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9 hours ago

1. Si comprimes un globo hasta reducir su volumen a un tercio de su valor original, ¿cuánto aumenta la presión en su interior?

Physics

1 answer:

ValentinkaMS [1.1K]9 hours ago

4 0

Answer:

See detailed explanation below.

Explanation:

1.-We know that the relationship:

P₁ * V₁ = P₂ * V₂

For a constant temperature, this must hold. Therefore, if the balloon is compressed to one third of its initial volume, the pressure must increase to three times its original value.

2.-Pressure is defined as force per unit surface area, which means that the force is determined by the height of the liquid column in the container, not by the total volume. Consequently, there is more pressure at the base of the vase due to the taller water column.

3.-The plumber's criterion cannot be accepted. His solution does not consider the increased height of the tank to actually achieve the necessary pressure increase.

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In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

ValentinkaMS [1138]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

$T\cos\theta=mg$

$T=\dfrac{mg}{\cos\theta}$

Vertical component

$T\sin\theta=m\omega^2 r$

Substituting the tension value

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

$\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}$

Substituting the value into the equation

$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

$\omega=1.34\ rad/s$

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0

12 days ago

Simone is walking her dog on a leash. The dog is pulling with a force of 32 N to the right and Simone is pulling backward with a

ValentinkaMS [1138]

Conclusion:

The total net force acting on the objects is 16 N, directed towards the right.

Clarification:

It is stated that,

The force exerted by the dog, $F_1 = 32\ N$ (to the right)

The force exerted by Simone, $F_2 = -16\ N$ (backward)

Here, assume the backward direction is negative and the right direction is positive.

The net force will move in the direction where the larger force is present. The net force can be calculated as:

$F=F_1+F_2$

$F=32+(-16)$

F = 16 N

Thus, the net force amounts to 16 N, acting towards the right.

6 0

3 days ago

Read 2 more answers

The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI un

serg [1189]

Answer:

The response to your inquiry is: 15 m/s²

Explanation:

Equation x = at³ - bt² + ct

a = 4.1 m/s³

b = 2.2 m/s²

c = 1.7 m/s

First we calculate x at t = 4.1 s

x = 4.1(4.1)³ - 2.2(4.1)² + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we calculate speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s²

6 0

1 day ago

A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa

Maru [1038]

Answer:

(a) 16.777 miles

(b) Yes, he exceeded the speed limit

Explanation:

(a)

We need to perform the necessary calculations to convert kilometers to miles:

$27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi$

Thus, the distance of the trip in miles is:

$d=16.77706389mi$

(b)

Next, we will compute the man's speed during the journey:

$v=\frac{d}{t}$

Before that, we must convert minutes to hours:

$16min*\frac{1h}{60min} =2.666666667h$

The resulting speed is:

$v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}$

Consequently:

$62.91398959\frac{mi}{h}>55\frac{mi}{h}$

Thus, it can be concluded that the driver was speeding

8 0

14 hours ago

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Sav [1081]

Answer: $F_{v} =\mu_{k} mg$

The force required has a magnitude of 2601.9 N

Explanation:

m = 450 kg

Static friction coefficient μs = 0.73

Kinetic friction coefficient μk = 0.59

The force necessary to initiate movement of the crate is $F_{s} =\mu_{s} mg$ .

Once the crate begins to move, the frictional force decreases to $F_{v} =\mu_{k} mg$ .

To maintain the motion of the crate at a steady velocity, we must lower the pushing force to $F_{v} =\mu_{k} mg$ .

Subsequently, the pushing force aligns with the frictional force stemming from kinetic friction, enabling balanced forces and consistent velocity.

<pMagnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

6 days ago