Consider a block sliding down a ramp whose motion is opposed by frictional forces. The total energy of this system is modeled by the equation:
Etotal = 1/2mv^2 + mgh + Ff(f is underscore)d

Which part of the equation represents the amount of energy converted to thermal energy?

A. mg
B. Ffd
C. mgh
D. 1/2 mv^2

Answer:

false ..........false

Answer:

FALSE                                            

Explanation:

Answer:

Q = PV(In 4)

Explanation:

We are told that the volume expands from V to a state with volume 4V.

Thus, initial volume is V and Final volume is 4V.

We want to find How much heat is added to the expanding gas.

For an isothermal process, the work done is calculated from;

W = nRT(In(V_f/V_i))

Where;

V_f is final volume

V_i is initial volume

Thus;

W = nRT(In(4V/V))

W = nRT(In 4)

Now, from ideal gas equation, we know that;

PV = nRT

Thus;

W = PV(In 4)

Now from first law of thermodynamics, we know that internal energy is zero and thus; Q = W

Where Q is quantity of heat

Thus;

Q = PV(In 4)

Answer:

[tex]J = 3~Kg.m/s[/tex]

Explanation:

Impulse and Momentum

The impulse-momentum theorem states that the change in momentum of an object equals the impulse applied to it.

The equation can be written as follows:

[tex]J =\Delta p = p_2-p_1[/tex]

Where:

J    = Impulse

p2 = Final Momentum

p2 = Initial Momentum

The momentum can be calculated as:

p = m.v

Where m is the mass of the object and v is the velocity.

The tennis ball with mass m=60 g = 0.06 Kg was served from rest (v1=0) to v2=50 m/s. The change in momentum is:

[tex]\Delta p = 0.06Kg~50~m/s-0[/tex]

[tex]\Delta p = 3~Kg.m/s[/tex]

Thus the impulse is:

[tex]\marhbf{J = 3~Kg.m/s}[/tex]

Acceleration of an object is the force divided by its mass. Hence, acceleration of the 100 Kg box with an applied force of 550 N is 5.5 m/s².

Acceleration is a physical quantity having both magnitude and direction. This vector measures the rate of change of velocity of a moving body. Thus, acceleration has the unit of m/s².

According to Newton's second law of motion, force acting on a body  is the product of its mass and acceleration. Thus, force is directly proportional to the mass and acceleration of the body.

Given the mass of the box = 100 Kg

 Force applied on it = 550 N.

Acceleration = 550 N / 100 Kg

                      = 5.5 m/s² ( 1 N  = 1 Kg m/s²)

Therefore, the acceleration of the box is 5.5 m/s².

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Answer:

3.32 atm

__________________________________________________________

We are given:

side of the cubical container = 28 cm

number of molecules in the container = 3 * Nₐ  

[where Nₐ is the Avogadro's number]

Temperature = 24°C  OR  297 K

We need to find the pressure exerted by the gas on the walls of the container

__________________________________________________________

Some Calculations:

Volume of the container

we are given the side of the cubical container = 28 cm

Volume of the cubical container = side³

Volume = 28³

Volume = 21952 cm³

We know that 1 cm³ = 1 mL

So,

Volume = 21952 mL

We also know that 1 L = 1000 mL

Volume = 21.952 L

Number of moles of Gas

We know that:

number of moles = number of molecules / Avogadro's number

number of moles = 3 * Nₐ / Nₐ                   [number of molecules = 3 * Nₐ]

number of moles = 3 moles

__________________________________________________________

Pressure Exerted by the Gas:

Using the ideal gas equation:

PV = nRT

Since the volume is in L, and Temperature is in K. R is equal to

0.082 L atm /mol K and the pressure will be in atm

P(21.952) = 3*(0.082)*(297)        

P = 3.32 atm

Hence, the gas will exert a pressure of 3.32 atm on the walls of the container

Answer:

Explanation:

a) Maximum height id expressed as;

H = u²sin² theta/2g

H = 36.6²(sin42.2)²/2(9.8)

H = 1,339.56(0.6717)/19.6

H = 899.81/19.6

H = 45.91m

Hence the maximum height is 45.91m

b) The Time of flight is the total time in air expressed as;

T = 2usin theta/g

T = 2(36.6)sin42.2/9.8

T = 73.2sin42.2/9.8

T = 49.17/9.8

T = 5.017secs

Hence the total time in air is 5.017scs

c) Range = U²sin2(theta)/g

Range = 36.6²sin2(42.2)/9.8

Range = 1,339.5(0.9952)/9.8

Range = 1,333.11/9.8

Range = 136.03m

Hence the range is 136.03m

d) USing the rime of flight formula;

T =2Usintheta/g

1.5 = 2Usin42.2/9.8

2Usin42.2 =1.5*9.8

2Usin42.2 = 14.7

U = 14.7/2sin42.2

U = 14.7/1.3434

U = 10.94m/s

Hence the speed of the projectile is 10.94m/s

Answer:100m west

Explanation:

Answer:

x = 1.26 R

Explanation:

For this exercise let's find the magnetic field using the Biot-Savart law

            B = μ₀ I/4π ∫ ds x r^ / r²

In the case of a loop or loop, the quantity ds is perpendicular to the distance r, therefore the vector product reduces to the algebraic product and the direction of the field is perpendicular to the current loop

suppose that the spiral eta in the yz plane, therefore the axis is in the x axis

         B = μ₀ I/4π ∫ ds / (R² + x²)

The total magnetic field has two components, one parallel to the x axis and another perpendicular, this component is annual when integrating the entire loop, so the total field is

            B = Bₓ i^

using trigonometry

            Bₓ = B cos θ

we substitute

            Bₓ = μ₀ I/4π ∫ ds cos θ / (x² + R²)

the cosine function is

            cos θ = R /√(x² + R²)

The differential is

            ds = R dθ

we substitute

             Bₓ = μ₀ I/4π ∫ (R dθ)  R /√( (x² + R²)³ )

we integrate from 0 to 2π

              Bₓ =μ₀ I/4π R² / √(x² + R²)³   2pi

therefore the final expression is

            B = μ₀ I R²/ 2√(x² + R²)³   i^

In our case the distance is requested where B is half of B in the center of the bone loop x = 0

Spire center field   x=0

              B₀ = μ₀ I/2R

Field at the desired point (x)

              B = B₀ / 2

we substitute

              R² /√(x² + R²)³ = ½  1 /R

              2R³ =√(x² + R²)³

              (x² + R²)³ = 4 (R²)³

              (x²/R² + 1)³ = 4

The exact result is the solution of this equation, but it is quite laborious, we can find an approximate result assuming that the distance x is much greater than R (x »R)

           B = μ₀ I/2x³ 

we substitute

            R² / x³ = 1/2   1 / R

            2R³ = x³

          x = ∛2  R

            x = 1.2599 R

The distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of  R is 0.766 R

Suppose we consider a magnetic field located at point z on the axis of the current loop with radius R carrying a current (I), then the magnetic field can be represented as:

[tex]\mathbf{B = \dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{^{\dfrac{3}{2}}}}}[/tex]

And the field situated at the center of the loop is:

[tex]\mathbf{B_{z=0} =\dfrac{\mu_o}{2} \dfrac{I}{R} }[/tex]

Let consider a distance (z) on the axis of the loop, in which the magnetic field as a result of the loop is equal to half the strength of the magnetic field at the center of the loop;

Then;

[tex]\mathbf{B(z) = \dfrac{1}{2}B_{z=0}}[/tex]

[tex]\mathbf{\dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{\frac{3}{2}}} =\dfrac{1}{2} \Big (\dfrac{\mu_o}{2} \dfrac{I}{R} \Big )}[/tex]

Multiply both sides by (2);

[tex]\mathbf{\dfrac{R^2}{(z^2+R^2)^{\frac{3}{2}}} = \dfrac{1}{2R}}[/tex]

Cross multiply;

[tex]\mathbf{2R^3 = (z^2 +R^2)^{\dfrac{3}{2}}}[/tex]

[tex]\mathbf{4R^6 = (z^2 +R^2)^3}[/tex]

[tex]\mathbf{z = \sqrt{4^{1/3} R^2 -R^2 }}[/tex]

[tex]\mathbf{z = R\sqrt{4^{1/3} -1 }}[/tex]

z = 0.766 R

Therefore, we can conclude that the distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of  R is 0.766 R

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Answer:

You will be very glad to know that my preparation for HSC examination is very good. I have prepared myself for the HSC Examination 2015. I am confident that I will obtain very good marks in the examination. I have revised them several times and I am hopeful of obtaining distinction marks in all the subjects.

Explanation:

This is from Brainly hope this person helped you

Answer:

penis

Explanation:

A spring balance measures the weight of an object by opposing the force of gravity acting with force of an extending spring. May be used to determine mass as well as weight by recalibrating the scale. Some spring balances are available in gram or kilogram markings and are used to measure the mass of an object. Spring balances consist of a cylindrical tube with a spring inside. One end (at the top) is fixed to an adjuster which can be used to calibrate the device. The other end is attached to a hook on which you can hang masses etc.

Answer:

[tex]v=\sqrt{26}~m/s[/tex]

Explanation:

Parametric Equation of the Velocity

Given the position of the particle at any time t as

[tex]r(t) = (x(t),y(t))[/tex]

The instantaneous velocity is the first derivative of the position:

[tex]v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))[/tex]

The speed can be calculated as the magnitude of the velocity:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

We are given the coordinates of the position of a particle as:

[tex]x=5t-3t^2[/tex]

[tex]y=5t[/tex]

The coordinates of the velocity are:

[tex]v_x(t)=(5t-3t^2)'=5-6t[/tex]

[tex]v_y(t)=(5t)'=5[/tex]

Evaluating at t=1 s:

[tex]v_x(1)=5-6(1)=-1[/tex]

[tex]v_y(1)=5[/tex]

The velocity is:

[tex]v=\sqrt{(-1)^2+5^2}[/tex]

[tex]v=\sqrt{1+25}[/tex]

[tex]\mathbf{v=\sqrt{26}~m/s}[/tex]

Answer:

5×10^4Pa

Explanation:

Given force of 108 lb is required to pull the lid off the box,

To convert "Ib"to Newton ,we use conversation rate below

1 pounds = 4.4482216282509 newtons

Then 108 lb=x Newton

Cross multiply we have

X= 480.41Newton

The force that is needed to open the lid is F and pressure P.

We know that Pressure= Force/Area

Area is given as 80.0 cm^2, we can convert to m^2 for unit consistency since 1cm^2= 0.001m^2 then

80.0 cm^2 = 80×10^-4m^2

Substitute to the equation of the pressure we have

P= 480.41Newton/(80×10^4m^2)

P=6×10^4 Pa

The pressure in the box will be difference between the initial pressure and final pressure

=( 1.01 ×10^5 Pa)-(6×10^4 Pa)

= 50100Pa

= 5×10^4Pa

Therefore, The pressure in the box was

5×10^4Pa

Answer:

You are a guest magician in a circus. One of your tricks is to place a football on an inclined plane without the football rolling over is explained below in details.

Explanation:

spinning ball halts after traveling some range due to friction energy act different direction of movement of the ball. you can observe in the figure.

Let any rolling ball of mass (m ) is traveling with velocity v ,

common effect on ball (N) = mg

because of motion, friction energy develops on the contact exterior and begins to resist the movement of the rolling ball.

hence,

fr = uN = umg act on communicating exterior, so, after any time due to friction energy rolling ball gets to rest.

Answer:

451 milliliters equals 15.25 fluid ounces

Explanation:

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three.

To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

So: [tex]x=\frac{c*b}{a}[/tex]

The direct rule of three is the rule applied in this case where there is a change of units.

In this case, the rule of three can be applied in the following way: if there are 29.57 milliliters in 1 fluid ounce, in 451 milliliters how many fluid ounces are there?

[tex]fluid ounces=\frac{451 mL*1 fluid ounce}{29.57 mL}[/tex]

fluid ounces= 15.25

451 milliliters equals 15.25 fluid ounces

Answer:

Explanation:

Given;

initial velocity of the ball, u = 23 m/s

time of motion, t = 2 s

The position of the ball after 2 s is given by;

h = ut - ¹/₂gt²

h = (23 x 2) - ¹/₂ x 9.8 x 2²

h = 46 - 19.6

h = 26.4 m

The velocity of the ball after 2 s is given by;

v² = u² + 2(-g)h

v² = u² - 2gh

v² = 23² - (2 x 9.8 x 26.4)

v² = 529 - 517.44

v² = 11.56

v = √11.56

v = 3.4 m/s

Answer:

horizontal component=fcostita

=150cos60

use calculator to evaluate it

for vertical=fsintita

=150sin60

Answer:

describe, explain, predict, and change/control behavior.

Explanation:

describe: What are they doing? -Pavlov noticed that dogs were salivating when they would see his lab assistant before food was presented to them. This observation acted as a description of what was happening to them.

explain: Why are they doing that?- Pavlov started to look into why they were doing it. There was a stimulus, the assistant giving them food in the past to where they started to salivate at the sight of the lab assistant

predict: What would happen if I responded in this way?- Pavlov predicted that he could get the same reaction if he used a bell as a stimulus. Using this he was able to condition dogs to salivate at the ring of the bell.

change/control: What can I do to get them to stop doing that? Because of this discovery we can use conditioning today. For example, in the classroom teachers can use conditioning with their students to make it easier, parents to teach their children right from wrong and to have good behavior. (you do this bad thing you get time out, do a good thing and I will praise you, etc) It can be used when training employees and many other places.

Answer:

Explanation:

step one:

given data

velocity v=34.2m/s

time t= 8.7s

Step two

Required is the distance the cheetah has covered on the condition

we know that speed= distance/time

make distance subject of formula we have

distance= velocity *time

distance= 34.2*8.7

distance = 297.54m

Therefore the displacement the cheetah covered at that velocity

is 297.54m

Answer:

v = 120 m/s

Explanation:

We are given;

earth's radius; r = 6.37 × 10^(6) m

Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s

Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.

The angle will be;

θ = ¾ × 90

θ = 67.5

¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.

The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:

v = r(cos θ) × ω

v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)

v = 117.22 m/s

Approximation to 2 sig. figures gives;

v = 120 m/s

Answer:

ill get back to this question once i find the answer to it

Answer:

contains many young stars

Explanation:

Irregular galaxies have no definite shape, which means that the first option is incorrect. They are definitely not round.

However, they contain many young stars because the degree of star formation is fast. They also contain old stars. Thus, the second choice is correct.

The "spiral galaxy" is the type of galaxy that has arms that extend from the center. These arms look "spiral," which influenced its name. This makes the last choice incorrect.

They are actually smaller than the other types of galaxies. This makes them prone to collisions. This makes the last choice incorrect.

Answer:

Contains many young stars

Explanation:

Answer:

v₁ = 0.375 m / s ,   x = 0.335 m

Explanation:

Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.

We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.

initial moment. Right at the start of the movement

       p₀ = m v₀ + 0

final moment. Just when it comes to equilibrium

      [tex]p_{f}[/tex] = (m + M) v₁

how the forces are internal

       p₀ =p_{f}

       m v₀ = (m + M) v₁

       v₁ = m /m+M    v₀

let's calculate

       v₁ = 0.4 /(0.4 + 2.8)  3

       v₁ = 0.375 m / s

Let's apply Newton's second law to the Block, to find the friction force

Y axis

       N - W = 0

       N = W

       N = m g

where m is the mass of the block

the friction force has the formula

      fr = μ N

      fr = μ m g

We apply Newton's second law to slab    

X axis

       fr = M a

where M is the mass of the slab

       μ m g = M a

       a = μ g m / M

let's calculate

       a = 0.15  9.8  0.4 / 2.8

       a = 0.21 m / s²

With kinematics we can find the position

       v²= v₀²+2 a x

as the slab is initially at rest, its initial velocity is zero

       v² = 2 a x

       x = v2 / 2a

let's calculate

        x = 0.375²/2 0.21

        x = 0.335 m

Answer:

Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.

Answer:

See the answer below

Explanation:

Average speed = total distance traveled/total time taken

In order to determine the total distance traveled by the coach, consider the attached image.

Distance covered:

30 yd. line to 10 yd. line (A to B)= 20 yds

10 yd. line to 50 yd. line (B to C) = 40 yds

50 yd. line to 20 yd. line (C to D) = 30 yds

Total distance covered = 20 + 40 + 30 = 90 yds

Time taken = 200 seconds

Average Speed = 90/200 = 0.45 yd/s

Velocity = speed with direction

Hence,

His Velocity = 0.45 yd/s to the left of his starting point.

Explanation:

Perspective- the art of drawing solid objects on a two-dimensional surface so as to give the right impression of their height, width, depth, and position in relation to each other when viewed from a particular point.

Answer:

D Magnesia

Explanation:

Ore particles Fe3O4 (magnetite) were found in the region called Magnesia.

Magnesia was an antic city in Asia, named after the inhabitants of Greek Magnesia.

Magnetite is used as an ore, abrasive, in paint production, electrophotography, as a micronutrient fertilizer, and as a high-density concrete aggregate.

Answer:

[tex]\displaystyle x=10\sqrt{3}\ m/s[/tex]

[tex]y=10\ m/s[/tex]

Explanation:

Rectangular coordinates of vectors in 2D

Given a vector with a magnitude v and angle θ with respect to the positive horizontal direction, the x and y components of the vector are given by:

[tex]x=v\cos\theta[/tex]

[tex]y=v\sin\theta[/tex]

The soccer ball is kicked upward at an angle θ = 30° and at a speed v=20 m/s.

The rectangular components of the vector are:

[tex]x=20\cos 30^\circ[/tex]

[tex]\displaystyle x=20\cdot \frac{\sqrt{3}}{2}[/tex]

Operating:

[tex]\mathbf{\displaystyle x=10\sqrt{3}\ m/s}[/tex]

[tex]y=20\sin 30^\circ[/tex]

[tex]\displaystyle y=20\cdot \frac{1}{2}[/tex]

Operating:

[tex]\mathbf{y=10\ m/s}[/tex]

Answer:

20.0V

SANA MAKATULONG

The voltage drop across the 10.0-ohm resistor would be 20.0 volt.

Resistance is the obstruction of electrons in an electrically conducting material.

The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

For calculating equivalent resistance in series combination.

Re = R1  + R2 + R3

For the given problem the total resistance of the circuit would be as all the three resistors are connected in the series combination.

Re= 10+20+30

Re=60 ohm

As given in the figure all the resistance are connected in the series combination therefore the current flowing through them would be the same.

For the given problem we have to design and construct a circuit that has two resistors connected in series.

By using Ohms law

V=IR

120 = 60×I

I = 2 ampere

Given that we have a 120 V battery, that will produce a current of 2 Ampere

By using Ohm's law we can calculate the voltage drop across a 10-ohm resistor

V=IR

  =2×10

  = 20 volt

Thus, the voltage drop across the 10.0-ohm resistor comes out to be 20.0 volt.

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Answer:

If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.

Explanation:

Given;

initial force exerted to hang on, F₁ = 100 N

The force exerted on the merry-go-round in order to hang on must be an inward force known as centripetal force.

Centripetal force is given by;

[tex]F_c = \frac{mv^2}{r} \\\\keeping \ "m" \ and \ "r" \ constant, we \ will \ have \ the \ following \ equation;\\\\\frac{F_c_1}{v_1^2} = \frac{F_c_2}{v_2^2} \\\\F_c_2 = \frac{F_c_1*v_2^2}{v_1^2}\\\\when \ the \ speed\ doubles \ i.e, v_2 = 2v_1\\\\ F_c_2 = \frac{F_c_1*(2v_1)^2}{v_1^2}\\\\ F_c_2 = \frac{F_c_1*4v_1^2}{v_1^2}\\\\F_c_2 = F_c_1 *4\\\\F_c_2 = 4(F_c_1)\\\\F_c_2 = 4 (100 \ N)\\\\F_c_2 = 400 \ N[/tex]

Therefore, If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.