is attached below

answer :

r2 = 4r1

Explanation:

Electric field strength = F / q

we will assume the rod has an infinite length

For an infinitely charged rod

E ∝ 1/ r

considering two electric fields E1 and E2 at two different locations as described in the question

E1/E2 = r1/r2 ----- ( 2 )

<u>Calculate for r2 when E2 = E1/4 </u>

back to equation 2

E1 / (E1/4) = r1 / r2

∴ r2 = 4r1

3 0
3 years ago
A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95
madam [21]

Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as  y=l'-l=41-30=11m

the mass is m=95 kg

the  force is  F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of  k is given by equating the energy at the equilibrium point which is given as

U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m

Now the force is

F=kx\\ or

x=\dfrac{F}{k}\\

So here F=380 N, k=630.92 N/m

x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m

So the distance is 0.602 m

6 0
3 years ago
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