In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are
?
A. the same
B. opposite
C. constantly changing.

Answer:

a) Find the work done by the resistive force during the roundtrip.

W=-30kJ

b) Is the resistive force a conservative force? explain.

The resistive force is not a conservative force since the work done during the round trip is not zero

Explanation:

The worf done on object y a constant force F is given by:

W= (F  cos ∅)S

Where S is the displacement and ∅ is the angle between the force and the displacement.

The displacement of the bicycle during each part of the trip is s=5000m and teh magnitude of teh resistance force is F=3.0N

∅1=180° he angle between the displacement and the force

W1=W2

W1 = (3.0 cos180) 5000m

W1=-15.O kJ

W=W1+W2

W=-30kJ

The resistive force is not a conservative force since the work done during the round trip is not zero

(a) The work done by the resistive force is 15,000 J

(b) The work done the resistive force is non-conservative since the resultant resistive force in not zero.

Work is said to be when an applied force displaces an object from its initial position.

The work done by the resistive force is calculated as follows;

W = FΔr

W = 3 x (5,000 - 0)

W = 15,000 J

Thus, the work done the resistive force is non-conservative since the resultant resistive force in not zero.

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Answer:

[tex]E=30.78\ J[/tex]

Explanation:

The force constant of the spring, k = 76 N/m

The extension in the spring, x = 0.9 m

We need to find the energy is stored in the extended spring. The energy stored in the spring is given by :

[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 76\times (0.9)^2\\\\E=30.78\ J[/tex]

So, 30.78 J of energy is stored in the spring.

Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.

Answer:

roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge

Explanation:

Answer:

They are the resultant vector.

Answer:

-the ratio of the speed of light

in air to the speed of light in the substance.

-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.

-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5

Explanation:

Answer:

c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.

Explanation:

Buoyancy can be defined as a force which is created by the water displaced by an object.

Simply stated, buoyancy is directly proportional to the amount of water that is being displaced by an object.

Hence, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.

The buoyancy of an object is given by the formula;

[tex] Fb = pgV [/tex]

[tex] But, \; V = Ah [/tex]

[tex] Hence, \; Fb = pgAh [/tex]

Where;

Fb = buoyant force of a liquid acting on an object.

g = acceleration due to gravity.

p = density of the liquid.

v = volume of the liquid displaced.

h = height of liquid (water) displaced by an object.

A = surface area of the floating object.

The unit of measurement for buoyancy is Newton (N).

In this scenario, you are standing on the bottom of a lake with your torso above water. Thus, there is a buoyant force that is proportional to the volume of your body that is below the level of the water.

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

a) The total electric potential is 2282000 V

b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in diagram 1 below is;

[tex]V_P=V_1+V_2[/tex]

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we know that; the Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

[tex]r_1^2=0.015^2+0.0125^2[/tex]

[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]

[tex]r_1 = \sqrt{0.00038125}[/tex]

[tex]r_1 = 0.0195[/tex]

Also

[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]

[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]

[tex]r_2 = \sqrt{0.000549[/tex]

[tex]r_2 = 0.0234[/tex]

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

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Answer:

POSITIVE CHARGE,  NEGATIVE CHARGE,  ZERO

Explanation:

To solve this completion exercise, we must remember that charges of the same sign repel each other and in a metallic object (frying pan) the charge is mobile.

Let's analyze the situation when we touch the pan, the charges are neutralized, therefore when we bring the pan to the plate that has a positive charge, it attracts the mobile negative charges in the pan, until it is neutralized, therefore on the opposite side of the pan. pan (edge ​​with a glued tape) is left with a positive charge; therefore the edge and the tape, which is very light, have positive charges and repel each other.

We must assume that the frying pan is insulated so that the net charge is zero, since the induction process.

Consequently the words to complete the sentence are

When the pan is on the plate, the edge of the plate has a _POSITIVE CHARGE_____.

This means that the base of the container is loaded NEGATIVE CHARGE_____ because the net charge of the container is ___ZERO_

Answer:a) - 0.4905 m/s²   b) distance = 35.48 m

Explanation:

Given that  

The initial velocity of the skater = 5.90 m/s

 kinetic friction coefficient = 0.0500

final velocity = 0 m/s(since it  comes to rest)

deceleration cause by the kinetic friction = ?

we know that  

F = μN

and N= mg

Therefore;

F = μ m g....................(1)

also  that

F = m a........................(2)

with our common Force, F, equating  (1) and (2), we have that

m a = - μ m g

a = - μ g

a = - 0.05 × 9.81

a = - 0.4905 m/s²

The deceleration cause by the kinetic friction is a = - 0.4905 m/s²

b)

The distance the skater  travels before stopping

is given as

    Vf² = v₀² - 2 a x

  final velocity = 0 m/s(since it  comes to rest)  

Therefore We have that

 0 = v₀² - 2 a x

 x = - v₀² / 2 a

 x = 5.90² / (2 x 0.4905  )

34.81/0.981  

x = 35.48 m

Or

using

v²-u² = 2aS final velocity = 0 m/s(since it  comes to rest)  

0²-5.90² = -2×0.4905×S

34.81=0.981S

S= 34.81/0.981

S=35.48m

Answer:

.50 M

Explanation:

5*.50=2.5 + 2*.25=.5 = 3n

6*.50= 3N

Final answer is .50M

Answer:

the impulse must be the same in these two cases    F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

Explanation:

For this exercise we use the relationship between momentum and momentum

         I = Δp

         F t = m v_f - m v₀

To know the speed we use the conservation of energy

starting point. Highest point

       Em₀ = U = m g h

fincla point. Just before the crash

      Em_f = K = ½ m v²

energy is conserved

        Em₀ = Em_f

        m g h = ½ m v²

         v = [tex]\sqrt{2gh}[/tex]

we substitute in the impulse relation

     F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases

Air resistance is exerted by an engine

Answer:

The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe should be filled with radiation that is literally the remnant heat left over from the Big Bang, called the “cosmic microwave background", or CMB.

Explanation:

The Big Bang theory suggest  that the universe in early stage was at very hot place and which can be expanded, the gas within it cools. It is in an infinite universe and it has no edge.

The Big Bang theory is a cosmological model which explain the existence of the observable universe from the earliest periods to the large-scale evolution.

The model describes the mechanism behind the universe expansion from an initial state of high density and temperature, it is very important concept as a lot of research is going on in this field to find out exactly how the universe began billions of years ago.

The universe began to cool down in order to allow the formation of particles become atoms after its initial phase of expansion,  Primordial elements  such as Hydrogen, Helium, and Lithium are condensed through gravity are formed early stars and galaxies.

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Answer:

a. Workdone = 44100 Joules

b. Power = 8820 Watts.

Explanation:

Given the following data:

Mass = 225kg

Distance = 20m

Time = 5 seconds

To find the workdone;

Workdone = force * distance

But force = mg

We know that acceleration due to gravity is equal to 9.8m/s²

Force = 225*9.8 = 2205N

Substituting the values into the equation, we have;

Workdone = 2205 * 20

Workdone = 44100 Joules

b. To find the power;

Power = workdone/time

Power = 44100/5

Power = 8820 Watts.

Answer:

A. v = √2gh

B. No! The final velocity does not depend on the mass of the car.

C. Yes! the final velocity depends on the steepness of the hill

D. 3.28 m/s

Explanation:

A. Determination of the final velocity.

½mv² = mgh

Cancel out m

½v² = gh

Cross multiply

v² = 2gh

Take the square root of both side

v = √2gh

B. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is no mass (m) in the formula.

Thus, the final velocity does not depend on the mass of the car.

C. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is height (h) in the formula.

Thus, the final velocity depends on the steepness of the hill

D. Determination of the final velocity.

Height (h) = 0.55 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = √2gh

v = √(2 × 9.8 × 0.55)

v = √10.78

v = 3.28 m/s

Answer: 0.435 m

Explanation:

Given

mass m=20 kg

initial speed u=2 m/s

coefficient of kinetic friction [tex]\mu_k=0.3[/tex]

deceleration which opposes the motion is given by

[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]

[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]

using [tex]v^2-u^2=2as[/tex]

[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]

Answer:

The speed of this particle is constantly [tex]c[/tex].

Explanation:

Position vector of this particle at time [tex]t[/tex]:

[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].

Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:

[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].

Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].

Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:

[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].

The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :

[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].

Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)

Answer:

1. Temperature= 869.35 K

2. Pressure of combustion = 12994.043 kpa

3. Thrust = 127x10⁶N

Explanation:

this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.

1.

The temperature = (273+2400k) - (3800)²/2(4003)

= 2673 - 14440000/8006

= 2673 - 1803.65

= 869.35 K

Approximately 869.4K

2. We first get mach number

= 3800/√1.3(923.8)(869.35)

= 3800/1021.78

= 3.719

Pressure = 100kpa[1+2.07464415]^1.3/0.3

= 12995.043kpa

C. Thrust

Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)

= 12678.621

= 126.781 kN

Thrust is approximately 127kN = 127x10⁶N

Answer:

Numerous studies have shown that the economic costs of divorce fall more heavily on women. After separation, women experience a sharper decline in household income and a greater poverty risk (Smock 1994; Smock and Manning

Answer:

sadness and stress...................

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

Answer:

mgh

Explanation:

Assume the height of the body is 1.8m.

The gravity?of the body is G=mg

the height of the gravity center is about 0.9m

E=mgh

=m*9.8m/s*0.9m

= 8.82mJ

Answer:

10 mm

Explanation:

We'll begin by calculating the spring constant of the spring. This can be obtained as follow:

Extention (e) = 5 mm

Force (F) = 125 N

Spring constant (K) =?

F = Ke

125 = K × 5

Divide both side by 5

K = 125 / 5

K = 25 N/mm

Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:

Force (F) = 250 N

Spring constant (K) = 25 N/mm

Extention (e) =?

F = Ke

250 = 25 × e

Divide both side by 25

e = 250 / 25

e = 10 mm

Thus, the spring will stretch 10 mm when a 250 N force is applied.

Answer:

C

Explanation:

Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.  

A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).

The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.

Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.

Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.  

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The simplest definition of a black hole is an object that is so dense that not even light can escape its surface. If we squished the Earth's mass into a sphere with a radius of 9 mm, the escape velocity would be the speed of light. Just a wee-bit smaller, and the escape velocity is greater than the speed of light.

HopeItHelps!

Answer:

solar energy

wind power

geothermal energy

hydraulic power

biomass energy

energy storage

(That's all I know).

Answer:

Change in momentum is zero.

Explanation:

The following data were obtained from the question:

Mass (m) = 5000 kg

Time (t) = 1 h

Net force (F) = 0

Change in momentum =?

Force = Rate of change of momentum

0 = change in momentum

Change in momentum = 0

We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.

Answer:

75N

Explanation:

a = v/t = 3/2

F = ma = 50(3/2) = 75