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2 years ago

Lesson 2 History of Physical Science

Physics

1 answer:

kolezko [41]2 years ago

4 0

The prompt above requires an explanatory essay. The focus of an explanatory essay is to give clarity to a concept. See the sample below.

<h3>What is the explanatory essay as per the prompt above?</h3>

From history, it is clear that Science builds on itself. For avoidance of doubt, this means that improvements in science are incremental with each improvement feeding off the last.

From the lesson we have examples of this phenomena such as that of Galileo. Recall that the text indicates that He improved the telescope; and while carrying out the same research that his predecessor Copernicus had done in relation to the stars and planets, Galileo used math to justify his position.

It is also on record that Johannes Kepler who lived in the same period as Galileo took interest in Astronomy. In this case, science built on itself because, Kepler's work served as the foundation for Sir Isaac Newton's work. This is also another example of how science builds on itself.

From the above textual evidence, it is given beyond reasonable double that science indeed builds on itself with the discovery of each generation serving as the foundation for the advancement of the next.

Learn more about Explanatory Essays at;
brainly.com/question/26984401
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Describe how a metamorphic rock might become a sedimentary rock over time.

andriy [413]

<span>Metamorphic rock undergoes weathering, erosion; the particles are deposited and undergo lithification.</span>

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3 years ago

A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.

NemiM [27]

Answer:

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, $\omega=2.8\ rad/s$

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

$K_t=\dfrac{1}{2}mv^2$

$K_t=\dfrac{1}{2}m(r\omega)^2$

$K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2$

The rotational kinetic energy of the ball is :

$K_r=\dfrac{1}{2}I \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2$

Ratio of translational to the rotational kinetic energy as :

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0

3 years ago

An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a uniform volume charge density p. Dete

xenn [34]

Answer: The electric field is: a) r<a , E0=; b) a<r<b E=ρ (r-a)/εo;

c) r>b E=ρ b (b-a)/r*εo

Explanation: In order to solve this problem we have to use the Gaussian law in diffrengios regions.

As we know,

∫E.dr= Qinside/εo

For r<a --->Qinside=0 then E=0

for a<r<b er have

E*2π*r*L= Q inside/εo in this case Qinside= ρ.Vol=ρ*2*π*r*(r-a)*L

E*2π*r*L =ρ*2*π*r* (r-a)*L/εo

E=ρ*(r-a)/εo

Finally for r>b

E*2π*r*L =ρ*2*π*b* (b-a)*L/εo

E=ρ*b* (b-a)*/r*εo

3 0

3 years ago

A gardener mows a lawn with an old-fashioned push mower. The handle of the mower makes an angle of 320 with the surface of the l

Virty [35]

Answer:

N = 337.96 N

Explanation:

∅ = 32º

F = 249 N

m = 21 Kg

N = ?

We can apply:

∑ F = 0 (↑)

- Fy - W + N = 0 ⇒ N = Fy + W

⇒ F*Sin ∅ + m*g = N

⇒ N = (249 N*Sin32º) + (21 Kg*9.81 m/s²)

⇒ N = 337.96 N (↑)

8 0

3 years ago

A bucket of mass M (when empty) initially at rest and containing a mass of water is being pulled up a well by a rope exerting a

Naily [24]

Answer:

$V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT$

Explanation:

Given that

Constant rate of leak =R

Mass at time T ,m=RT

At any time t

The mass = Rt

So the total mass in downward direction=(M+Rt)

Now force equation

(M+Rt) a =P- (M+Rt) g

$a=\dfrac{P}{M+Rt}-g$

We know that

$a=\dfrac{dV}{dt}$

$\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g$

$\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt$

$V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT$

$V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT$

This is the velocity of bucket at the instance when it become empty.

6 0

3 years ago