a system absorbs heat from the surroundings and has work done on it by the surroundings. what are the signs of qsystem and wsystem?

If an early frost destroys most of the apple crop, the equilibrium price of an apple is likely to increase.

An early frost that damages a significant portion of the apple crop would result in a decrease in the supply of apples. When the supply decreases, while demand remains constant or even increases, it creates a situation where the quantity demanded exceeds the quantity supplied. This imbalance in the market typically leads to an increase in the equilibrium price of apples.

With a reduced supply of apples, consumers would still desire to purchase them at the same or even higher quantities, especially if there are no close substitutes available. As a result, buyers may be willing to pay a higher price to secure the limited supply of apples. This increased demand coupled with the decreased supply causes the equilibrium price to rise.

Moreover, the extent of the increase in price will depend on the severity of the frost damage and the elasticity of demand. If the damage is severe and there are no alternative sources of apples, the price increase may be significant. However, if there are other regions or countries with unaffected apple crops that can fulfill the demand, the price increase may be relatively moderate.

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Answer:

venus

Explanation:

its the closest to the sun

could you plz vote me brainliest? thx :)

Answer:

Venus because it really close to the sun.

Explanation:

Answer:

\( \boxed{ \sf{see \: below}}\)

Explanation

The unit of force, Newton is a derived unit because it depends on three fundamental units. i.e kilogram , meter and second.

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A) Total energy is: Em = 0.1058 J ; B) Maximum speed is: v = 0.650 m / ;  C) Maximum acceleration is: a = 15.1 m / s².

An oscillatory movement where the restoring force is proportional to the displacement is called as the simple harmonic motion.

Given amplitude, A = 2.8 cm = 2.8 10⁻² m

Given spring constant(k) = 270 N / m

m = 0.50 kg

As we know, x = A cos (wt + Ф)

w² = k / m

x is displacement, A is amplitude w is angular velocity, t is time, Ф --phase constant, k -- spring constant and m -- mass.

A) Mechanical energy is given as

 Em = ½ k A²

 Em = ½ 275 (2.8 10⁻²) ²

Em = 0.1058 J

B) V= dx/dt = - Aw sin ( wt + fi)

To obtain maximum velocity, sine function must be ±1

 Vmax = w A

w = √ 270/0.5

 w = 23.23 rad / s

Vmax = 23.23 *2.8 * 10⁻²

 = 0.650 m / s

C) a = dv/dt  = - A w² cos (wt + fi)

For maximum value, cosine function must be maximum, that is ±1    

a = A w²

a = 2.8 10⁻²*  23.23²

a = 15.1 m / s²

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The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

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Answer:

The Sun transfers in each second and amount of 1367J per square meter, so in hour it can transfer:

E = 60*60*1367 J/h*m2 = 4,921,200 J/h*m^2.

in one year, we have 3000h*4,921,200 J/h*m^2. = 14,763,600,000 Joules per square meter.

Now, only 20% of this can be used as energy, so the amount that can be used is:

E = 0.2* 14,763,600,000 J/m^2 = 2,952,720,000 J/m^2

Now, we want to find the number X of square meters such:

X*2,952,720,000 J/m^2 = 4x10^11 J

X = ( 4x10^11 J)/(2,952,720,000 J/m^2) = 135.5 m^2

So you only need around 135.5 square metters per house.

The general solution is y(t) = c1t + c2t ln(t) - (1/3)ln(t), where c1 and c2 are arbitrary constants.

To find the general solution of the differential equation\(t^2\)y'' + 3ty' + y = ln(t), where t > 0, we can use the method of variation of parameters.

First, we find the complementary solution by solving the associated homogeneous equation\(t^2\)y'' + 3ty' + y = 0. This equation can be rewritten as \(t^2\)y'' + 3ty' + ty' + y = 0, and factoring out y' gives t(ty')' + (ty)' = 0. Integrating this equation, we obtain ty' + ty = c1, where c1 is a constant.

Now, we assume the particular solution as y_p = u(t)y1 + v(t)y2, where y1 and y2 are linearly independent solutions of the homogeneous equation. Differentiating y_p twice and substituting it into the original equation, we can solve for u'(t) and v'(t) using the formulas:

u'(t) = -(y2 ln(t))/(W(t)y1)

v'(t) = (y1 ln(t))/(W(t)y2)

Here, W(t) is the Wronskian of y1 and y2. After integrating u'(t) and v'(t), we can substitute the obtained expressions back into the particular solution equation to find the general solution.

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Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

Given the production rules below, is the plus operator (+) left-associative, right-associative, or neither a parse tree of it is in the explanation part below.

In the provided production rules, the addition operator (+) is left-associative.

Consider the statement "1 + 2 + 3" to show this.

This sentence's parse tree is as follows:

The addition operator's left-associativity is shown in this parse tree. The operands 1 and 2 are combined by the leftmost plus operator, and the resultant total is merged with the operand 3 by another plus operator. This depicts the assessment of the statement "1 + 2 + 3" from left to right.

If the plus operator were right-associative, the parse tree would be different, with the rightmost plus operator first combining operands 2 and 3, and then combining the resultant sum with operand 1.

Thus, the parse tree is attached below as image.

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The distance of the object is 12.27m

We are given that,

The net force = F = 6.6N

The mass = m = 9.0kg

The velocity of the object = v = 3.0m/s

Therefore , the hamster is travelled distance ,that can be calculated by Newton's second law of motion,

F = ma

F = (mv²)/r

r = (mv²)/F

Putting the above values in the given equation,

r = (9.0kg)(3.0m/s)²/6.6N

r = 12.27m

Therefore, the hamster travelled 12.27m.

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Answer:

El automóvil recorre una distancia de 11.806 metros antes de deternerse por completo.

Explanation:

De acuerdo con el enunciado, el conductor nota la luz roja, empieza a decelerar 0.5 segundos después y decelera hasta detenerse. La distancia total recorrida por el automóvil desde el instante en que el conductor nota la luz roja (\(\Delta s_{T}\)), medida en metros:

\(\Delta s_{T} = \Delta s_{1}+\Delta s_{2}\) (1)

Donde:

\(\Delta s_{1}\) - Distancia recorrida a velocidad constante, medida en metros.

\(\Delta s_{2}\) - Distancia recorrida hasta alcanzar el reposo, medida en metros.

Si suponemos que la segunda etapa describe un movimiento uniformemente acelerado, entonces la distancia recorrida total que representada por la siguiente fórmula:

\(\Delta s_{T} = v_{o}\cdot \Delta t_{o} + \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a}\) (2)

Donde:

\(v_{o}\) - Velocidad inicial del automóvil, medida en metros por segundo.

\(v_{f}\) - Velocidad final del automóvil, medida en metros por segundo.

\(\Delta t_{o}\) - Tiempo de reacción del conductor, medido en segundo.

\(a\) - Aceleración, medida en metros por segundo al cuadrado.

Si conocemos que \(v_{o} = 16.667\,\frac{m}{s}\), \(v_{f} = 0\,\frac{m}{s}\), \(\Delta t_{o} = 0.5\,s\) y \(a = -40\,\frac{m}{s^{2}}\), entonces la distancia recorrida total es:

\(\Delta s_{T} = \left(16.667\,\frac{m}{s} \right)\cdot (0.5\,s)+\frac{\left(0\,\frac{m}{s} \right)^{2}-\left(16.667\,\frac{m}{s}\right)^{2}}{2\cdot \left(-40\,\frac{m}{s^{2}} \right)}\)

\(\Delta s_{T} = 11.806\,m\)

El automóvil recorre una distancia de 11.806 metros antes de deternerse por completo.

Answer:

1428  m/s

Explanation:

Given that,

The frequency hear by a dolphin, f = 280 kHz

We need to find the speed of sound in water if a wave with this frequency has a wavelength  of 0.51 cm.

The formula for the speed in terms of wavelength and frequency is given by :

\(v=f\lambda\\\\v=280\times 10^3\times 0.51\times 10^{-2}\\\\v=1428\ m/s\)

So, the speed of sound in water is 1428  m/s.

Answer:

They are standing 3 feet away from each other

Explanation:

If Nathan Is standing 2 feet away from the mirror. Then Hiva is standing 7 feet away from the mirror is she is standing 5 feet further than him. They are standing 3 feet away from each other.

Answer:

Coefficient of friction (COF) is a dimensionless number that is defined as the ratio between friction force and normal force (Eqn (2.1)). Materials with COF smaller than 0.1 are considered lubricous materials. COF depends on the nature of the materials and surface roughness.

Explanation:

The displacement of the military jet during the time it was accelerating was approximately 284.59 meters.

the formula to find displacement of an object is:  

`s=ut + 1/2 at^2`

Here, the initial velocity is u, the time taken is t and acceleration is a.

Let's find the acceleration of the military jet. Using the formula,

v = u + at345

= 195 + a(1.65) 345 - 195

= 1.65a150 = 1.65a

a = 150/1.65

a ≈ 90.91 m/s²

We can substitute these values in the formula to find the displacement of the military jet during the time it was accelerating.s =

ut + 1/2 at^2s

= 195(1.65) + 1/2 (90.91)(1.65) s

≈ 284.59 meters

The displacement of the military jet during the time it was accelerating was approximately 284.59 meters.

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The trailer's mass is 0.10 kg.

The period of a mass-spring system is directly related to the square root of the mass but inversely related to the square root of the spring constant.

Thus, the period T is expressed as;

T = 2π × √ m ÷ k

The period (T) is reciprocally related to the frequency (f)

Expressed in an equation as;

T = 1 ÷ f

Therefore, T = 2π × √ m ÷ k = 1 ÷ f

Making the mass 'm' the subject of the formula, the above equation becomes;

m = k ÷ 2π^2 × f^2

Where

m represents the mass

k represents the spring constant

f is the frequency

π = 3. 14 (constant)

Replacing the values, the mass equation becomes;

Mass, m = 1. 0 × 10^ 0 ÷ 2× 3.14 × 3.14 × 0.71 × 0. 71

= 1.0 × 1 ÷ 9. 94

= 0. 10 kg

Therefore, the mass of the trailer is 0. 10kg.

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Answer:

Hypermetropia (hyperopia, long-sightedness or far- sightedness) is a form of refractive error in which parallel rays of light coming from infinity are focused behind the light sensitive layer of the retina, when the eye is at rest.5 Dec 2017

Explanation:

thank me later

Answer:

Explanation:

a) work=Force times distance. given force is 888N and distance travelled is 22M

work= 888 x 22 = 19536J or 1.954x10^5J

b) work = 1111 x 22= 24442J or 2.444x10^5J

c) putting on a larger engine means that there are gonna be more force applied to the go cart. therefore the cart will travel the same distance faster

a. The amount of work done by the go cart engine is 19,536 Nm.

b. The amount of work done when a bigger engine is used is 24,442 Nm.

c. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.

Given the following data:

a. To determine the amount of work done by the go cart engine:

Mathematically, the work done by an object is given by the formula;

\(Work\;done = Force \times distance\)

Substituting the given parameters into the formula, we have;

\(Work\;done = 888 \times 22\)

Work done = 19,536 Nm.

b. To determine the amount of work done when a bigger engine is used:

\(Work\;done = Force \times distance\)

\(Work\;done = 1111 \times 22\)

Work done = 24,442 Nm.

c. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.

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Answer:

147 newton.

Explanation:

Force is the product of mass and acceleration. The formula of force is given below: Force =  mass x acceleration, but due to height we put gravity instead of acceleration. So by putting values of mass i. e. 15 kg and gravity i. e. 9.8 meter/ second in the formula we get the net force of the ball which is 147 newton or 147 kg meter/second square.

The pressure reading on the upper portion of the pipe can be determined by using Bernoulli's equation. Bernoulli's equation relates the pressure, velocity, and height of a fluid in a steady state and incompressible flow.

The Bernoulli's equation states that:

P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2

Where:

P1 = pressure at point 1 (232 kPa)

v1 = velocity at point 1 (2.04 m/s)

h1 = height at point 1 (0 m)

P2 = pressure at point 2 (unknown)

v2 = velocity at point 2 (3.26 m/s)

h2 = height at point 2 (12.28 m)

ρ = density of the fluid (842 kg/m3)

g = acceleration due to gravity (9.81 m/s^2)

We can plug in the given values and solve for P2:

P2 = P1 + 1/2ρv1^2 + ρgh1 - 1/2ρv2^2 - ρgh2

P2 = 232 + 1/28422.04^2 + 8429.810 - 1/28423.26^2 - 8429.8112.28

P2 = 232 + 8424.16 + 0 - 84210.66 - 10,205.9

P2 = -9,973.7 Pa

Therefore, the pressure reading on the upper portion of the pipe is -9,973.7 Pa which is approximately -99.74 kPa.

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Answer:

How do X rays use electromagnetic waves to function?

X-rays are very high-frequency waves and carry a lot of energy. They will pass through most substances, making them useful in medicine and industry to see inside things. An X-ray machine works by firing a beam of electrons at a "target." If we fire the electrons with enough energy, X-rays will be produced.

Explanation:

Answer:

C

Explanation:

The distance that the pig moves up will be less than the distance that the strongman pulls down.

Explanation: The strongman is further from the fulcrum than the pig is, so the distance that he pulls down will be greater than the distance that the pig moves up.

The distance that the pig moves up will be less than the distance that the strongman pulls down. Hence, option (C) is correct.

The given problem is based on the work done by the applied force. When some magnitude of force is applied on any object, then there will be some obvious displacement of object. This is known as work done by the object. The expression for the work done by the object is,

w = Fd

here, F is the applied force and d is the displacement.

In the given problem, the applied force will be due to the weight of man and pig. So, the modified form of work done is,

work = Wd

W = work/d

here, W is the weight.

Clearly, more the weight, less will be the displacement and vice versa. So due to heavy weight of man, he will pull the pig easily. Such that the upward distance covered by pig will be less than the distance that the strongman pulls down.

Thus, we can conclude that the distance that the pig moves up will be less than the distance that the strongman pulls down. Hence, option (C) is correct.

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The lighter one travel with the speed of 0.574 m/sec. Force is defined as the product of mass and acceleration. Its unit is Newton.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

The two skaters push off against each other on frictionless ice, then torque act by the one skater on the other is equal.

P₁₂=P₂₁

F₁₂V₁ =F₂₁V₂

625 ×V₁= 725 N×  1.5 m/s

V₁=0.574 m/sec

Hence, the lighter one travel with the speed of 0.574 m/sec.

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Answer:

1.74 m/s

Explanation:

use the formula

     \(F_{1} v_{1} = F_{2}v_{2}\)

the heavier skater being \(F_{1}\) and the lighter skater being \(F_{2}\)

    (725 N)(1.5 m/s) = (625 N) \(v_{2}\)

solve for \(v_{2}\)

  \(v_{2} =\) 1.74 m/s

The diagram is labelled to DRAW the RESULTANT wave and LABEL the areas of interference with CI, DI and/or points of TDI.

When a light ray falls on the interface of two medium like air and water, then the portion of light is reflected and another portion of the light is refracted into the water medium. The angle of incidence is equal to the angle of reflection but the angle of incidence is larger then the angle of refraction.

Open the attached file. It shows 2 waves (a blue one and an orange one) that are moving through the same medium at the same time.

Thus, the diagram is labelled.

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All of the above statements are true.

Both yellow and orange colors (as visible light) travel at the same speed, approximately 2.99 X 10^8 m/s, in a vacuum. The speed of light is constant for all colors within the visible spectrum.

The frequency of orange color is lower than that of yellow color. In the electromagnetic spectrum, orange light has a lower frequency compared to yellow light. Lower frequency corresponds to a longer wavelength.

The wavelength of yellow color is shorter than that of orange color. In the visible spectrum, yellow light has a shorter wavelength compared to orange light. Shorter wavelength corresponds to a higher frequency.

Therefore, all three statements are true for yellow and orange colors.

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Answer:

by obtaining the total mass of the dimes present:

d = 27.22 g / dozen       the density of dimes

M = n * d = 5 dozen * 27.22 g / dozen = 126.1 g

nd ko ka balo mag anser sina ya doy

So, you know how we are able see things around us since light bounces off of them and goes into our eyes? Well, an infrared camera works a small bit in an unexpected way. Rather than seeing obvious light like our eyes do, it sees something called infrared radiation.

Everything around us gives off a small bit of this infrared radiation, indeed things that we can't see with our eyes. When an infrared camera looks at something, it's really detecting the warm that's given off by that question.

The camera encompasses a uncommon focal point that lets this infrared radiation in and it encompasses a sensor that can "see" it. At that point, it turns that heat vitality into an picture that we are able see on a screen.

Hence, in case you were to point an infrared camera at a individual or a pooch, you'd be able to see the warm coming off of them, and it would see like a colored picture where distinctive colors appear distinctive temperatures

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