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Advocard [28]
2 years ago
8

A spring (k = 802 N/m) is hanging from the ceiling of an elevator, and a 5.0-kg object is attached to the lower end. By how much

does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at a = 0.41 m/s2?
Physics
1 answer:
love history [14]2 years ago
6 0

Answer:

0.00256 m

Explanation:

For this case, we should use the Hook Law for a spring:

F= -Kx , the negative sign indicates that the force that tries to “restore” the original status of the spring, and is against the force that causes the spring´s displacement

This is, the force F required to stretch an elastic object (for example, a metal spring), is directly proportional the extension “x” of the spring

“x” can be the extension or compression of the spring, and “K” is a constant (in units of N/m)

We also know that, according to Newton´s Law:

F=m*a

m= mass

a= acceleration

Then:

m*a= -Kx

Finally, the spring has an original length of L₀, so:

- If the spring is compressed, the final strength will be: L = L₀ – x

- If the spring is extended, the final strength will be: L = L₀ + x

According to the statement:

K = 802 N/m

a = 0.41 m/s2

m = 5 Kg

Then:  

x =- (m*a)/K = (5x0.41)/802 = 0.00256 m (we do not consider the negative sign, due to above explanation: it only indicates that the restoration force is always against the force imposed on the spring )

So, if the spring has an original length of L₀, when the elevator is accelerating upwards, the spring will stretch from L₀ to (L₀ – 0.00256) m

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g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
2 years ago
Turner's treadmill runs with a velocity of -1.3 m/s and speeds up at regular intervals during a half-hour workout. after 25 min,
Travka [436]

Given : Initial velocity = -1.3 m/s

            Final Velocity = -6.5 m/s.

            Time = 25 minutes.

To find : Average acceleration.

Solution: We are given units in meter/second (m/s).

So, we need to convert time 25 minutes in seconds.

1 minute = 60 seconds.

25 minutes = 60*25 = 1500 seconds.

Formula for average acceleration is given by,

\frac{Final \ velocity -Initial \ velocity}{Final \ time - Initial \ time }

We are not given intial time, so we can take initial time =0.

Plugging values in the above formula.

Average \ acceleration = \frac{-6.5 -(-1.3)}{1500-0}

= \frac{-5.2}{1500}

= -0.003467

or Average \ acceleration = -3.467 \times 10^{-3}\ m/s^2..


3 0
2 years ago
Read 2 more answers
The speed of sound in air is 320 ms-1 and in water it is 1600 ms-1. It takes 2.5 s for sound to reach a certain distance from th
Nonamiya [84]

Answer:

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

Explanation:

Given:

Speed of sound in air = 320 m/s

Speed of sound in water = 1600 m/s

Time taken to reach certain distance in air = 2.5 sec

a.

We have to find the distance traveled by sound in air.

Distance = Product of speed and time.

⇒ Distance = Speed\times time\ taken

⇒ Distance = 320\times 2.5

⇒ Distance = 800 meters.

b.

Now we have to find how much time the sound will take to travel in water.

⇒ Time = Ratio of distance and speed

⇒ Time =\frac{distance}{speed}

⇒ Time =\frac{800}{1600}   <em>   ...distance = 800 m and speed = 1600 m/s</em>

⇒ Time =\frac{1}{2}

⇒ Time =0.5 seconds.

Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.

7 0
2 years ago
A zebra runs across a field at a constant speed of 14m/s how far does the zebra go in 8 seconds?
Ratling [72]

Answer:

112m/s

Explanation:

14x8=112 therefore meaning the zebra would run 112m/s

3 0
2 years ago
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A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the
tankabanditka [31]

Answer

given,

F_L= 1.52\ N

\theta_L= 31^0

mass of book = 0.305 Kg

so, from the diagram attached  below

F_L cos {\theta_L} + F_b sin {\theta_b} = m g

1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8

F_b sin {\theta_b} = 2.989 -1.303

F_b sin {\theta_b} = 1.686

computing horizontal component

F_b cos {\theta_b} = F_L sin {\theta_L}

cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}

cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}

cos {\theta_b} = 0.464

θ = 62.35°

5 0
2 years ago
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