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lozanna [386]
3 years ago
15

You are holding a string tied to a lamp post 15 m away. You pull the string so that it has a tension of 1.5 N and a mass density

of 0.0045 kg/m. You shake the end of the string at 4 Hz. What is the wavelength of the wave you generate in m
Physics
1 answer:
Dvinal [7]3 years ago
6 0

Answer:

Explanation:

Given that

L= 15 m

T= 1.5 N

μ =0.0045 kg/m

f= 4 Hz

We know that

Velocity in string V is given as

V=\sqrt{\frac{T}{\mu}}

Now putting the values in the above equation we get

V=\sqrt{\frac{1.5}{0.0045}}=18.25\ m/s

We know that

V= f λ

\lambda =\frac{V}{f}

\lambda =\frac{18.25}{4}=4.5 m

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A sprinf has a potential energy of 84.08 J and a constant of 342.25 N/m. How far it been stretched? Use potential energy elastic
valina [46]

The spring has been stretched 0.701 m

Explanation:

The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the elongation of the spring with respect to its equilibrium position

For the spring in this problem, we have:

E = 84.08 J (potential energy)

k = 342.25 N/m (spring constant)

Therefore, its elongation is:

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(84.08)}{342.25}}=0.701 m

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

8 0
3 years ago
A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary
Archy [21]

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

  • m = 10 kg
  • g = 9.8 m/s²
  • T = ?

Use formula:

  • \boxed{\bold{T=m*g}}

Replace and solve:

  • \boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}
  • \boxed{\boxed{\bold{T=98\ N}}}

The tension in the rope is <u>98 Newtons.</u>

Greetings.

5 0
3 years ago
The force acting between two charged particles a and b is
Nina [5.8K]

Answer: The correct answer is B. trust me, I just took that test. :)

6 0
3 years ago
A 1kg sphere rotates in a circular path of radius 0.2m from rest and it reaches an angular speed of 20rad/sec in 10 second calcu
Len [333]

Answer:

0.4 m/s²

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 1 kg

Radius (r) = 0.2 m

Angular speed (w) = 20 rad/sec

Time (t) = 10 s

Tangential acceleration (aₜ) =?

Next, we shall determine the angular acceleration (a) of the sphere. This can be obtained as follow:

Angular speed (w) = 20 rad/sec

Time (t) = 10 s

Angular acceleration (a) =?

a = w/t

a = 20/10

a = 2 rad/s²

Finally, we shall determine the tangential acceleration (aₜ) of the sphere. This can be obtained as follow:

The tangential acceleration (aₜ) and the angular acceleration (a) are related according to the equation:

Tangential acceleration (aₜ) = Angular acceleration (a) × Radius (r)

aₜ = ar

With the above formula, we can obtain the tangential acceleration (aₜ) as follow:

Radius (r) = 0.2 m

Angular acceleration (a) = 2 rad/s²

Tangential acceleration (aₜ) =?

aₜ = ar

aₜ = 2 × 0.2

aₜ = 0.4 m/s²

Therefore, the tangential acceleration is 0.4 m/s²

6 0
3 years ago
Which two options are forms of potential energy?
dezoksy [38]
C and E Gravitational and Chemical energy
4 0
3 years ago
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