A mass with mass 4 is attached to a spring with spring constant 24 and a dashpot giving a damping 20. The mass is set in motion with initial position 4 and initial velocity 2. (All values are given in consistent units.) Find the position function

Answer:

d) Both blocks have had equal losses of energy to friction

Explanation:

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces

Moreover, the block A is twice as fast than block B at the time of crossing the finish line

So based on the above information,  it contains the losses of identical friction

And we also know that

Friction energy loss is

[tex]= \mu \times m \times g \times D[/tex]

It would be the same for both the blocks

hence, the option d is correct

The correct answer will be both blocks have had equal losses of energy to friction.

Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.

As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.

Moreover, the block A is twice as fast as block B at the time of crossing the finish line.

So based on the above information,  it contains the losses of identical friction.

And we also know that

Friction energy loss is

[tex]E_f=\mu m g D[/tex]

It would be the same for both the blocks

Hence both blocks have had equal losses of energy to friction.

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Answer:

v = Square Root of StartFraction 2 times K E over m EndFraction.

Explanation:

Kinetic Energy ,K.E = 1/2 × mass × velocity squared

K.E = 1/2 × mv2

Hence v2 = (2× K.E)/m

v = √ 2K.E/m

Expressing it in computer language we have :

v = Square Root of StartFraction 2 times K E over m EndFraction.

Answer:

v = Square Root of StartFraction 2 times K E over m EndFraction

Explanation:

also d on edgenuity:)

Answer:

The constant 0.091 in the astronauts' equation of the best fit line is equal to   [tex]\frac{L}{T^2}[/tex]

The value of  g on Mars is  [tex]g = 3.593 \ m/s^2[/tex]

Explanation:

From the question we are told that

     The line of best fit is defined by the equation  [tex]y = 0.091 x \ and \ R^2 = 1[/tex]

Now the equation of a straight line is defined as

       [tex]y = mx + c[/tex]

Now comparing the given equation to this we have that

        [tex]m = slope = 0.091[/tex]

Now from the graph the formula for the slope is  

          [tex]m = \frac{L}{T^2}[/tex]

=>      [tex]0.091 = \frac{L}{T^2}[/tex]

Now from the question we are told that

        [tex]T = 2 \pi \sqrt{\frac{L}{g} }[/tex]

=>     [tex]\frac{g}{4\pi r^2} = \frac{L}{T^2} = 0.091[/tex]

=>     [tex]g = 4\pi^2 * 0.091[/tex]

=>      [tex]g = 3.593 \ m/s^2[/tex]

Answer: 56.72 ft/s

Explanation:

Ok, initially we only have potential energy, that is equal to:

U =m*g*h

where g is the gravitational acceleration, m the mass and h the height.

h = 50ft and g = 32.17 ft/s^2

when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Then we have:

K = U

m*g*h = (m/2)*v^2

we solve it for v.

v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s

Answer:

Q = 5.9 nC (Approx)

Explanation:

Given:

Further distance = 10 cm

Electric potential(V) = 190 v

Potential difference(V1) = 140 v

Find:

Magnitude of the charge Q

Computation:

V = KQ / r

190 = KQ / r.............Eq1

V1  = KQ / (r+10)

140 = KQ / (r+10) ............Eq2

From Eq2 and Eq1

r = 28 cm = 0.28 m

So,

190 = KQ / r

190 = (9×10⁹)(Q) / 0.28

53.2 = (9×10⁹)(Q)

5.9111 = (10⁹)(Q)

Q = 5.9 nC (Approx)

Answer:

126 mWb

Explanation:

Given that:

length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.

We assume that the magnetic field in the solenoid is constant.

The magnetic flux is given as:

[tex]\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb[/tex]

Answer:

V₂ = 800 Volts (rms)

Explanation:

The turns ratio of an ideal transformer is given by the following formula:

N₁/N₂ = V₁/V₂

where,

N₁ = No. of turns in primary coil of the transformer = 1000 N

N₂ = No. of turns in secondary coil of the transformer = 8000 N

V₁ = rms Voltage on primary side of the transformer = 100 V

V₂ = rms Voltage on secondary side of the transformer = ?

Therefore, using these values in equation, we get:

1000 N/8000 N = 100 V/V₂

V₂ = (100)(8) Volts

V₂ = 800 Volts (rms)

Answer:

We know that the density of the ice is smaller than the density of the water (and this is why the ice floats in water).

Dw > Di

Da is the density of the water and Di is the density of the ice

Since in Bowl A we have a volume V, only of water, then the mass of the bowl A is:

Dw*V.

Now, in the bowl B we have a combination of water and ice, suppose that Vw is the volume of water and Vi is the volume of ice, and we know that:

Vw + Vi = V.

Then the mass in this second bowl is:

Dw*Vw + Di*Vi = Dw*(V - Vi) + Di*Vi = Dw*V + (Di - Dw)*Vi

and we know that Dw > Di, then the left term is a negative term, then the mass of bowl B is smaller than the mass of bowl A.

Answer:

This is an inelastic collision. This means, unfortunately, that KE cannot save you, at least in the problem's current form.  

Let's see what conservation of momentum in both directions does ya:

Conservation in the x direction:

Only 1 object here has a momentum in the x direction initally.  

m1v1i + 0 = (m1 + m2)(vx)

3.09(5.10) = (3.09 + 2.52)Vx

Vx = 2.81 m/s

Explanation:

Conservation in the y direction:

Again, only 1 object here has initial velocity in the y:

0 + m2v2i = (m1 +m2)Vy

(2.52)(-3.36) = (2.52 + 3.09)Vy

Vy = -1.51 m/s

++++++++++++++++++++

Now that you have Vx and Vy of the composite object, you can find the final velocity by doing Vf = √Vx^2 + Vy^2)

Vf = √(2.81)^2 + (-1.51)^2

Vf = 3.19 m/s

Answer:

Explanation:

Given that:

The mass of the cell is 9.0 x 10^-14 kg

The charges of the cell is -2.5pC and the other -3.30 pC

[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]

Radius is  3.75 × 10-6 m

The final distance is twice the radius

i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]

The formula for the velocity of the cell is

[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]

[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]

[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]

The maximum acceleration of the cells as they move toward each other and just barely touch is

[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]

[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]

[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]

The answers obtained are;

1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].

2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].

The answer is explained as shown below.

We have, the mass of the red blood cell;

Also, the charges of the cells are;

The distance between the charges when they barely touch will be two times the radius of each charge.

1. As both the cells are negatively charged they will repel each other.

Now, substituting all the known values in the equation, we get;

[tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]

2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.

Substituting the known values, we get;

[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]

[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]

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Answer:

The answer is 3.48 seconds

Explanation:

The kinematic equation

y= y0+V0*t+1/2*a*(t*t)

-50=0+(0)t+1/2(-9.8)*(t*t)

t=3.194 seconds

During ribbons ball,

x=x0+ Vt+1/2*a*(t*t)

x= 0+(15)*(3.194)+1/2*(0)* (3.194*3.194)

x= 47.9157m

So, distance (D) = 100-47.9157= 52.084m

52.084m=0+15(t)+1/2*(0)(t*t)

t=52.084/15=3.472286= 3.48seconds

Answer:

A) v = 148,242.72 m/s

B) v = 6,328,025.58 m/s

Explanation:

To solve this, we will equate electric potential to kinetic energy.

Formula for Electric potential is qV where q is charge and V is potential difference.

While formula for kinetic energy is ½mv² where m is mass and v is velocity

Thus;

qV = ½mv²

Let us make the velocity the formula;

v = √(2qV/m)

A) PROTON

Charge of proton has a constant value of 1.6 × 10^(-19) C

Mass of proton has a constant value of 1.66 × 10^(-27) kg

We are given that potential difference = 114 V.

So, v = √(2qV/m)

Thus; v = √(2*1.6 × 10^(-19)*114/(1.66 × 10^(-27)))

v = 148,242.72 m/s

B) ELECTRON

Charge of electron has a constant value of 1.6 × 10^(-19) C

Mass of electron has a constant value of 9.11 × 10^(-31) kg

v = √(2qV/m)

Thus;

v = √(2*1.6*10^(-19)*114)/(9.11 × 10^(-31)))

v = 6,328,025.58 m/s

Answer:

Explanation:

Gravitational force between two objects having mass m₁ and m₂ at a distance R

F = G m₁ m₂ / R²

Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²

= 1.01 x 10⁻⁶ N

b )

Force between baby and Jupiter

F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²

= 1.31 x 10⁻⁶  N

c )

Ratio = 1.01 / 1.31

= .77

Water requires more energy to raise its temperature than sand does.  In fact, of all the common substances that we see around us every day, water is one of the BEST at storing heat energy.

This is a big part of the reason why we use frozen water to cool our soda, instead of cold wood or cold steel balls.  

It's also a big part of the reason why we warm up the bed in the Winter with a hot water bag, instead of a bag of hot rocks or hot BBs.

On a hot summer day, the temperature of the sand is much higher than the temperature of the water. The same amount of energy was supplied by the sun to both the sand and the water, but the water required more energy to raise its temperature than the sand.

The specific heat of any substance is explained by the amount of heat required to increase the temperature by 1 degree; here, the specific heat of water is much higher than that of sand. The sand needs 670 joules of energy to raise the temperature, while the water needs nearly 3800 joules of energy to raise one degree of temperature.

Despite the fact that the sun cast the same amount of light on both water and sand, sand heated up faster than water. The water has a high latent heat of vaporization, which means it needs more energy to vaporize. The animal body maintains homeostasis as a result of this water.

Hence, water requires more energy to raise the temperature due to its high specific heat.

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Answer:

t = 166 years

Explanation:

In order to calculate the amount of years that electrons take to cross the complete transmission line. You first calculate the drift speed of the electrons by using the following formula:

[tex]v_d=\frac{I}{nqA}[/tex]             (1)

I: current on the wire = 1,010A

n: free charge density = 8.50*10^28 electrons/m^3

A: cross-sectional area of the transmission line = π*r^2

r: radius of the cross-sectional area = 2.00cm = 0.02m

You replace the values of the parameters in the equation (1):

[tex]v_d=\frac{1,010A}{(8.50*10^{28}electron/m^3)(1.6*10^{-19}C)(\pi (0.02m)^2)}\\\\v_d=5.9*10^{-5}\frac{m}{s}[/tex]

Next, you use the following formula:

[tex]t=\frac{x}{v_d}[/tex]                     (2)

x: length of the line transmission = 310km = 310,000m

You replace the values of vd and x in the equation (2):

[tex]t=\frac{310,000m}{5.9*10^{-5}m/s}=5.24*10^9s[/tex]

Finally, you convert the obtained t to seconds

[tex]t=5.24*10^9s*\frac{1\ year}{3.156*10^7s}=166.03\ years[/tex]

The electrons take approximately 166 years to travel trough the complete transmission line

Answer:

12m/s

Explanation:

[tex]v_f=v_o+at[/tex]

Let's call the velocity that the car maintains for 10 seconds [tex]v_f_1[/tex], and the final velocity [tex]v_f_2[/tex].

[tex]v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s[/tex]

Hope this helps!

A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. The final speed of the car is 12 m/s.

The velocity of an object is usually referred to as the change rate of the object in a specified direction.

Given that:

Using the first equation of motion;

Let v₁ be the initial speed of the car

v₁ = u₁ + a₁t₁

Since the car starts from rest, the initial speed = 0 m/s

v = 0 + (4.0 m/s²) × 5.0 s

v = 0 + 20 m/s

v = 20 m/s

Using the same first equation of motion;

Let v₂ be the final speed of the car

v₂ = u₂ + a₂t₂

v₂ = 20 m/s + (-2 m/s²) × (4.0 s)

v₂ = (20 - 8) m/s

v₂ = 12 m/s

Therefore, we can conclude that the final speed of the car is 12 m/s.

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Answer:

The capillary rise of the glycerin is most nearly  [tex]y = 0.0204 \ m[/tex]

Explanation:

From the question we are told that

  The diameter of the glass tube is  [tex]d = 1 \ mm = 0.001 \ m[/tex]

   The density of glycerin is  [tex]\rho = 1260 \ kg /m^3[/tex]

   The surface tension of the glycerin is [tex]\sigma = 6.3 *10^{-2} \ N /m[/tex]

The capillary rise of the glycerin is mathematically represented as

       [tex]y = \frac{4 * \sigma * cos (\theta )}{ \rho * g * d}[/tex]

substituting value  

       [tex]y = \frac{4 * 6.3 *10^{-2} * cos (0 )}{ 1260 * 9.8 * 0.001}[/tex]

      [tex]y = 0.0204 \ m[/tex]

Therefore the height  of the glass tube  the glycerin was able to cover is

[tex]y = 0.0204 \ m[/tex]  

Answer:

The value of the potential energy of the particle at point B is 85 joules.

Explanation:

According to the Principle of Energy Conservation, the energy cannot be created nor destroyed, only transformed. The particle at point A has kinetic and potential energy and receives a work due to an external conservative force (Work-Energy Theorem), whose sum is equal to potential energy at point B. Mathematically speaking, the expression that describes the phenomenon is:

[tex]K_{A} + U_{A} + W_{A \rightarrow B} = U_{B}[/tex]

Where:

[tex]K_{A}[/tex] - Kinetic energy at point A, measured in joules.

[tex]U_{A}[/tex] - Potential energy at point A, measured in joules.

[tex]W_{A \rightarrow B}[/tex] - Work due to conservative force from A to B, measured in joules.

[tex]U_{B}[/tex] - Potential energy at point B, measured in joules.

The initial kinetic energy of the particle is:

[tex]K_{A} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v[/tex] - Velocity, measured in meters per second.

If [tex]m = 0.4\,kg[/tex] and [tex]v = 10\,\frac{m}{s}[/tex], then:

[tex]K_{A} = \frac{1}{2}\cdot (0.4\,kg)\cdot \left(10\,\frac{m}{s} \right)^{2}[/tex]

[tex]K_{A} = 20\,J[/tex]

Finally, the value of the potential energy at point B is:

[tex]U_{B} = 20\,J + 40\,J + 25\,J[/tex]

[tex]U_{B} = 85\,J[/tex]

The value of the potential energy of the particle at point B is 85 joules.

The potential energy of the particle at point B is 85 J.

Given to us:

Mass of the particle, [tex]m=0.40\ kg[/tex]

velocity of the particle, [tex]v= 10\ m/s[/tex]

potential energy of the particle, [tex]PE= 40\ J[/tex]

Workdone from pt. A to B, [tex]WD_{(A\rightarrow B)} = 25\ J[/tex]

Calculating the kinetic energy of the particle,

[tex]\begin{aligned}KE&= \frac{1}{2}mv^2 \\\\&=\frac{1}{2}\times0.40\times (10)^2\\\\&=20 J\\\end{aligned}[/tex]

According to the  Principle of Energy Conservation,

The energy cannot be created nor be destroyed, it can only be transformed from one form to another.Therefore,

Also,

Total Energy at point A ,

[tex]\begin{aligned}(TE)_A &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_A+ PE_A+UE_A+ WD_{(0\rightarrow A)}\\&=20+40+0+0\\&=60\ J\end{aligned}[/tex]

Total Energy at point B,

[tex]\begin{aligned}(TE)_B &= Kinetic\ Energy+ Potential\ Energy+ Internal Energy+ Workdone\\ &=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\end{aligned}[/tex]

As the total energy is conserved from point A to B and also an external work is done on the particle. we can write the above equation as,

[tex]\begin{aligned} TE_B&=KE_B+ PE_B+UE_B+ WD_{(A\rightarrow B)}\\&=(KE_B+ PE_B+UE_B)+ WD_{(A\rightarrow B)}\\&= TE_A+ WD_{(A\rightarrow B)}\\&=60+25\\&=85\ J\end{aligned}[/tex]

Therefore, the total energy for the particle at point B is 85 J but as the particle is not moving neither work is done at point B, the total energy of the particle is the potential energy of the particle.

Hence, the potential energy of the particle at point B is 85 J.

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Answer:

I = 7.5*10^-10 kg m/s

Explanation:

In order to calculate the impulse you first take into account the following formula:

[tex]I=m\Delta v=m(v-v_o)[/tex]       (1)

m: mass of the pollen grain = 1.0*10^-7g = 1.0*10^-10 kg

v: final speed of the pollen grain = ?

vo: initial speed of the pollen grain = 0 m/s

Next, you calculate the final speed of the pollen grain by using the information about the acceleration and time. You use the following formula:

[tex]v=v_o+a t[/tex]       (2)

a: acceleration = 2.5*10^4 m/s^2

t: time = 0.30ms = 0.30*10^-3 s

[tex]v=0m/s+(2.5*10^4m/s^2)(0.30*10^{-3}s)=7.5\frac{m}{s}[/tex]

Next, you replcae this value of v in the equation (1) and calculate the impulse:

[tex]I=m(v-v_o)=(1.0*10^{-10}kg)(7.5m/s-0m/s)=7.5*10^{-10}kg.\frac{m}{s}[/tex]

The impulse delivered to the pollen grain is 7.5*10^-10 kg m/s

Answer:

The astronauts are separated by 28 m.

Explanation:

The separation of the astronauts can be found by conservation of linear momentum:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f} [/tex]

[tex] m_{1}*0 + m_{2}*0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex]

[tex] m_{1}v_{1f} = -m_{2}v_{2f} [/tex]

[tex] v_{1f} = -\frac{m_{2}v_{2f}}{m_{1}} = -\frac{80v_{2f}}{60} [/tex]

Now, the distance (x) is:      

[tex] x = \frac{v}{t} [/tex]  

The distance traveled by the astronaut 1 is:

[tex] x_{1} = v_{1f}*t = -\frac{80v_{2f}}{60}*t [/tex]    (1)

And, the distance traveled by the astronaut 2 is:

[tex] x_{2} = v_{2f}*t [/tex]  (2)

From the above equation we have:  

[tex] t = \frac{x_{2}}{v_{2f}} [/tex]    (3)                                    

By entering equation (3) into (1) we have:    

[tex] x_{1} = -\frac{80v_{2f}}{60}*(\frac{x_{2}}{v_{2f}}) [/tex]

[tex] x_{1} = -\frac{4*12}{3} = -16 m [/tex]    

The minus sign is because astronaut 1 is moving in the opposite direction of the astronaut 2.      

Finally, the separation of the astronauts is:

[tex] x_{T} = |x_{1}| + x_{2} = (16 + 12)m = 28 m [/tex]

Therefore, the astronauts are separated by 28 m.

I hope it helps you!

The total separation between the two astronauts is 28m.

The given parameters:

Apply the principle of conservation of momentum to determine the final velocity of each astronauts as follows;

[tex]m_1u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\60(0) + 80(0) = 60(v_1) + 80(v_2)\\\\0 = 60v_1 + 80v_2\\\\-60v_1 = 80v_2\\\\v_1 = \frac{-80v_2}{60} \\\\v_1 = -1.333v_2[/tex]

Let the time when astronaut 2 moved 12 m = t

The distance traveled by astronaut 1 is calculated as;

[tex]x_1 = v_1 t\\\\x_1 = -1.333v_2t[/tex]

The  distance traveled by astronaut 2 is calculated as;

[tex]x_2 = v_2 t\\\\12 = v_2t\\\\t = \frac{12}{v_2}[/tex]

Now solve for the distance of astronaut 1

[tex]x_1 = - 1.333v_2 \times t\\\\x_1 = -1.333 v_2 \times \frac{12}{v_2} \\\\x_1 = -16 \ m[/tex]

The total separation between the two astronauts is calculated as follows;

[tex]d = |x_1| + x_2\\\\d = 16 + 12\\\\d = 28 \ m[/tex]

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Answer:

The correct answer is option E

Explanation:

Relative density of the different phases of the same compound like water are basically determined by their number of molecules per volume when each of the molecules have the same mass in each of their phases.

Now, for the water vapor phase, it's molecules have very little interaction with themselves and so they are at large distance apart, whereas in ice(solid), molecules are in continuous contact with each other because they are at close distance between each other. Therefore, it's obvious that there are less molecules per liter in water vapour than in ice, and thus the density is smaller.

The correct answer is option E

Answer:

See the answer below.

Explanation:

"Latent Heat", also called the "Heat of Vaporization", is the amount of energy necessary to change a liquid to a vapour at constant temperature and pressure.

Best Regards!

Answer:

Position of object is;

s(t) = 4t³/3 + 3t + 1

Explanation:

We are told that the velocity has an expression;

v(t) = 3.00 m/s + ( 4.00 m/s³)t²

Now, to get the expression for the position(s(t)) of the object, we have to integrate the velocity expression. Thus;

s(t) = ∫3 + 4t²

s(t) = 3t + 4t³/3 + c

Now, we were told that at x = 1.00 m, time t = 0.000 s

Thus, plugging the values in;

1 = 3(0) + 4(0³/3) + c

c = 1

Thus,the expression for the position of the object is;

s(t) = 4t³/3 + 3t + 1

Answer:

46648 J

Explanation:

mass m= 85 Kg

velocity v = 56 m/s

distance covered s =40 m

According to Question,

frictional drag force to him that matched his weight

[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]

Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J

W=Q= F_d×s

=833×56 = 46648 J

Answer:

According to Newtons 2nd law of motion ;

  The acceleration an object experiences is as a result of the net force which is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

Explanation:

This law is simply saying ;

Force = Mass ×Acceleration

I Hope It Helps  :)

if the velocity of the car reduces from 70km/h to 50km/h then the speed of the car will be equal to the speed of the lorry...

thus the relative velocity will be 0

Answer:

The energy dissipated is  [tex]E = 9.8 J[/tex]

Explanation:

From the question we are told that

    The distance covered at constant velocity d =  2 m  

      The velocity is  [tex]v = 1.5 \ m/s[/tex]

Generally at constant velocity the magnetic  force on the bar is mathematically represented as

         [tex]F = m * g[/tex]

substituting values

        [tex]F = 0.5 * 9.8[/tex]

        [tex]F = 4.9 \ N[/tex]

The energy dissipated is mathematically evaluate as

         [tex]E = F * d[/tex]

 substituting the value

         [tex]E = 4.9 * 2[/tex]

        [tex]E = 9.8 J[/tex]

Answer:

[tex]\theta=65.18rad[/tex]

Explanation:

The angle in rotational motion is given by:

[tex]\theta=\frac{w_o+w_f}{2}t[/tex]

Recall that the angular speed is larger than regular frequency (in rpm) by a factor of [tex]2\pi[/tex], so:

[tex]\omega_f=2\pi f\\\omega_f=2\pi*250rpm\\\omega_f=1570.80 \frac{rad}{min}[/tex]

The wheel spins from rest, that means that its initial angular speed is zero([tex]\omega_o[/tex]). Finally, we have to convert the given time to minutes and replace in the first equation:

[tex]t=5s*\frac{1min}{60s}=0.083min\\\theta=\frac{\omega_f}{2}t\\\theta=\frac{1570.800\frac{rad}{min}}{2}(0.083min)\\\theta=65.18rad[/tex]

Answer:

   h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

       Em₀ = K = ½ m v₁²

final point. Highest point of the curl

        [tex]Em_{f}[/tex] = U = m g y

Find the height y = 2R

      Em₀ = Em_{f}

      ½ m v₁² = m g 2R

       v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

    -fr = m a                      (1)

Y Axis  

      N - W = 0

      N = mg

the friction force has the formula

     fr = μ  N

     fr = μ m g

    we substitute 1

    - μ mg = m a

     a = - μ g

having the acceleration, we can use the kinematic relations

    v² = v₀² - 2 a x

    v₀² = v² + 2 a x

the length of this zone is x = 2R

    let's calculate

     v₀ = √ (4 gR + 2 μ g 2R)

     v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

     Em₀ = U = m g h

final point. Just before reaching the area with rubbing

     [tex]Em_{f}[/tex] = K = ½ m v₀²

      Em₀ = Em_{f}

     mgh = ½ m 4gR(1 + μ)

       h = ½ 4R (1+ μ)

       h = 2 R (1 +μ)