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lisabon 2012 [21]
2 years ago
11

Hold a spoon such that the water will hit the back of the spoon. The handle of the spoon will be horizontal. Turn on the water,

but not full strength though.
Physics
1 answer:
antoniya [11.8K]2 years ago
6 0

This is an experiment that explains the Bernoulli principle. According to this principle, at the bottom of the spoon, there will be a reduction in pressure.

<h3>Explanation of the Bernoulli principle</h3>

So the pressure at the bottom of the spoon reduces because the velocity of the water is larger and the height of the column of water is smaller at this part.

So the spoon is attracted to the water at the point of lowest pressure because objects gravitate to low-pressure points.

Learn more about the Bernoulli Principle at:

brainly.com/question/24623919

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3 years ago
An object is allowed to fall freely near the surface of a planet. The object has an acceleration due to gravity of 24 m/s2. How
Alborosie

Answer:

12 m

Explanation:

The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For this problem,

g=24 m/s^2

and

u = 0, since we are considering the first second of motion

So, substituting t = 1 s, we find

s=0+\frac{1}{2}(24)(1)^2=12 m

6 0
3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

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The correct answer is 2.8s

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Answer:

Radiative zone

4 0
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