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Anna [14]
2 years ago
14

The upper end of a 3.80-m-long steel wire is fastened to the ceiling, and a 54.0-kg object is suspended from the lower end of th

e wire. you observe that it takes a transverse pulse 0.0492 s to travel from the bottom to the top of the wire. what is the mass of the wire?
Physics
1 answer:
Novosadov [1.4K]2 years ago
3 0
Given that the lenth write steel l =3.80m fastened celling mass m= 54.0kg  
v = s/t 
 v = 3.8/0.0492 
 v = 77.23 m/s  
 now, the formula for speed in a wire is 
 v = ( T/ÎĽ )^1/2  
 where T is the tension in the string and ÎĽ is the mass pee unit length.  
 v = ( Tl/m )^1/2 .......(replace ÎĽ with m/l) 
 now T = 550N  
 v = [ 550(3.8)/m ]^1/2 
 squaring both sides  
 v^2 = 550(3.8)/m 
 v^2 = 2090/m 
 putting the value of v calculated above and solving 
  m =0.35 kg
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Answer:

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Explanation:

Mass\ of\ electron=m_e=9.11\times10^{-31} kg\\ Charge\ of\ the\ electron=q_e=-1.60\times10^{-19} C\\

We need to find how much charge is contained  in the electron per unit of mass, to do this we divide the charge in an electron and the mass of an electron:

\frac{Charge\ of\ electron}{Mass\ of\ electron}=\frac{q_e}{m_e}\frac{C}{kg}=\frac{-1.60\times10^{-19} C}{9.11\times10^{-31} kg}=-1.76\times10^{11} \frac{C}{kg}

4 0
2 years ago
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Novay_Z [31]

Answer:

Mr Myers and his son use the same force to pull the bags between the gates

Explanation:

The work done by Mr. Myers in pulling the carryon bags = The work done by his 13 year old grandson in pulling the identical bag

Let F₁ represent the force used by Mr Myers, and let F₂ represent the force F₂ used by his grandson

Let d represent the distance through the gate

Therefore, given that Work done, W = Force, F × Distance, we have;

The work done by Mr Myers between the gates, W₁ = F₁ × d

The work done by his grandson between the gates, W₂ = F₂ × d

Where, the work done by both Mr Myers and his grandson are equal, we have;

W₁ = W₂ and therefore, F₁ × d = F₂ × d, which gives;

F₁ = F₂, the force used by both Mr Myers and his son between the gates are equal.

5 0
2 years ago
Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater.
solong [7]

Answer:u=97.41m/s

Explanation:

Given

inclination \theta =58.7^{\circ}C

Horizontal distance travel by Particle d=1200 m

Vertical height h=780 m

Let u be the initial velocity

calculating vertical distance

y=u\sin \theta +\frac{at^2}{2}

y=u\sin \theta t-\frac{gt^2}{2}-------1

Calculating horizontal distance

x=u\cos \theta \times t+0

t=\frac{x}{u\cos \theta }

put value of t in equation 1

y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{g}{2}\times (\frac{x}{u\cos \theta })^2

y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2\theta }

\frac{gx^2}{2u^2\cos ^2\theta }=x\tan \theta -y

u^2=\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}

u=\sqrt{\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}}

at y=-780\ m\ x=1200 m

u^2=\frac{18.154}{2753.65}\times 1200^2

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6 0
2 years ago
Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant sp
Savatey [412]

km x h = km/h

First trial: 6 x 1 = 6km/h

Second trial: 9 x 2 = 18km/h

6 + 18 = <u>24km/h</u> (Total)

Or

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2 + 1 = 3h

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30 - 6 = <u>24km/h</u>

8 0
2 years ago
Consider a 4-mg raindrop that falls from a cloud at a height of 2 km. When the raindrop reaches the ground, it won't kill you or
docker41 [41]

Answer:

The work done by the air resistance is -0.0782 J

Explanation:

Hi there!

The energy of the raindrop has to be conserved, according to the law of conservation of energy.

Initially, the raindrop has only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = acceleration due to gravity (9.8 m/s²)

h = height.

Let´s calculate the initial potential energy of the drop:

(convert 4 mg into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000  m

PE = 0.0784 J

When the drop starts falling, some of the potential energy is converted into kinetic energy and some energy is dissipated by the work done by the air resistance. On the ground all the initial potential energy has been either converted into kinetic energy or dissipated by the resistance of the air:

initial PE = final KE + W air

Where:

KE = kinetic energy.

W air = work done by the air resistance.

The kinetic energy when the raindrop reaches the ground is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

Then:

KE = 1/2 ·  4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can calculate the work done by the air resistance:

initial PE = final KE + W air

0.0784 J = 2 × 10⁻⁴ J + W air

W air = 0.0784 J - 2 × 10⁻⁴ J

W air = 0.0782 J

Since the work is done in the opposite direction to the displacement, the work is negative, then, the work done by the air resistance is -0.0782 J.

5 0
2 years ago
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