Which statement is the best interpretation of the ray diagrem shown below

Answer:

1. T₁ is approximately 100.33 N

T₂ is approximately -51.674 N

2. 230°F is 383.15 K

3. Part A

The total torque on the bolt is -4.2 N·m

Part B

Negative anticlockwise

Explanation:

1. The given horizontal force = 86 N

The direction of the given 86 N force = To the left (negative) and along the x-axis

(The magnitude and direction of the 86 N force = -86·i)

The state of the system of forces = In equilibrium

The angle of elevation of the direction of the force T₁ = 31° above the x-axis

The direction of the force T₂ = Downwards, along the y-axis (Perpendicular to the x-axis)

Given that the system is in equilibrium, we have;

At equilibrium, the sum of the horizontal forces = 0

Therefore;

T₁ × cos(31°) - 86 = 0

T₁ = 86/(cos(31°)) ≈ 100.33

T₁ ≈ 100.33 N

Similarly, at equilibrium, the sum of the vertical forces = 0

∴ T₁×sin(31°) + T₂ = 0

Which gives;

100.33 × sin(31°) + T₂ = 0

T₂ = -100.33 × sin(31°) ≈ -51.674

T₂ ≈-51.674 N

2. 230° F to Kelvin

To convert degrees Fahrenheit (°F) to K, we use;

[tex]Degrees \ in \ Kelvin, K = (x^{\circ} F + 459.67) \times \dfrac{5}{9}[/tex]

Pluggining in the given temperature value gives;

[tex]Degrees \ in \ Kelvin, K = (230^{\circ} F + 459.67) \times \dfrac{5}{9} = 383.15[/tex]

230°F = 383.15 K

3. Part A

Torque = Force × perpendicular distance from the line of action of the force

Therefore, the clockwise torque = 9 N × 0.4 m = 3.6 N·m (clocwise)

The anticlockeisre torque = 13 N × 0.6 m = 7.8 N·m (anticlockwise)

The total torque o the bolt = 3.6 N·m - 7.8 N·m = -4.2 N·m (clockwise) = 4.2 N·m anticlockwise

Part B

The torque is negative anticlockwise.

Answer:

Input power, Ip = 4 Watts

Explanation:

Given the following data;

Power output = 0.8 Watts

Efficiency = 20%

To find the power input of the sunlight onto the solar panel;

Mathematically, the efficiency of a machine is given by the formula;

[tex] Efficiency = \frac {Out-put \; power}{In-put \; power} * 100 [/tex]

Substituting into the formula, we have;

[tex] 20 = \frac {0.8}{Ip} * 100 [/tex]

[tex] 20 = \frac {80}{Ip} [/tex]

Cross-multiplying, we have;

[tex] 20Ip = 80 [/tex]

[tex] Ip = \frac {80}{20} [/tex]

Input power, Ip = 4 Watts

Answer:

1.1 m/s

Explanation:

Applying,

v = u+at.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, t = time.

From the question,

Given: u = 0.3 ms, a = 0.4 m/s², t = 2 seconds

Substitute these values into equation 1

v = 0.3+0.4(2)

v = 0.3+0.8

v = 1.1 m/s

Hence the velocity at the end of 2 seconds is 1.1 m/s

Answer:

4 capacitors

Explanation:

Given

[tex]n = 5[/tex] --- conducting plates

Required

The number of capacitor (c)

This is calculated as:

[tex]c = n - 1[/tex]

So, we have:

[tex]c = 5 - 1[/tex]

[tex]c = 4[/tex]

Answer:

Explanation:

When a light bulb connects to an electrical power supply, an electrical current flows from one metal contact to the other. As the current travels through the wires and the filament, the filament heats up to the point where it begins to emit photons, which are small packets of visible light.

Explanation:

Uniform velocity is when an object goes an equal amount of space in an equal amount of time whereas non uniform velocity is when the object covers an unequal amount of distance in an equal amount of time.

Explanation:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. ... The gravitational potential energy is equal to its weight times the height to which it is lifted. PE = kg x 9.8 m/s2 x m = joules. The 9.8 us the gravitational acceleration constant.

so the answer is "the gravitational potential energy is decreasing"

Solution :

Let us consider the Gaussian surface that is in the form of a cylinder having a radius of r and a length of A which is [tex]$\text{coaxial with the charged cylinder}$[/tex].

The charged enclosed by the cylinder is given by,

[tex]$q=\rho V$[/tex]       (here, V = [tex]$\pi r^2l$[/tex]  is the volume of the cylinder)

  [tex]$=\pi r^2lp$[/tex]    

If [tex]$\rho$[/tex] is positive, then the electric field lines moves in the radial outward direction and is normal to Gaussian surface which is distributed uniformly.

Therefore, total flux through Gaussian cylinder is :

[tex]$\phi=EA_{cyl}$[/tex]

   [tex]$=E(2\pi rl)$[/tex]

Now using Gauss' law, we get

[tex]$2\pi \epsilon_0rlE = \pi r^2lp$[/tex]

or [tex]$E=\frac{\rho r}{2 \epsilon_0}$[/tex]

Therefore, the electric field is [tex]$E=\frac{\rho r}{2 \epsilon_0}$[/tex]

Hence, option (d) is correct.

Answer:

Explanation:

This is a classic Law of Momentum Conservation problem. For us the equation will look like this:

[tex][(m_yv_y+m_bv_b)]_b=[(m_y+m_b)v_{both}]_a[/tex] Filling in with our given info:

[tex][(70.0)(0)+(.40)(10.0)]_b=[(70.0+.40)v_{both}]_a[/tex] and

4.0 = 70.4v and

v = .06 m/s

third down, chris is correct

Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of

a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²

There are two forces acting on the car in this situation:

• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n

• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)

By Newton's second law, the net forces in the perpendicular and parallel directions are

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) = ma

==>   sin(θ) = a/g   ==>   θ = arcsin(a/g) ≈ 17.8°

(Notice that in the paralell case, the positive direction points toward the center of the curve.)

When rounding the curve at 31.8 m/s, the car's radial acceleration changes to

a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²

and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) + f = ma

The first equation gives

n = mg cos(θ)

and substituting into the second equation, we get

mg sin(θ) + µmg cos(θ) = ma

==>   µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12

Answer:

Explanation:

You are in the chapter on Physics about uniform circular motion and gravity. This is a centripetal force problem in particular, and the equation for that is

[tex]F_c=\frac{mv^2}{r}[/tex] where

[tex]F_c[/tex] is the centripetal force needed to keep the car moving in its circular path,

m is the mass of the car,

v is the velocity with which the car is moving, and

r is the radius of the circle that the car is moving around.

For us, the centripetal force is supplied by the friction keeping the car on the road, altering the equation to become

[tex]f=\frac{mv^2}{r}[/tex] and friction is defined by

f = μ[tex]F_n[/tex] (the coefficient of friction multiplied by the weight of the car).

Going on and getting buried even deeper,

[tex]F_n=mg[/tex] which says that the weight of the car is equal to its mass times the pull of gravity. Putting all that together, finally, we have the equation we need to solve this problem:

μ·m·g = [tex]\frac{mv^2}{r}[/tex] and we solve this for μ:

μ = [tex]\frac{mv^2}{mgr}[/tex] and it just so happens that the mass of the car cancels out. (I'll tell you why the mass of the car doesn't matter at the end of this problem). Filling in and solving for the coefficient of friction:

μ = [tex]\frac{31.8^2}{(9.8)(75.0)}[/tex] to 2 significant figures is

μ = 1.4

The mass of the car doesn't affect whether or not the car can stay on the curve. Even though a car with a greater mass will have a greater frictional force, that doesn't mean that it's easier for that car to stay on the road; a larger mass only means that a larger centripetal force is needed to keep it moving in a circle. This makes the gain in friction become offset by the fact that a larger centripetal force is necessary. Thus,

On a flat curve, the mass of the object experiencing circular motion does not affect the velocity at which it can stay on the curve.

Answer:

The correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.

Explanation:

The inter-stimulus interval (ISI) is the temporal interval between two successive stimuli, measured from the offset of the first stimulus to the commencement of the second.

A cell's refractory period is the time during which it is unable to replicate an action potential. Therefore, the absolute refractory period is the amount of time it takes for a second action potential to be initiated, regardless of how large a stimulus is applied repeatedly.

A second action potential is generated when the gap between the stimuli decreases until the interval reaches the absolute refractory period.

Therefore, the correct option is A. a second action potential is generated until the interval reaches the absolute refractory period.

That is, when the interval between the stimuli decreases, a second action potential is generated until the interval reaches the absolute refractory period.

Answer:

Explanation:

[tex]h=-v^2 /2g[/tex]

[tex]with\\g = 9,8 m/s^2 or 10 m/s^2[/tex]

[tex]h= (-13)^2 / 2 * 9,8 = 8,6[/tex]

Explanation:

hope it helps

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Fundamental

forming a necessary base or core; of central importance.

"the protection of fundamental human rights"

Answer:

D

Explanation:

Let's analyze each option.

A "Multiple copies of the stored data take up very little space."

This is actually true, and it is a positive aspect, as stored digital data does not need any "physical space", it only needs memory and not a lot of it.

So storing data digitally is way more efficient than storing analog data.

B: "They lose quality when they are copied several times."

The data shouldn't change when it is copied, so you should not see a lose in quality.

C: "Stored data are made up of only two different values."

True, but as we know, we can define a lot of things with only two values (zeros and ones), so this is not really a disadvantage.

D: "They are vulnerable to hackers and viruses."

This is true, when you store your data digitally you become vulnerable to hackers stealing your data, so you need to get informatic security in order to protect your data. The same thing with viruses, if you have all your data stored in a given device, and the device becomes infected, there is a chance that you just lost all your data, so you need to have multiple backups of your important information, and again, some protection against viruses.

The correct option is D.

Answer:

D

trust me i just did it

Explanation:

Answer:

a. 0.41 m

b. 5.72 m/s

c. i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

d. 5.72 m/s

Explanation:

a. Use energy to find how high the penny goes above your hand before stopping.

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy at the hand, E equals the total mechanical energy when the penny stops in the air, E'.

E = E'

U + K = U' + K' where U = initial potential energy at hand level = mgh where h = height at hand level = 0, K = initial kinetic energy at hand level = 1/2mv² where v = speed at hand level = 2.85 m/s, U' = final potential energy at stopping level = mgh' where h' = height at stopping level, K = final kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops)

So, U + K = U' + K'

mgh + 1/2mv² = mgh' + 1/2mv'²

substituting the values of the variables into the equation, we have

mg(0) + 1/2m(2.85 m/s)² = mgh' + 1/2m(0 m/s)²

0 + 1/2m(8.1225 m²/s²) = mgh' + 0

m(4.06125 m²/s²) = mgh'

h' = 4.06125 m²/s² ÷ g

h' = 4.06125 m²/s² ÷ 9.8 m/s²

h' = 0.41 m

(b) The penny then falls to the floor, 1.26 m below your hand. Use energy to find its speed just before it hits the floor.  

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh" where h' = height at stopping level = height of penny above hand, h' + height of hand above ground = 0.41 m + 1.26 m = 1.67 m, K = initial kinetic energy at stopping level = 1/2mv'² where v = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₁ = height at ground level = 0, K = final kinetic energy at ground level = 1/2mv"² where v" = speed at ground level,

So, U' + K' = U' + K'

mgh" + 1/2mv'² = mgh₁ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(1.67 m) + 1/2m(0 m/s)² = mg(0) + 1/2mv"²

1.67mg + 0 = 0 + 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

(c) Explain your choice of reference level for parts (a) and (b).

i. For part (a), I chose the hand as the reference level since the penny was thrown from the hand and the height of the penny at the hand is zero and also, it is easier to calculate from a zero reference level.

ii. For part (b), I chose the ground as the reference level since the height of the penny above the ground is positive and the height of the penny when the penny hits the ground is zero and also, it is easier to calculate from a zero reference level.

(d) Choose a different reference level and repeat part (b)

Taking the hand as the ground level, and from the law of conservation of energy, the total mechanical energy when the penny stops in the air, E'  equals the total mechanical energy on the ground, E"

E' = E"

U' + K' = U" + K" where U' = initial potential energy at stopping level = mgh' where h' = height at stopping level = 0.41 m, K = initial kinetic energy at stopping level = 1/2mv'² where v' = speed at stopping level = 0 m/s (since the penny momentarily stops), U = final potential energy at ground level = mgh₁ where h₂ = height of hand above the ground level = height of ground below hand = -1.26 m(it is negative since the ground is below the hand), K = final kinetic energy at ground level = 1/2mv"² where v = speed at ground level,

So, U' + K' = U' + K'

mgh' + 1/2mv'² = mgh₂ + 1/2mv"²

substituting the values of the variables into the equation, we have

mg(0.41 m) + 1/2m(0 m/s)² = mg(-1.26 m) + 1/2mv"²

0.41mg + 0 = -1.26 mg + 1/2mv"²

0.41mg + 1.26mg = 1/2mv"²

1.67mg = 1/2mv"²

1.67g = 1/2v"²

v"² = 2(1.67g)

v" = √[2(1.67g)]

v" = √[2(1.67 m × 9.8 m/s²)]

v" = √[2(16.366 m²/s²)]

v" = √[32.732 m²/s²)]

v" = 5.72 m/s

A fixed mass of gas has a volume of 25 [tex]cm^3[/tex], the pressure of the gas is 100 kPa, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.

PV = nRT (P= pressure of the gas, V =volume, n = number of moles of gas, R = gas constant, and T =temperature of the gas in kelvin)

Suppose the gas is an ideal gas and that the temperature is constant,

P1V1 = P2V2

Here P1 = 100 kPa, V1 = 25 [tex]cm^3[/tex], V2 = 10 [tex]cm^3[/tex],

100 kPa x 25 [tex]cm^3[/tex] = P2 x 10 [tex]cm^3[/tex]

P2 = (100 kPa x 25 [tex]cm^3[/tex]) / 10 [tex]cm^3[/tex]

P2 = 250 kPa

the change in pressure of the gas is,

ΔP = P2 - P1 = 250 kPa - 100 kPa = 150 kPa

The reason is that when the volume of a fixed mass of gas is decreased, the pressure of the gas increases proportionally, so here assuming that the temperature is constant it is calculated.

Hence, the volume of the gas is slowly decreased by 15 [tex]cm^3[/tex] at a constant temperature, and the change in pressure of the gas is 150 kPa, which is option b.

Learn more about the calculation of the change in pressure here.

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Answer:

• All Of The Above

Hope it helps uh

Mark me as brainliest

Answer:

Yes

Explanation:

If there were to be a supernova of a star or a planet/meter that was directed at earth, then we would know in advance and make a plan to stop it

Answer:

Figure is not there

Explanation:

Answer:

I am not sure if this is the answer

acceleration: 2.2m/s

Explanation:

here

initial velocity(u): 11m/s

Final velocity(v): 33m/s

time taken(t): 10 s

now

a:v-u/t

or

acceleration:final velocity-initial velocity/time taken

or

a: 33-11/10

or

a:22/10, divide it

: a=2.2m/s#

Answer:

a sidereal day is the time it takes for the earth to rotate about it axis so that the distant star appears In the same position in the sky while a solar day is the time it takes for the earth to rotate about it axis so that the sun appears in the same position in the sky

Answer:

Speed = 1.6 m/s

Explanation:

Formula,

Speed = Distance ÷ Time

Answer:

Oh, umm…

Jump! (The higher, the better!)

Drop your pencil! (or pen or ruler, whichever you prefer!)

Throw a ball of paper! (and make sure you pick it back up later!)

Explanation:

if you satisfied to my answer , just put brainliest please , and ur welcomeee♥️♥️

Answer:

313kelvin

Explanation:

40 degree celcius plus 273=313K

Answer:

The water table is the upper surface of the zone of saturation. The zone of saturation is where the pores and fractures of the ground are saturated with water. It can also be simply explained as, the upper level, below which the ground is saturated.

Answer:

Explanation:

The equation for Coulomb's Law is

[tex]F=\frac{kq_1q_2}{r^2}[/tex] where k is Coulomb's constant, q1 is one of the charges, q2 is another one of the charges (1 of these has to be the one in question, so we will let that be q2) and r is the distance between them squared.

First thing we are going to do is convert those microCoulombs to Coulombs (C).

q1: [tex]-53.0\mu C*\frac{.000001C}{1\mu C}=-5.3*10^{-5}C[/tex] and

q2: [tex]105\mu C*\frac{.000001C}{1\mu C}=1.05*10^{-4}C[/tex] and

q3: our main particle that we will put in for q2 in the formula converts as follows:

q3: [tex]-88.0\mu C*\frac{.000001C}{1\mu C} =-8.8*10^{-5}C[/tex]  

First we will find the charge between q1 and the main particle:

[tex]F_1=\frac{(9.0*10^9)(5.3*10^{-5})(8.8*10^{-5})}{(1.45)^2}[/tex] Notice that we did not use the negative charges here. We take the negative charge into account depending upon whether or not the charges are repelled or attracted. Both of these charges are negative, so they will repel and the answer will be made negative. Finding the first force:

[tex]F_1=-2.0*10^1N[/tex] (negative because they repel so q1 will move away from the charge in question, which is also negative)

[tex]F_2=\frac{(9.0*10^9)(1.05*10^{-4})(8.8*10^{-5})}{(.95)^2}[/tex] and the charge between these is

[tex]F_2=92N[/tex] and that is to the right, so positive. These charges are opposite, so they attract. The net force is the sum of the forces, so:

[tex]F_1+F_2:[/tex] -2.0 × 10¹ + 92 = 72N (to the right)