A 7.8 µF capacitor is charged by a 9.00 V battery through a resistance R. The capacitor reaches a potential difference of 4.20 V at a time 3.21 s after charging begins. Find R.

Answer:

Answer to the question:

Explanation:

Differences between ionic bond and covalent bond:

The ionic bond occurs between two different atoms (metallic and non-metallic), while the covalent bond occurs between two equal atoms (non-metallic).

In the covalent bond there is an electron compartment, while in the ionic bond there is an electron transfer.

Ionic bonds have a high melting and boiling point, while covalent bonds usually have a low point.

Final answer:

An ionic bond involves the transfer of electrons, resulting in a bond between a positive ion and a negative ion. A covalent bond involves the sharing of electrons between atoms, creating a stable electron arrangement for both atoms.

Explanation:

An ionic bond is formed by the transfer of electrons from one atom to another, resulting in a bond between a positive ion (cation) and a negative ion (anion). On the other hand, a covalent bond is formed by the sharing of electrons between atoms, creating a bond in which electrons are shared rather than transferred.

In an ionic bond, one atom gains electrons and becomes negatively charged, while the other atom loses electrons and becomes positively charged. This attraction between oppositely charged ions forms the bond. In a covalent bond, atoms share one or more pairs of electrons, resulting in a stable electron arrangement for both atoms.

Examples of compounds with ionic bonds include sodium chloride (NaCl) and magnesium oxide (MgO), while examples of compounds with covalent bonds include water (H2O) and methane (CH4).

Answer:

a) Travel time = 60.56 s

b) Maximum speed = 36.33 m/s

Explanation:

a)  Distance = 1.1 km = 1100 m

 A subway train accelerates at +1.2 m/s² from rest  through the first half of the distance and decelerates at −1.2 m/s² through the second half.

 So half the distance is traveled at an acceleration of +1.2 m/s².

 We have equation of motion [tex]s=ut+\frac{1}{2}at^2[/tex]

 Substituting

         [tex]\frac{1100}{2}=0\times t+\frac{1}{2}\times 1.2t^2\\\\t=30.28s[/tex]

 Travel time = 2 x 30.28 = 60.56 s

b) We have equation of motion v = u+at

   Substituting t = 30.28 s and a = 1.2 m/s²

   v = 0 + 1.2 x 30.28 = 36.33 m/s

   Maximum speed = 36.33 m/s

c) Photos of graphs are given

Answer:

a) ttotal=60.58 s

b) v^2=36.3 m/s

c) the attached photo shows the graphics.

Explanation:

a) we have the following:

S1=1.1x10^-3/2=550 m

a=1.2 m/s

we have the following formula:

v^2=u^2+2as

replacing values:

v^2=0+(2*1.2*550)=36.33 m/s

The time is equal to:

t1=v/a=36.33/1.2=30.3 s

we have the following:

u=36.33 m/s

v=0

S=550 m/s

using the following expression:

v=u+at2

clearing t2:

t2=(v-u)/a=(0-36.33)/-1.2=30.28 s

ttotal=t1+t2=30.3+30.28=60.58 s

b) the mass speed is equal to:

v^2=u^2+2as=36.3 m/s

Answer:

250 N

Explanation:

m = Mass of cart = 50 kg

F = Pushing Force = 250 N

g = Acceleration due to gravity = 9.81 m/s²

μ = Coefficient of friction

[tex]F_s =\text{Static frictional force between the floor and the crate}= 275\ N[/tex]

N = Weight of cart = mg

N = 50×9.81

N = 490.5 N

[tex]F_s =\mu N\\\Rightarrow \mu=\frac{F_s}{N}\\\Rightarrow \mu=\frac{275}{490.5}\\\Rightarrow \mu=0.56[/tex]

Here it can be seen that the pushing force is less than the static frictional force so the crate will not move

∴ The frictional force on the crate while the person pushes on it is 250 N

Answer:

Speration between the slits [tex]d=2.86\ \mu m[/tex]

Explanation:

Given that,

Wavelength of light, [tex]\lambda=612\ nm=612\times 10^{-9}\ m[/tex]

It is incident on a double slit produces a second-order maximum at an angle of 25 degree.

The condition for maxima is given as :

[tex]dsin\theta=m\lambda[/tex]

d = seperation between the slits

m = order, m = 2

[tex]d=\dfrac{m\lambda}{sin\theta}[/tex]

[tex]d=\dfrac{2\times 612\times 10^{-9}\ m}{sin(25)}[/tex]

d = 0.00000289

or

[tex]d=2.86\mu m[/tex]

So, the seperation between the silts is [tex]2.86\ \mu m[/tex]. Hence, this is the required solution.

Answer:

W = 627.2 J

Explanation:

Given:

[tex]\rho_{chain}[/tex] = 2kg/m

length of chain = 12 m

length pulled will be = 8 m

We know

Work done (W) = mgh

where

m = mass of the object

g= acceleration due to gravity

h = displacement

For a small length dy of the chain, the work done can be written as:

dW  = (mass of the small length pulled)× g×dy

dW = 2kg/m ×dy×9.8×y

where, y is the distance from the ground level of the end of chain

integerating the above equation

W = [tex]\int\limits^8_0 {19.6y} \, dx[/tex]

W =[tex][19.6\frac{y^2}{2}]_{0}^{8}[/tex]

W = 627.2 J

To find the mass needed to produce equilibrium, we perform a torque balance about the pivot point. This involves adding up the torques on each side of the pivot point and setting them equal to each other, then solving for the unknown mass at 10 cm.

In this question, we are dealing with torque and equilibrium. For a system to be in equilibrium, the sum of the torques about any pivot point must be zero. Given a 0.5-kg uniform meter stick suspended by a single string at the 30-cm mark, with a 0.2-kg mass hanging at the 80-cm mark, we want to find the mass that should be hung at the 10-cm mark to ensure equilibrium.

To find this, we set up the following equation for the torques about the pivot point (the 30-cm mark):

Torque_due_to_0.5kg_rod + Torque_due_to_hanging_mass_at_10_cm = Torque_due_to_0.2kg_mass_at_80cm

Each torque can be calculated as the product of the force (which is the weight, or mass times gravity) and the distance from the pivot. After plugging in the given values and solving for the unknown mass at 10 cm, you will find the mass that should be hung at the 10 cm mark to produce equilibrium.

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Answer:

peak value of magnetic field is

[tex]B_o = 2.9 \times 10^{-6} T[/tex]

peak value of electric field is

[tex]E_o = 868 N/C[/tex]

Explanation:

As we know that the intensity of electromagnetic waves is given by the formula

[tex]I = \epsilon_0 E_{rms}^2 c[/tex]

here we have

[tex]I = 1000 W/m^2[/tex]

[tex]\epsilon_0 = 8.85 \times 10^{-12}[/tex]

[tex]c = 3\times 10^8[/tex]

now we have

[tex]1000 = (8.85 \times 10^{-12})E_{rms}^2(3 \times 10^8)[/tex]

[tex]E_{rms} = 614 N/C[/tex]

now peak value of electric field is given as

[tex]E_o = \sqrt2 E_{rms}[/tex]

So the peak value of electric field is given as

[tex]E_o = 868 N/C[/tex]

Now we have relation between electric field and magnetic field as

[tex]B_o = \frac{E_o}{c}[/tex]

[tex]B_o = \frac{868}{3\times 10^8}[/tex]

[tex]B_o = 2.9 \times 10^{-6} T[/tex]

Answer:

Part a)

[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]

Part b)

[tex]C = 4.96 \times 10^{-11} farad[/tex]

Part c)

[tex]Q = 0.69 \mu C[/tex]

Part d)

[tex]E = 4.86 \times 10^{-3} J[/tex]

Explanation:

Part a)

As we know that dielectric constant of pyrex glass is 5.6 and its dielectric breakdown strength is given as

[tex]E = 14 \times 10^6 V/m[/tex]

now we have

[tex]E . d = \Delta V[/tex]

[tex](14 \times 10^6)(0.001) = \Delta V[/tex]

so we have

[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]

Part b)

Capacitance is given as

[tex]C = \frac{k\epsilon_0 A}{d}[/tex]

[tex]C = \frac{5.6(8.85 \times 10^{-12})(10 \times 10^{-4}}{0.001}[/tex]

[tex]C = 4.96 \times 10^{-11} farad[/tex]

Part c)

Now we have

[tex]Q = C\Delta V[/tex]

[tex]Q = (4.96 \times 10^{-11})(14 \times 10^3)[/tex]

[tex]Q = 0.69 \mu C[/tex]

Part d)

[tex]Energy = \frac{1}{2}CV^2[/tex]

[tex]E = \frac{1}{2}(4.96 \times 10^{-11})(14 \times 10^3)^2[/tex]

[tex]E = 4.86 \times 10^{-3} J[/tex]

Answer:

B = 12.46 T

Explanation:

At n = 1 state we know that radius is given as

[tex]R = 5.29 \times 10^{-11} m[/tex]

now we have

[tex]T = \frac{2\pi R}{v}[/tex]

here we know that speed is given in that

[tex]v = 2.18 \times 10^6 m/s[/tex]

now the time period is given as

[tex]T = \frac{2\pi R}{v}[/tex]

[tex]T = \frac{2\pi (5.29 \times 10^{-11})}{2.18 \times 10^6}[/tex]

[tex]T = 1.52 \times 10^{-16} s[/tex]

Now the electric current due to revolution of charge is given by

[tex]i = \frac{e}{T}[/tex]

[tex]i = \frac{1.6 \times 10^{-19}}{1.52 \times 10^{-16}}[/tex]

[tex]i = 1.05 \times 10^{-3} A[/tex]

now magnetic field at the center position is given as

[tex]B = \frac{\mu_0 i}{2R}[/tex]

[tex]B = \frac{4\pi \times 10^{-7} (1.05 \times 10^{-3})}{2(5.29 \times 10^{-11}}[/tex]

[tex]B = 12.46 T[/tex]

The speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.

The speed of a body is the rate at which it covers the total distance is in the time taken.

If the speed of the body has direction, then it is known as the velocity. The velocity is the vector quantity.

The constant force applied on the object is  <200, 460, -150> N. It can be written in the vector form as,

[tex]\vec F=200\hat i+460\hat j-150\hat k[/tex]

The initial position vector of the object for the location <7, -34, -7> can be given as,

[tex]d_i=7\hat i-34\hat j-7\hat k[/tex]

The final position vector of the object is for the location <12, -42, -11> m can be given as,

[tex]d_f=12\hat i-42\hat j-11\hat k[/tex]

Thus, the total distance traveled by the constant force is,

[tex]d=(12\hat i-42\hat j-11\hat k)-(7\hat i-34\hat j-7\hat k)\\d=5\hat i-8\hat j-4\hat k[/tex]

The work done by a object is the product of force and displacement. Thus, work done on the object is,

[tex]W=\vec F.\vec d \\W=(200\hat i+460\hat j-150\hat k).(5\hat i-8\hat j-4\hat k)\\W=1000-3680+600\\W=-2080 \rm J[/tex]

Now, the work done on a object is equal to the kinetic energy of the body. Thus, work done on the object is,

[tex]W=\dfrac{1}{2}m(v^2_f-v^2_i)[/tex]

Here, v(f) is the final speed and v(i) is the initial speed of the object. As the mass of the object is 80 kg and initial speed is 14 m/s. Thus, put the values as,

[tex]-2080=\dfrac{1}{2}(80)(v^2_f-14^2)\\v_f=12\rm m/s[/tex]

Thus, the speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.

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To find the speed of the object at the final location, we need to calculate the net force acting on it using the equation F = ma. By plugging in the values, we can calculate the final speed of the object.

To determine the speed of the object at the final location, we need to calculate the net force acting on the object using the equation F = ma, where F is the force, m is the mass, and a is the acceleration.

Given that the force acting on the object is <200, 460, -150> N, and the mass of the object is 80 kg, we can calculate the acceleration using the formula a = F/m. After calculating the acceleration, we can use it to find the final speed of the object using the equation v^2 = u^2 + 2as, where v is the final speed, u is the initial speed, a is the acceleration, and s is the displacement.

By plugging in the given values, we can calculate the final speed of the object at the location <12, -42, -11> m.

Hey there!:

subject 1

Mechanical work =mgh

= 130 *9.8 *4.4

= 5605.6 J

Mechanical power =Work/time

= 5605.6 / 3.4

= 1648.706 W

Subject 2

Mechanical work =mgh

= 75 * 9.8 * 4.4

= 3234 J

Mechanical power =Work/time

= 3234/2.8

= 1155 W

Athlete 1 is more powerful in absolute sense

relative to mass

athlete 1 power =12.68 W/kg

athlete 2 power =15.4 W/kg

thus athlete 2 is more power in relative to mass power  .

metabolic energy spent by athlete 1 :

= (5605.6 / 0.25)*0.000239 kCal

= 5.359 kCal

metabolic energy spent by athlete 2=(3234/0.25)*0.000239 kCal

= 3.092 kCal

Hope this helps!

Answer:

[tex]T = 1.26 \times 10^8 K[/tex]

Explanation:

As we know that rms speed of ideal gas is given by the formula

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

here we know that

[tex]v_{rms} = 2.67 \times 10^5 m/s[/tex]

molecular mass of gas is given as

[tex]M = 44 g/mol = 0.044 kg/mol[/tex]

now from above formula we have

[tex]2.67\times 10^5 = \sqrt{\frac{3(8.31)T}{0.044}}[/tex]

now we have

[tex]T = 1.26 \times 10^8 K[/tex]

The root mean square speed of a molecule in an ideal gas can be used to determine the temperature of the gas when vrms is known. This involves solving the formula for temperature using the given vrms and mass of a molecule. We need to ensure that the molar mass is properly converted to the mass of a molecule and that we use SI units.

The subject question is related to the concept of root mean square speed (vrms) in gas physics and involves calculations using the ideal gas model. According to this concept, we know that vrms is given by the equation vrms = sqrt(3kB T / m), where:

Given that vrms = 2.67 x 105 m/s and molar mass (M) = 44.0 g/mol (which must be converted to kg as SI units are used in this context), we can solve the formula for T to find the temperature T = vrms^2×m / (3×kB) after substituting the known values into this equation. Note that one must convert the molar mass to mass of a molecule by dividing it by Avogadro's number (6.022 x 10²³ mol⁻¹).

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The induced current in a loop inside a sinusoidally-driven solenoid can be determined from Faraday's law of electromagnetic induction. The time derivative of the changing flux through the loop, when divided by the resistance of the loop, provides the value of the induced current.

The question asks about the current induced in a loop placed inside a solenoid driven by an alternating current. This is a physics problem related to electromagnetic induction. The alternating magnetic field in the solenoid induces an electric field that in turn induces a current in the loop, by Faraday's Law.

In this case, the magnetic field B inside the solenoid is oscillating sinusoidally with time as B(t) = B cos(ωt). Thus, the magnetic flux Φ through the loop is changing with time. Using Faraday's law, this flux change induces an emf in the loop, which induces a current I. The emf -dΦ/dt is equal to I*R where R is the resistance of the loop. Hence the induced current I as a function of time can be given by I = -(1/R) * (d/dt) (B cos(ωt) * π * (a/2)^2).

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The best way to reduce the force of friction between the box and the floor while the box is moving is 8%.

If two surfaces are in contact and moving relative to one another, then the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on ice.

In static friction, the frictional force resists the force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object.

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Answer:

a) After 26.67 seconds it returns back to the origin

b) Velocity when it returns to the origin = 60 m/s in the -x direction

Explanation:

a) Let the starting position be origin and time be t.

  After time t displacement, s = 0 m

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

 We have equation of motion s = ut + 0.5 at²

 Substituting

        s = ut + 0.5 at²

        0 = 60 x t + 0.5 x (-4.5) x t²        

        2.25t² - 60 t = 0

        t² - 26.67 t = 0

        t (t-26.67) = 0

      t = 0s or t = 26.67 s

So after 26.67 seconds it returns back to the origin

b) We have equation of motion v = u + at

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

  Time , t = 26.67

 Substituting

        v = 60 - 4.5 x 26.67 = -60 m/s

Velocity when it returns to the origin = 60 m/s in the -x direction

At the frequency of 5 MHz, the period of the oscillations is 1/5meg. That's a period of 1/5 microsecond.

There are 5 full cycles in one full microsecond, and there are 2.5 full cycles in a 0.5 us pulse.

You'll have to decide for yourself how damped a pulse of 2.5 cycles is, because the parameters of the definition are corrupted in the question.

Answer:

G1

Explanation:

Some specialized cells such as red blood cells, neurons lose their ability to replicate in the G1 phase of cell cycle. G1 phase is indeed the terminal phase of the cell division as now cell do not divide any more. These cells do not have the necessary nucleus for further cellular replication in the G1 phase. They become mature need not divide anymore.

The terminal phase for specialized cells such as neurons and red blood cells that lose their ability to replicate upon maturity is the G0 phase. In this phase, these cells exit the cell cycle and cease active preparation for division. They continue to perform their functions despite the cessation of division.

The terminal phase for these specialized cells such as neurons and red blood cells, which lose their ability to replicate when they mature, is usually referred to as the G0 phase, also known as the resting or quiescent phase. This is as the cell cycle consists of the interphase (G1, S, and G2 phases) and the mitotic (M) phase. After the M phase, cells usually enter the G1 phase of interphase. However, some cells exit the cycle and enter the G0 phase, in which they are not actively preparing to divide.

For non-dividing cells like neurons and red blood cells, they stay in G0 permanently. This means they stop dividing but continue to perform their functions. Such cells have lost the capacity for cell division either temporarily or permanently. This mechanism helps prevent unregulated cell division which can lead to cancers.

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Answer:

30.63

Explanation:

v=2x/t

t=2x/v

t=2*1220.5/8.0

t=30.63//

Answer:

1.25 v

Explanation:

m₁ = mass of first car = m

m₂ = mass of the second car = (0.517) m₁ = (0.517) m

v₁ = initial velocity of first car before collision = v

v'₁ = final velocity of first car after collision = (0.353) v₁ = (0.353) v

v₂ = initial velocity of second car before collision = 0

v'₂ = final velocity of second car after collision = ?

Using conservation of momentum

m₁ v₁ + m₂ v₂ = m₁ v'₁ + m₂ v'₂

(m)(v) + ((0.517) m)(0) = (m)((0.353) v) + ((0.517) m) v'₂

v = (0.353) v + (0.517) v'₂

(0.517) v'₂ = v - (0.353) v

(0.517) v'₂ = (0.647) v

v'₂ = 1.25 v

Answer:

Density (φ) = 0,8827 Kg/L

Specific weight (Ws) = 8,65 N/L

Specific gravity (Gs) = 0,8827 (without unit)

Explanation:

The density formula: φ = [tex]\frac{m}{V}[/tex]

I know the mass "m", I need to find out the volume of the cylinder (V)

V = π* r²*h

The radius "r" is equal to half the diameter (150mm) = 75mm

Now I can find out the density (φ)

φ = [tex]\frac{1,56Kg}{1,767145L}[/tex] = 0,8827 Kg/L

The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:

Ws = φ*g

(g = gravity = 9,8 m/s²)

Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :

Gs = φ(o) /  φ(w)

(φ(w) = 1 Kg/L

Hope this can help you !!

Final answer:

To calculate gauge pressure in a tapered water pipe, apply Bernoulli's equation considering changes in radius, height, and flow rate.

Explanation:

The gauge pressure at a point in a tapered water pipe can be calculated using Bernoulli's equation. Given the information provided, we can determine the pressure difference between two points in the pipe based on the change in radius and height along the pipe.

In this scenario, the gauge pressure at a point 2.65 m downstream from the upper end can be calculated by considering the change in radius and height along the pipe as well as the flow rate of water through the pipe.

Understanding fluid dynamics and applying relevant equations helps determine the pressure variation in tapered pipes based on given parameters.

Answer:

When the separation between the capacitor plates is increased to 2d, and if the battery is disconnected, the charge between the condenser plates does not change, it is the same, and this is represented by the following expression: Q'= Q

Explanation:

Final answer:

The charge (Q) on a capacitor remains constant when its plates are pulled apart after being disconnected from a circuit, even though the capacitance decreases due to the increased separation.

Explanation:

When a capacitor is charged and then disconnected from any circuit, the charge Q on the capacitor's plates remains constant despite changes in plate separation or other geometrical modifications. This is because, once disconnected, there are no paths available for the charge to leave or enter the capacitor. Therefore, when the plates of a charged capacitor are pulled apart to increase the separation from d to 2d, the charge Q does not change. The process of separating the plates requires work to be done against the attractive force between the charges on the capacitor plates. Capacitance (C), which is defined as C = Q/V, where V is the potential difference, will decrease because the capacitance of a parallel plate capacitor is inversely proportional to the separation between the plates (d), according to the formula C = ε0A/d, where ε0 is the permittivity of free space and A is the plate area.

Answer:

option (C)

Explanation:

EMF stand for electro motive force. the emf of a battery is the potential between the two electrodes when it is not use in the circuit.

The terminal potential difference is the potential difference between the electrodes of a cell when it is in use.

EMF is only when the current is very large in the battery.

The emf of a battery is equal to its terminal potential difference only when a large cwrrent is in the battery.

C) only when a large cwrrent is in the battery.

Answer:

Velocity of golf ball≅ 12.87mph    

Explanation:

Using the theory of conservative momentum to both the club head and the golf ball  we have;

Qi1 + Qi2= Qf1 + Qf2

Qi1: initial momentum for the club head

Qf1: final momentum for the club head

Qi2: initial momentum for the golf ball

Qf2: final momentum for the golf ball

momentum (Q) = Mass x velocity

which means the sum of the momentums of both club head and golf ball has to be the same before and after they have collided.

using the Coeficient of restitution e= 0.83 allows us to know what kind of collision we are dealing with, which is a partially elastic collision since

0> e=0.83 >1.

Qi1= 190 x 94= 17,860 mph.g

Qf1= 190 x Vf1

Qi2= 46 x 0= 0 mph.g

Qf2= 46x Vf2

Using the value of e to determine Vf2 as the final velocity of the golf ball:

e= [tex]\frac{Vf2- Vf1}{Vi1 - Vi2}[/tex]

0.83= [tex]\frac{Vf2 - Vf1}{94 - 0}[/tex]

Vf1= (0.83 x 94)+ Vf2

Vf1= 78.02 + Vf2

Qi1 + Qi2= Qf1 + Qf2

17,860 + 0 = 190xVf1 + 46xVf2    

17,860= 190x (78.02+Vf2) + 46xVf2

17,860= 14,823+ 190xVf2+46xVf2

Vf2≅ 12.87mph                

Answer:

0.68 kg-m²

Explanation:

F = Force applied by the muscle = 2615 N

r = effective perpendicular lever arm = 2.85 cm = 0.0285 m

α = Angular acceleration of the forearm = 110.0 rad/s²

I  = moment of inertia of the boxer's forearm = ?

Torque is given as

τ = I α                                   eq-1

Torque is also given as

τ = r F                                   eq-2

using eq-1 and eq-2

r F =  I α

(0.0285)(2615) = (110.0) I

I = 0.68 kg-m²

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

[tex]a(t)=(5-10t)\ m/s^2[/tex]

(a) Since, [tex]a=\dfrac{dv}{dt}[/tex]

v = velocity

[tex]dv=a.dt[/tex]

[tex]v=\int(a.dt)[/tex]

[tex]v=\int(5-10t)(dt)[/tex]

[tex]v=5t-\dfrac{10t^2}{2}=5t-5t^2[/tex]

(b) [tex]v=\dfrac{dx}{dt}[/tex]

x = position

[tex]x=\int v.dt[/tex]

[tex]x=\int (5t-5t^2)dt[/tex]

[tex]x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3[/tex]

(c) Velocity function is given by :

[tex]v=5t-5t^2[/tex]

[tex]5t-5t^2=0[/tex]

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

To find the velocity function of the particle, integrate the acceleration function, which results in 5t - 5t^2. The position function is found by integrating the velocity function, yielding 1/2 * 5t^2 - 1/3 5t^3. The velocity is zero at t = 0 and t = 1 second.

We can solve for the velocity function and the position function by integrating the acceleration function.

Velocity Function

The velocity function is the integral of the acceleration function. Since the acceleration function given is a(t) = 5 - 10t m/s2, the velocity function v(t) is obtained by integrating a(t) with respect to time:

v(t) = ∫ (5 - 10t) dt = 5t - 5t2 + C

Since the particle starts from rest, the initial velocity is 0, which means C = 0. Therefore, the velocity function is v(t) = 5t - 5t2.

Position Function

The position function s(t) is the integral of the velocity function. We integrate v(t) to obtain the position function:

s(t) = ∫ (5t - 5t2) dt = ½ 5t2 - ⅓ 5t3 + K

Since the particle starts from position zero, K = 0 and the position function is s(t) = ½ 5t2 - ⅓ 5t3.

Zero Velocity

To find when the velocity is zero, we set the velocity function equal to zero and solve for t:
5t - 5t2 = 0t = 0 or t = 1 s.

Answer: A. Hot air balloon

Explanation: A hot air balloon is the heaviest, at around 800 pounds, because it carries lots of gas and is very large.

Explanation:

First, convert 87 mi/h to ft/min.

87 mi/h × (5280 ft/mi) × (1 h / 60 min) = 7656 ft/min

The time to reach the home plate is:

t = 60 ft / 7656 ft/min

t = 0.00784 min

The number of revolutions made in that time is:

n = 1710 rev/min × 0.00784 min

n = 13.4 rev

Rounding to 2 significant figures, the ball makes 13 revolutions.

Answer:

The engine's efficiency is 80%.

Explanation:

Given that,

Heat [tex]Q_{in}=75000\ J[/tex]

Remove heat [tex]Q_{out}=15000\ J[/tex]

We need to calculate the engine's efficiency

The efficiency is equal to the difference Input energy and output energy and divided by the input energy.

Using formula of efficiency

[tex]\eta=\dfrac{Q_{in}-Q_{out}}{Q_{in}}[/tex]

[tex]\eta=\dfrac{75000-15000}{75000}[/tex]

[tex]\eta\times100=80\%[/tex]

Hence, The engine's efficiency is 80%.

Answer:

Linear Perspective.

Explanation:

Linear perspective cue to depth and it is in respect to both texture gradient or the next depth cue and relative size. In linear perspective, parallel lines that go along seems to have converged at distance very far, eventually reaching a vanishing point at the horizon. For example the railway track appears to be converging at a point very distant from the observer but it didn't.

Answer:

d. linear perspective

Explanation:

According to a different source, these are the options that come with this question:

a. visual acuity

b. texture gradient

c. retinal disparity

d. linear perspective

Linear perspective refers to a way in which depth is created. This term is particularly employed by artists who create an illusion of depth on a flat surface. This perspective leads all parallel lines to be lost in a single vanishing point on the horizon line. In this example, Gina is using the linear perspective cue to understand the depth of the scene she is observing.