The work of warren washington would be most likely to explain which of the following phenomena

To do your homework correctly you should (5 points)

pychu [463]

Answer:

C according to me

Explanation:

ITS HARD

4 0

3 years ago

Read 2 more answers

17. A volleyball weighs about 300 grams.

atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

1 kilograms = 1000 grams.

We convert it using a rule of 3, replacing, simplifying units and solving:

$\boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}$

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Earth's gravity is known to be 9.8 m/s², so we have:

Data:

m = 0.3 kg
g = 9.8 m/s²
h = 15 m
PE = ?

Use formula of potencial energy:

$\boxed{\bold{PE=m*g*h}}$

Replace and solve:

$\boxed{\bold{PE=0.3\ kg*9.8\frac{m}{s^{2}}*15\ m}}$

$\boxed{\boxed{\bold{PE=44.1\ J}}}$

Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

Greetings.

8 0

3 years ago

Predict whether the changes in enthalpy, entropy, and free energy will be positive or negative for the melting of ice, and expla

BlackZzzverrR [31]

<span>Melting of ice is an endothermic process, meaning that energy is absorbed. When ice spontaneously melts, ΔH (change in enthalpy) is "positive". ΔS (entropy change) is also positive, because, becoming a liquid, water molecules lose their fixed position in the ice crystal, and become more disorganized. ΔG (free energy of reaction) is negative when a reaction proceeds spontaneously, as it happens in this case. Ice spontaneously melts at temperatures higher than 0°C. However, liquid water also spontaneously freezes at temperatures below 0°C. Therefore the temperature is instrumental in determining which "melting" of ice, or "freezing" of water becomes spontaneous. The whole process is summarized in the Gibbs free energy equation:
ΔG = ΔH – TΔS</span>

4 0

2 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

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

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

3 years ago

An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh

lara [203]

Answer:

$a_y = 4.9\ m/s^2$

Explanation:

Given,

Width of rectangular tank, b = 1 m

Length of the tank, l = 2 m

height of the tank, d = 1.5 m

Depth of gasoline on the tank, h = 1 m

$\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}$

$\dfrac{dz}{dy}=-0.5$

The differential form with the acceleration

$\dfrac{dz}{dy}=\dfrac{-a_y}{a_z + g}$

$-0.5=-\dfrac{a_y}{a_z + g}$

acceleration in z-direction = 0 m/s²

g = 9.8 m/s²

a_y is the horizontal acceleration of the gasoline.

$0.5=\dfrac{a_y}{0 + 9.8}$

$a_y = 9.8\times 0.5$

$a_y = 4.9\ m/s^2$

Hence, Horizontal acceleration of the gasoline before gasoline would spill is equal to 4.9 m/s²

3 0

3 years ago