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Mariana [72]
3 years ago
15

Introduction This section describes the principles and concepts related to the experiment. It is used to help the person reading

your report to understand the information that serves as foundation for the experiment. It also includes a brief statement about the purpose of the lab. Example: Surface tension is the attractive force exerted upon the surface molecules of a liquid by the molecules beneath. Water has a high surface tension because there are strong attractive forces between water molecules. Adding surfactants, like soap, can reduce the surface tension of a liquid. The purpose of the experiment was to determine the effect of soap on the surface tension of water.
Physics
1 answer:
ArbitrLikvidat [17]3 years ago
6 0

Answer:

See explanation

Explanation:

It is a common observation that when water drops on a clean glass surface, it appears as little bags of water.

Surface tension makes water surface act like a stretched elastic skin bag.

Surface tension is caused by attractive forces between the molecules at the surface and the molecules in the bulk of a liquid. Remember that the forces of cohesion between water molecules are very strong.

When surfactants such as soap are added to water, the surface tension of water is decreased, hence the experiment.

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What evidence supports the conclusion that a chemical reaction has occurred?
GalinKa [24]

Answer:

Precipitation formed

Temperature change

fizzing and hissing

bubbles

change in state

Explanation:

3 0
2 years ago
Can you please answer this question?
Sidana [21]
I think its b too but i may be wrong
5 0
3 years ago
a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d
USPshnik [31]

Answer:

The time is t = 2.595 \  s

The speed is v = 25.43 \ m/s

Explanation:

From the question we are told that

    The height of the cliff is  h =  33 \  m

 Generally from kinematic equation we have that

       h  =  ut + \frac{1}{2} gt^2

before the jump the persons initial velocity is  u =  0 m/s

 So

        33   =  0 * t + \frac{1}{2} 9.8 * t^2

=>      t = 2.595 \  s

Generally from kinematic equation

     v= u + gt

=>  v= 0 + 9.8 * 2.595

=>  v = 25.43 \ m/s

3 0
3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f
Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

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

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m

The height h above the ground at which the acorn was is given by,

h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0
3 years ago
A Ford Mustang weighs about 3500 pounds, and can accelerate from 0-60 MPH in about 5 seconds. What force is responsible for this
Hunter-Best [27]

We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

m = 3500lb = 1587.57kg

The values of the velocities are,

\text{Initial Velocity} = V_i = 0

\text{Final Velocity} = V_f = 60mph = 26.822m/s

We know that the acceleration is equivalent to the change of the speed in a certain time therefore

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

a = \frac{26.822-0}{5}

a = 5.36m/s^2

Now applying the Newton's second law we have,

F= ma

F = (1587.57)(5.36)

F = 8516.36N

Therefore the approximate magnitude is 8516.36N

5 0
3 years ago
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