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3 years ago

4. Analyze: What can you say about the acceleration of dividers when the pressure increases

Physics

1 answer:

Pavlova-9 [17]3 years ago

5 0

Answer:

if increase the pressure in a system the acceleration should increase

Explanation:

The definition of pressure is

P = F / A

Where F is the force and A the area

F = P A

Let's write Newton's second law

F = ma

We substitute

P A = m a

So we see that the pressure is directly proportional to the acceleration, if increase the pressure in a system the acceleration should increase

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A football is thrown due north across a 40 meter river and it takes 2 seconds to cover that distance.

Alexxx [7]

Answer:

I. Speed = 20m/s

II. Velocity = 20m/s due North.

Explanation:

<u>Given the following data;</u>

Distance = 40m

Time = 2secs

To find the speed;

Mathematically, speed is given by the formula;

$Speed = \frac{distance}{time}$

Substituting into the equation, we have;

$Speed = \frac{40}{2}$

<em>Speed = 20m/s.</em>

In physics, we use the same formula for calculating speed and velocity. The only difference is that speed is a scalar quantity and as such has magnitude but no direction while velocity is a vector quantity and as such it has both magnitude and direction.

$Velocity = \frac{distance}{time}$

<em>Therefore, the velocity is 20m/s due North</em>.

6 0

3 years ago

20 points and brainiest

DIA [1.3K]

Answer:

electrical energy sometimes.

Explanation:

125.0m

300 degree Fahrenheit

7 0

3 years ago

In billiards, the 0.165 kg cue ball is hit toward the 0.155 kg eight ball, which is stationary. The cue ball travels at 5.8 m/s

Damm [24]

Answer:

another ball velocity = 3.92 m/s and with 30° clockwise from initial direction

Explanation:

given data

mass m1 = 0.165 kg

mass m2 = 0.155 kg

before collision velocity v1 = 5.8 m/s

before collision velocity v2 = 0

angle = 35.0° from initial direction

after collision 1st ball velocity v3 = 3.2 m/s

to find out

after collision another ball velocity v4

solution

we consider here ball move in x axis and after collision 1st ball move upside of x axis with angle 35 degree and other ball move downside with x axis with angle θ

so from conservation of momentum we say

m1v1 = m1v3cos35 + m2v4cosθ with x axis .............1

m1v3sin35 = m2v4sinθ with y axis .............2

so from 1 equation

0.165 × 5.8 = 0.165(3.2)cos35 + 0.155(v4)cosθ

v4 cosθ = 3.38 .................3

form 2 equation

0.165(3.2)sin35 = 0.155(v4)sinθ

v4 sinθ = 1.95 ......................4

so magnitude of another ball velocity is square and adding equation 3 and 4

another ball velocity = √(3.39²+1.96²)

another ball velocity = 3.92 m/s

and direction is tanθ = 1.96/3.39

θ = 30° clockwise from initial direction

3 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

The spectrum of light emitted by helium atoms is different than that emitted by hydrogen atoms because

MaRussiya [10]

<h3><u>Answer;</u></h3>

The different atoms have different quantized energy levels

<h3><u>Explanation;</u></h3>

The atoms of different elements have different energy levels because they have different nuclear charges and spins, and different numbers of electrons.
Each different kind of atom, like hydrogen or radon, has a distinct nuclear charge and number of electrons. This makes the potential energy function different for each atom, and therefore results in different energy levels.
In an emmission spectra, each bright band corresponds to a difference between energy levels within the atom.

4 0

3 years ago