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2 years ago

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Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on

the ball as long as it is in the tube. Part A How high does the ball go above the top of the tube

1 answer:

tekilochka [14]2 years ago

6 0

Answer:

2.87 m

Explanation:

Given:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Upward force by compressed air on the ball (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assume)

Final velocity of the ball at the end of tube (v) = ?

Acceleration of the ball (a) = ?

Weight of the ball is equal to the product of mass and gravity. So,

Weight (W) = $mg=0.06\times 9.8=0.588\ N$

Therefore, the net force acting on the ball is equal to the difference of upward force and downward force. So,

Net force = Force by air - Weight

$F_{net}=F-mg\\F_{net}=3.0-0.588 = 2.412\ N$

Now, from Newton's second law, the net force is equal to the product of object's mass and acceleration. So,

$F_{net}=ma\\\\a=\frac{F_{net}}{m}=\frac{2.412\ N}{0.06\ kg}=40.2\ m/s^2$

Now, acceleration (a) = 40.2 m/s²

Using equation of motion, we have:

$v^2=u^2+2ah\\\\v^2=0+2\times 40.2\times 0.7\\\\v=\sqrt{56.28}=7.5\ m/s$

Now, let the maximum height reached be 'H'.

So, applying energy conservation from the top of pipe to the maximum height.

Decrease in kinetic energy = Increase in potential energy.

$\frac{1}{2}mv^2=mgH\\\\H=\frac{v^2}{2g}$

Plug in the given values and solve for 'H'. This gives,

$H=\frac{56.28}{2\times 9.8}\\\\H=\frac{56.28}{19.6}=2.87\ m$

Therefore, the ball reaches a height of 2.87 m above the top of the tube.

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x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
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2 years ago

A plane increases its speed from 450 km/hr to 750 km/hr in 0.30 hour. What is its acceleration?

ryzh [129]

Answer:

The acceleration is found as:

a = 1000 km/h²

Explanation:

Initial speed of the plane = 450 km/h

Final speed of the plane = 750 km/h

Time taken = 0.3 h

Acceleration can be defined as the change of speed of the object divided by the time taken to bring that change

Acceleration is given as:

$a=\frac{v_f-v_i}{t}$

where v(f) = 750 km/h , v(i) = 450 km/h , t = 0.3 h

Substitute the values in the equation of acceleration

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2 years ago

A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0

Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

$F_{lift}=F_{up} - F_{dw}=0.5*p*V^2$

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

$F_{lift}=0.5*p*V^2=W$

By replacing the known values it is possible to find the plate's weight:

$F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W$

$W=72.6 N$

When the plate kept to 30° from the vertical the moment equation balance is written as:

$F_{lift}=0.5*p*V^2=W*sen(30\°)$

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

$V=\sqrt(W*sen(30\°)/0.5p)$

$V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})$

$V=\sqrt(60.5 \frac{N}{kg/m^3})$

$V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})$

$V=\sqrt(60.5 \frac{m^2}{s^2})$

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2 years ago

you want to compare brands of paper towels to see which holds the most liquid. the independent variable in your experiment would

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2 years ago

Батискаф витримує тиск 60 МПа. Чи можна провести дослідження

Elanso [62]

1) Yes

2) $6.34\cdot 10^9 N$

Explanation:

1)

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

$p=p_0+\rho g h$

where

$p_0 = 101,300 Pa$ is the atmospheric pressure

$\rho$ is the fluid density

$g=10 m/s^2$ is the acceleration due to gravity

h is the height of the column of fluid

Here we have:

$\rho=1030 kg/m^3$ is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

$p=101,300+(1030)(10)(5440)=56.1\cdot 10^6 Pa = 56.1 MPa$

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

2)

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

$p=\frac{F}{A}$

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

$p=56.1\cdot 10^6 Pa$ is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

$A=4\pi r^2$

Therefore, we can find the force exerted on it by re-arranging the previous equation:

$F=pA=4\pi pr^2 = 4\pi (56.1\cdot 10^6)(3)^2=6.34\cdot 10^9 N$

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2 years ago