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Ilya [14]
3 years ago
9

A phone cord is 2.89 m long. The cord has a mass of 0.258 kg. A transverse wave pulse is produced by plucking one end of the tau

tcord. The pulse makes four trips down and back along the cord in 0.737 s. What is the tension in the cord
Physics
1 answer:
Aleksandr [31]3 years ago
8 0

Answer:

T =87.87\ N

Explanation:

Given,

Length , l=2.89 m

Mass , M=0.258 Kg

Now, displacement d=4(2 \times 2.89) m

                                    =23.12 m

Velocity of waves

v=\sqrt{T / \mu}

\mu=\frac{0.258}{2.89 }

=0.0893

Now, velocity, v=\frac{d}{t}=\frac{23.12}{0.737 s}

=31.37 m/s

The tension

T=v^{2} \mu=31.37 \times 31.37 \times 0.0893

T =87.87\ N

Tension in the chord is equal to  T =87.87\ N

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Kieran takes off from rest down a 50 m high, 10° slope on his jet-powered skis. The skis have a thrust of 280 N parallel to the
xz_007 [3.2K]

Answer:0.46

Explanation:

Given

Initial height h=50 m

inclination \theta =10^{\circ}

Thrust=280 N

combined mass of kieran and skis m=50 kg

Speed at the bottom v=40 m/s

From Work Energy Theorem

Work done by all the force is equal to change in kinetic Energy

W_{gravity}+W_{friction}+W_{thrust}=\frac{1}{2}mv^2-0------------1

distance traveled along the slope x=\frac{50}{\sin 10}=287.93 m

W_{gravity}=mgh=50\times 9.8\times 50=24500 J

W_{thrust}=F\times x=280\times 287.93=80,620.4 J

W_{friction}=-\mu mg\cos 10

substitute in 1

24,500+80,620.4+W_{friction}=\frac{1}{2}\times 50\times 1600

W_{friction}=40,000-24,500-80,620.4

-\mu \cdot 50\times 9.8\times 287.93=-65,120.4

\mu =\frac{65,120.4}{141,085.7}=0.46

8 0
3 years ago
A feather is dropped on the moon from the height of 1.4m. The acceleration of gravity on the moon is 1.67ms-1. Determine the tim
Zinaida [17]

1.3s

Explanation:

Given parameters:

Height = 1.4m

Gravity on moon = 1.67ms⁻¹

Unknown:

Time for feather to fall = ?

Solution:

To solve this problem, we are going to use one of the motion equation that relates time, gravity and height.

    H = ut + \frac{1}{2} g t^{2}

Sine the body was dropped from rest, initial velocity is zero;

 H = height

  u = initial velocity

  t = time

  g = acceleration due to gravity

since u = 0;

H = \frac{1}{2} g t^{2}

 1.4 = \frac{1}{2} x 1.67 x t²

  t = 1.3s

learn more:

Gravity brainly.com/question/10934170

#learnwithBrainly

8 0
3 years ago
A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the
MissTica

Answer : The value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

q=n\times c_v\times \Delta T

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

\Delta T = Change in temperature = 15.7 K

c_v = heat capacity at constant volume of N_2 (diatomic molecule) = \frac{5}{2}R

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K

q=286.2J

Now we have to calculate the change in internal energy of the system.

\Delta U=q+w

As we know that, work done is zero at constant volume. So,

\Delta U=q=286.2J

Therefore, the value of q\text{ and }\Delta U is 286.2 J and 286.2 J respectively.

8 0
4 years ago
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irina [24]
First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.
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3 years ago
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When ultra violets lights shine on glass what does it do to electrons in the glass structure?
andreev551 [17]

Answer:

No

Explanation:

7 0
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