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3 years ago

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A small hole is made at the bottom of a plastic cup. If it is filled with water and allowed to fall freely, will waterfall down

from the hole?

2 answers:

Alex_Xolod [135]3 years ago

6 0

Answer:

Yes, because of both gravity and the fact that nothing is obstructing the hole.

klemol [59]3 years ago

6 0

Answer:

It depends on how big the hole is.

Explanation:

If the hole is too big, the water will fall freely out of the cup. However, if it is just a small opening (from a needle, pencil/pen tip, etc.) the water pressure will be able to hold in the water.

You might be interested in

In 2002, a gargantuan iceberg broke away from the Ross Ice Sheet in Antarctica. It was approximately a rectangle 218 km long, 25

Musya8 [376]

Answer:

a) 1.25e15 kg

b) 4.17e20 J

c) 44.55 years

Explanation:

To find the volume you need to multiply 218 km * 25 km * 250 m (be careful with units), so the volume is 1.3625e12 m^3, if you multiply this value by the density you will obtain the mass, that is 1.25e15 kg.

To find the energy needed to melt the ice, you use the latent heat, in this case, it is 3.34e5 J/kg. Now you multiply this value by the mass, so you need 4.17e20 J to melt the iceberg.

The surface area of the iceberg is 545e7 m^2, so the ice absorbs 594e9 W, one W is one J/s, so in 12 hours the iceberg absorbs 2.56e16 J, so in 365 days absorbs 9.36e18 J. Now you just divide 4.17e20 J by the amount f energy per year, and obtain 44.55 years.

5 0

3 years ago

A 1200-kg car is travelling east at a rate of 9 m/s. A 1600-kg truck is travelling south at a rate of 13 m/s. The truck accident

weeeeeb [17]

Answer:

Around 3.57m/s

Explanation:

p=mv

Let's denote the momentum, mass, and velocity of the car with the subscript 1, and for the truck use 2. After the collision, the combined momentum can be denoted with the subscript 3.

$p_1=1200\cdot 9=10800 \\\\p_2=1600\cdot 13=20800 \\\\20800-10800=10000 \\\\1200kg+1600kg=2800kg \\\\10000=2800v_3 \\\\v_3\approx 3.57m/s$

Hope this helps!

3 0

3 years ago

A source charge of 5.0 µC generates an electric field of 3.93 × 105 at the location of a test charge. How far is the test charge

kap26 [50]

Answer:

B. ) 0.34 m

I definitely guessed and got the right answer so :))

6 0

3 years ago

Read 2 more answers

How many planets in the solar system

Nataliya [291]

Answer:

There are eight planets in our Solar System.

Explanation:

7 0

3 years ago

Read 2 more answers

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago