the highest theoretical efficiency of a certain engine is 30% if the engine uses the atmosphere which has a temperature of 300k as its cold reservoir what is temperature of its hot reservoir

The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

The acceleration of the box is calculated as follows;

vf² = vi² + 2as

a = (vf² - vi²)/2s

a = (11.5² - 13²) / (2 x 8.5)

a = -2.16 m/s²

The time taken for the box to travel is calculated as follows;

a = (vf - vi)/t

t = (vf - vi) / a

t = (11.5 - 13) / (-2.16)

t = 0.69 s

P = Fv

P = (ma)(vf - vi)

P = (1 x -2.16) x (11.5 - 13)

P = 3.24 W

Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

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The time it takes for the ball to hit the ground from the platform is determined as 1.56 seconds.

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

where;

h = 0 + ¹/₂gt²

h = ¹/₂gt²

t = √(2h/g)

t = √(2 x 12 / 9.8)

t = 1.56 seconds

Thus, the time it takes for the ball to hit the ground from the platform is determined as 1.56 seconds.

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The effective height of the water for Smith's house will be 24.61m.

Based on the information given, the volume of the water in sphere will be:

= 4/3πr³ = (5.80 × 10^5)/1000

= 4.18r³ = 580

r³ = 138.7

r = 5.18m

The effective height of the water will be:

= 18.0 + 2(5.18)

= 28.36

The gauge pressure at Faucet of Jones house will be:

= pgh

= 1000(9.8)(28.36)

= 277.9kPa

The effective height of the water for Smith's house will be:

= 18.0 + 2(5.18) - 3.75

= 24.61m

The gauge pressure at Faucet of Jones house will be:

= 1000 × 9.8 × 24.61

= 241.2kPa

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Answer:

The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.

Answer:

each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first nonzero digit.

Explanation:

(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.

(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.

The INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;

Increase in sound level = final sound level - original sound level

Increase in sound level = 86 dB - 85 dB = 1 dB

from the graph at 1.5 hours/day, sound level = 97 dB

Increase in sound intensity = final sound level - original sound level

Increase in sound intensity = 116 dB - 97 dB = 19 dB

Factor increase = 19/97 = 0.196 = 19.6%

Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.

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1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s

In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.

What are the parameters to consider ?

The parameters are;

Speed = distance S ÷ Time t

Convert kilometer to meter by multiplying it by 1000

C = S/t

3 × [tex]10^{8}[/tex] =  3.85 × [tex]10^{8}[/tex] / t

Make t the subject of formula

t = 3.85 × [tex]10^{8}[/tex] / 3 × [tex]10^{8}[/tex]

t = 1.2833

t = 1.3 s

Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

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A thin-walled hollow sphere has a radius of 4 cm from the center

of the sphere, the electric field points radially inward and

has a magnitude of 1.5 × 10⁴ N/C then the charge on the surface would be 2.6690×10⁻⁹ C

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.

The electric field inside a spherical shell is given by the formula

E = q/4πεr²

where q is the charge on the surface

r is the radius of the sphere

ε is the electrical permeability

By substituting the respective values in the given formula

1.5 × 10⁴ = q/4π(8.85✕ 10⁻⁻¹²)(.04²)

q = 2.6690×10⁻⁹ C

Thus, the charge on the surface would be2.6690×10⁻⁹ C

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The statement that best explains the type of chemical reaction represented by Maya's picture is that it is neither a synthesis reaction nor a decomposition reaction because two reactants form two products. That is option B.

A chemical reaction is the combination of two elements to yield a new product through the formation of bonds.

A chemical reaction is said to be a synthesis reaction when when two different atoms or molecules interact to form a different molecule or compound.

A chemical reaction is said to be a decomposition reaction when one reactant breaks down into two or more products.

Therefore, from the picture, the chemical reaction is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.

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The center of mass of the three-mass system is 0.433m.

A position established in relation to an object or set of objects is the center of mass. It represents the system's average location as weighted by each component's mass. In a collection of unconnected items, the center of mass can also be established.

Mass of the object, m₁ = 1.03kg

Mass of the object, m₂= 1.50kg

Mass of the object, m₃= 1.10

Taking the location of m as the origin and towards right as positive X-axis.

x₁=0

x₂=0.50m

x₃=0.25+0.50 = 0.75m

The X-coordinate of the center of mass [tex]x_c[/tex] of a system of three masses  m₁, m₂ and m₃ located at the positions x₁, x₂ and x₃ on X-axis is given by,

[tex]x_c=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}[/tex]

[tex]x_c=\frac{0+1.50*0.50+1.10*0.75}{1.03+1.50+1.10}\\x_c= 0.433 m[/tex]

Therefore,  the center of mass of the three-mass system is 0.433m.

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Answer:

B. 2.5 Ω

Explanation:

[tex]\frac{1}{5} +\frac{1}{5}=0.4[/tex]

Put a one on top of the 0.4

[tex]\frac{1}{0.4} =2.5[/tex]

Answers are:

a. The work done by the weight lifter W = 25000 Joule

b. The work done by the gravity W = 12250 Joule

c. The net work done on the object W = 12750 Joule

What is Work ?

Work is the product of force and distance in the direction of the force applied. That is, W = F x h

The S.I unit is Joule.

Given that a weight lifter lifts a mass of 250 kg with a force of 5000N to a height of 5m.

The given parameters are;

a. The work done by the weight lifter will be

W = F x h

W = 5000 x 5

W = 25000 Joule

b. The work done by the gravity will be

W = mg x h

W = 250 x 9.8 x 5

W = 12250 Joule

c. The net work done on the object will be

W = 25000 - 12250

W = 12750 Joule

Since the work done by the gravity is in opposite direction to the work done by the weight lifter, the net work done on the object will be the difference between the two.

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It is calculated that a) The angular velocity of the wheel is 272.13 rad/s,

b) On the edge of the grinding wheel, the linear speed is 47.62 m/s,

and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².

Calculation of angular velocity, linear speed & acceleration:

Provided that,

the diameter of the wheel = 0.35 m

So, the radius, r = 0.35/2 = 0.175 m

As 1 revolution = 2π rad

(a) the angular velocity, ω = 2600 rpm = [tex]\frac{2600 * 2\pi }{60}[/tex] rad/s

⇒ω = 272.13 rad/s

So, the angular velocity is 272.13 rad/s.

(b) The linear speed, v = r * ω

⇒v = 0.175 * 272.13

⇒v= 47.62 m/s

(c) The angular acceleration, [tex]a=\frac{v^{2} }{r}[/tex]

[tex]a = \frac{(47.62)^{2} }{0.175}[/tex]

⇒[tex]a[/tex] = 12958.08 m/s²

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The center of mass of the car with all people seated will be 3.20 m far from the front of the car.

Taking the front of the car as reference:

m₁x₁ + m₂x₂ + m₃x₃ = mx

where,

m₁ = mass of empty car = 1300 kg

x₁ = distance of center of mass of empty car from front = 2.45 m

m₂ = mass of 2 people sitting in front = 2 x 60 kg = 120 kg

x₂ = distance of center of mass of 2 people sitting in front from the front    = 2.7 m

m₃ = mass of 3 people sitting in back seat = 3 x 60 kg = 180 kg

x₃ = distance of center of mass of 3 people sitting in the back seat from the front = 3.65 m

m = total mass of car with all people = 1300 kg + 120 kg + 180 kg = 1600kg

x = distance of center of mass of the car from the front when all are seated

Therefore,

(1300 kg)(2.45 m) + (120 kg)(2.7 m) + (180kg)(3.65 m) = (1300 kg)x

x = 4166 kg.m/1300 kg

x = 3.20 m

Therefore, the center of mass of the car with all people seated will be 3.20 m far from the front of the car.

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Work done on the system is smaller than heat absorbed.

If the internal energy of a system is decreased, which of the following is impossible:

C) Work done on the system is smaller than heat absorbed.

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MAGNITUDE AND DIRECTION OF A VECTOR

Given a position vector →v=⟨a,b⟩,the magnitude is located by |v|=√a2+b2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tanθ=(ba)⇒θ=tan−1(ba)

The formula to determine the extent of a vector (in two dimensional space) v = (x, y) is: |v| =√(x2 + y2). This formula is derived from the Pythagorean theorem. the procedure to determine the magnitude of a vector (in three dimensional space) V = (x, y, z) is: |V| = √(x2 + y2 + z2)

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main answer is given below,

The minimum necessary height of the IV bag above the position of the needle is 0.37 m.

The minimum necessary height of the IV bag above the position of the needle is calculated as follows;

P = ρgh

where;

Substitute the given parameters and solve for minimum height

h = P/ρg

h = (4759.609) / (9.8 x 1308)

h = 0.37 m

Thus, the minimum necessary height of the IV bag above the position of the needle is 0.37 m.

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[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]

We would calculate the magnitude by applying pythagorean theorem:

[tex]\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}[/tex]

[tex]\longrightarrow \sf{Magnitude = 33.12}[/tex]

[tex]\longrightarrow \sf{The \: vector \: is \: (- 14, 30)}[/tex]

The angle between two vectors is given by the formula:

[tex]\sf{\longrightarrow \small \cos \emptyset = \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }[/tex]

In two dimensional, the x axis of vector form is:

[tex]\small\sf{\longrightarrow (b1, b2) = (1, 0) }[/tex]

[tex]\sf{\longrightarrow \small \cos \: \emptyset = \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }[/tex]

[tex]\small\longrightarrow \sf{ \dfrac{14}{33.12} }[/tex]

[tex]\small\longrightarrow \sf{\emptyset \: = arcCos (\dfrac{ - 14}{33.12} )}[/tex]

[tex]\small\longrightarrow \sf{\emptyset= 115^\circ}[/tex]

[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]

[tex] \small\bm{The \: angle \: between \: the \: vector \: }[/tex]

[tex]\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}[/tex]

The net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.

The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;

F = kq₀q₁/r²

where;

F(01) = (9 x 10⁹ x 4 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.2²)

F(01) = 1.44 x 10⁻¹⁶ N

The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;

F = kq1q2/r²

where;

F(12) = (9 x 10⁹ x 6 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.6²)

F(12) = 2.4 x 10⁻¹⁷ N

F(net) = 2.4 x 10⁻¹⁷ N +  1.44 x 10⁻¹⁶ N

F(net) = 1.68 x 10⁻¹⁶ N

Thus, the net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.

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Answer:

X component= 46 N

Y component= 80 N

Magnitude =( 46 ^2 + 80^2 )^0.5 = 92.3 N

The orbital speed of an ice cube in the rings of Saturn is 355358.97m/s

According to the gravitation law, the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between them. Mathematically;

F = GMm/r²

where

m = mass of ice cube and

s = Gm1/r^2

Hence,

F = sm2

On rearranging,

s = m2/F

let V = orbital speed

centripetal acceleration = V^2/r

Such that;

V²/r = Gm/r²

V² = Gm/r

V = √Gm/r

Substitute the given parameters

V =  √6.67×10^-11 *  5.68 x 10^26 / 3.00 x 10^5

V = 355358.97m/s

Hence the orbital speed of an ice cube in the rings of Saturn is 355358.97m/s

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Answer:

Explanation:

Answer:

See below

Explanation:

vf = vo + at        when it reaches top vf = 0    vo = 10m/s      a = -9.81 m/s^2

0 =   10  + (-9.81) t  

t = 1.02 seconds to reach max

time up = time down   so    roundtrip =  2.04 seconds

Answer:

The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)

Explanation:

Let [tex]u[/tex] denote the initial velocity of the vehicle ([tex]20\; \text{mph}[/tex] or [tex]60\; \text{mph}[/tex]) and let [tex]v[/tex] denote the velocity of the vehicle after braking ([tex]0\; \text{mph}[/tex]). Let [tex]x[/tex] denote the braking distance.

Assume that the acceleration during braking are both constantly [tex]a[/tex] in both scenarios. The SUVAT equations would apply. In particular:

[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}[/tex].

Since [tex]v = 0[/tex] (the vehicle has completely stopped), the equation becomes [tex]x = (-u^{2}) / (2\, a)[/tex].

Assuming that [tex]a[/tex] (braking acceleration) stays the same, the braking distance [tex]x[/tex] would be proportional to [tex]u^{2}[/tex], the square of the initial velocity.

Hence, increasing the initial speed from [tex]20\; \text{mph}[/tex] to [tex]60\; \text{mph}[/tex] would increase the braking distance by a factor of [tex]3^{2} = 9[/tex].

Answer:

9 times

Explanation:

A piston-work cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the steam is cooled. compression work is 44.32 KJ,

The amount of labor put into the piston to cause its reciprocating motion is known as the piston work. It is calculated by multiplying the piston's displacement by the net force.

An expanding gas cylinder's force output is transferred by pistons to the crankshaft, which then drives the flywheel's rotation. A reciprocating engine is a device like this.

Piston work is the effort made by the piston to make its reciprocating motion. The piston's displacement is calculated by multiplying it by the net force.

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Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.

Mass of piece B = Mb

Ma×Va = Mb×Vb

So, 1.9Mb × Va = Mb×Vb

=> 1.9Va = Vb

=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900

=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900

=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j

=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule

Thus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.

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a.

b.

c.

Pressure is the force per unit area on a surface.

Speed is the distance moved per unit time.

Pressure

Since pressure, P = hρg where

Since ρ and g are constant

P ∝ h

So, we see that pressure is directly proportional to depth.

Since U is lower than Q, Pressure at U is greater than pressure at Q.

So,PU is greater than PQ.

Using the continuity equation

VUAU = VTAT where

So, VUAU = VTAT

VUπ(dU)² = VTπ(dT)²

VU = VT(dT)²/(dU)²

VU = VT(2.0)²/(1.0)²

VU = VT(4)

VU = 4VT

Since VU = 4VT,VU is Greater than 2VT

Since R is at the same depth as U, Pressure at R is equal to pressure at U.

So,PR is Equal to PU.

Using the continuity equation

VRAR = VSAS where

So, VRAR = VSAS

VRπ(dR)² = VSπ(dS)²

VR = VS(dS)²/(dS)²

VR = VS(2.0)²/(2.0)²

VR = VS(1)

VR = VS

Since VR = VS,VR is Equal to VS

Using the continuity equation

VQAQ = VUAU where

So, VQAQ = VUAU

VQπ(dQ)² = VUπ(dU)²

VQ = VU(dU)²/(dQ)²

VQ = VU(1.0)²/(1.0)²

VQ = VU(1)

VQ = VU

Since VQ = VU, VQ is Equal to VU

Since R is lower than S, Pressure at R is greater than pressure at S.

So,PR is Greater than PS.

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Answer:

No because a pot plant system must interact with its surroundings (air, water, etc) and these are not closed systems.

Only if the pot plant were completely enclosed, say in a large glass jar would the system be closed.

Radius of the circular loop is 0.0091m.

Magnetic field is the area around a magnet where the magnetism influence is felt .

B =(μ)I/2r

μ= Magnetic permeability =4(π)*10^(-7)

I= current flowing through the loop

r= radius of the loop

=0.0091m

Thus, we can conclude that the radius of the loop is 0.0091m .

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Each side has to have at least 44 horses

F61160 N. This is further explained below.

Generally, We are only interested in the component that operates horizontally since the vertical components all cancel each other out. The pressure difference works on the hemisphere to generate a normal force all over the surface, but we are only concerned with that force's horizontal component. This may be determined by supposing the hemispheres to be two flat circular plates of the same radius as the hemispheres that have been forced together.

Therefore, force is equal to pressure multiplied by area, which is

F= (970 -15 )( * (0.45 m)2)

F=60754 N for each side.

Therefore, each side has to have at least 44 horses

44* 1390 = 61160 N

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