42 cal of heat are required to increase the temperature of a 30 g piece of an unknown substance from 20 to 27 °c and the specific heat capacity of the substance is _____ cal/g deg•°c.

Answer:

wind power can be tracked back to sunlight since sunlight is one of the causes of moving air which is wind .

High temperature causes air to expand causing it to spread over a wider area decreasing it's pressure compared to places of lower temperature which causes air to contract causing a concentration of air molecules which increases its pressure . This two processes causes a pressure difference which leads to movement of air (wind) whose energy can be used to generate wind power.

Answer:

Explanation:

magnetic field B = (3 i + 8 x 2 j )x 10⁻³ T

= (3 i + 16 j )x 10⁻³ T

L = - i ( unit length of conductor )

Force F = I ( L x B ) , I is current

= 5 [ - i x ( 3i + 16 j ) 10⁻³]

= 5 ( - 16 k ) x 10⁻³

F = - 80 x 10⁻³ k

Answer:

sunlight

Explanation:

Photosynthesis is made possible by energy from the sun.

Hi there!

We can use the following equation for a point charge in a magnetic field:

\(\large\boxed{F_B = qv \times B}\)

\(F_B\) = Force due to magnetic field (7.6 × 10⁻¹³N)
\(q\) = Charge of particle (1.6 × 10⁻¹⁹ C)

\(v\) = velocity of particle (6.6 × 10⁶ m/s)

\(B\) = Magnetic field strength (1.8 T)

Or, without the cross product:
\(F_B = qvBsin\theta\)

θ = angle between particle's velocity and field

We can rearrange to solve for theta:
\(\frac{F_B}{qvB} = sin\theta\\\\\theta = sin^{-1} (\frac{F_B}{qvB})\)

Solve for theta:
\(\theta = sin^{-1} (\frac{7.6*10^{-13}}{(1.6*10^{-19})(6.6*10^6)(1.80)}) = \boxed{23.57^o}\)

The current through the nichrome wire of the heater when it is turned on is: 12.5 A

The current through the nichrome wire of the heater when it is turned on is calculated by dividing the power rating of the heater (1500 W) by the potential difference (120 V), resulting in a current of 12.5 A.

To explain further, power (P) is the rate of energy transfer, and is equal to the product of the potential difference (V) and the current (I). Therefore, P = VI. In this example, 1500 W = 120 V x I. Rearranging the equation, we find that I = P/V, resulting in I = 1500 W/120 V, which is equal to 12.5 A.

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Answer:

The correct answer to the following question will be "An ecosystem".

Explanation:

So that the above seems to be the right answer.

Answer:

Roots are basically hooked into the ground?

Explanation:

Maybe you could comment here the choices if it's multiple choice.

v_a = 178.74 m/s

using Bernoulli's theorem :-

P_a + ½ρ(v_a)² = P_b + ½ρ(v_b)²

Where;

P_a is pressure above wings

P_b is pressure below wings

v_a is speed above wings

v_b is speed below wings

ρ is density of air

We want to find V_a, so let's make V_a the subject;

v_a = √[(2(P_b - P_a)/ρ) + (v_b)²]

Now, we don't know (P_b - P_a)

(P_b - P_a) = Force/Area

(P_b - P_a) = mg/Area

We are given m = 2.5 × 10^(6) kg and area = 1600 m²

Thus, (P_b - P_a) = (2.5 × 10^(6) × 9.81)/1600 = 15328.125 N/m²

Density of air will be taken as 1.2 kg/m³

Thus;

v_a = √[(2(15328.125)/1.2) + (80)²]

v_a = √31946.875

v_a = 178.74 m/s.

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So acceleration of the bike which started from rest, reaches at speed of 7.10 m/s by travelling distance 35.4 m will be 0.712 m/ sec^2.

Answer:

10 s

Explanation:

Given:

Δx = 100 m

v₀ = 0 m/s

a = 2 m/s²

Find: t

Δx = v₀ t + ½ at²

100 m = (0 m/s) t + ½ (2 m/s²) t²

t = 10 s

The work required to move a charge q from the other vertex to the center of the line joining the fixed charges is zero.

The charge on two vertices of the equilateral triangle are Q and -Q which is fixed. On the other vertex is a point charge q. The work done to move the point charge q from the vertex to the midpoint of other two charges Q and -Q is the change in potential energy of the charge q.

We know that,

U = q V

V = k q / r

where,

U = Electric potential energy of a charge

V = Electric potential at a point due to a point charge    

k = Coulomb's constant

r = Distance

Change in potential energy of charge q, ΔU = q ΔV

ΔU = q ( \(V_{f}\) - \(V_{i}\) )

Electric Potential at the midpoint of charges Q and -Q,

\(V_{f}\) = k Q / ( a / 2 ) + k ( - Q ) / ( a / 2 ) = 0

Electric Potential at the vertex of charges q,

\(V_{i}\) = k q / a + k ( - q ) / a = 0

ΔU = 0

The given question is incomplete. The complete question is:

Two particle with charges Q and -Q are fixed at the vertices of an equilateral triangle with sides of length a, the work required to move a particle with a charge q from the other vertex to the center of the line joining the fixed charges is?

Therefore, the work required to move a charge q from the other vertex to the center of the line joining the fixed charges is zero.

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The frictional force exerted on the block when it slides along a rough surface is a non-zero force.

When the block comes to a stop, it can be concluded that the frictional force is equal in magnitude to the block's applied force but opposite in direction. This means that the frictional force is doing negative work since it is resisting the motion of the block. In other words, the frictional force is in the opposite direction of the motion and reduces the kinetic energy of the block until it stops.

The magnitude of the frictional force can be determined by the equation:

Ff = μFn, where Ff is the frictional force, μ is the coefficient of friction and Fn is the normal force.

The coefficient of friction is determined by the type of surfaces the block and the ground have. For example, if both the block and the ground are made of steel, the coefficient of friction would be higher than if the block was made of rubber and the ground was made of marble.

Therefore, when a block slides along a rough surface and comes to a stop, we can conclude that a non-zero frictional force is exerted on the block, which is equal in magnitude to the applied force but opposite in direction.

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The time available on Machine II must lie between 1 minute and 3 minutes. The shadow price for Resource 2 (associated with constraint 2) is $3 per minute.

(a) To determine the range of available time on Machine II, we need to consider the constraints provided. The time available on Machine II must be between the time required for Type A souvenirs and the time required for Type B souvenirs.

Time required for Type A souvenir on Machine II: 1 minute

Time required for Type B souvenir on Machine II: 3 minutes

Therefore, the time available on Machine II must lie between 1 minute and 3 minutes.

The meaningful solution for the available time on Machine II is 1 min ≤ Machine II ≤ 3 min.

(b) To maximize the profit, we need to determine the optimal production quantities for Type A and Type B souvenirs given a change in the available time on Machine II.

Let's assume the change in available time on Machine II is represented by k.

To maximize the profit, we need to find the production quantities that maximize the total profit. Let's denote the production quantity for Type A souvenirs as x and the production quantity for Type B souvenirs as y.

The objective function for the profit can be expressed as:

Profit = 0.80x + 1.60y

Subject to the following constraints:

2x + y ≤ 120 (Machine I constraint)

x + 3y ≤ (300 + k) (Machine II constraint)

Using linear programming techniques, the optimal solution will depend on the value of k.

The statement "Ace Novelty's profit is maximized by producing Type A souvenirs 540 5 and 2(223+ *). 3 Type B souvenirs, where -225 1x ** $ 150 X X" seems to be incomplete and unclear. The specific production quantities and profit cannot be determined without knowing the value of k.

(c) To find the shadow price for Resource 2 (associated with constraint 2), we can perform sensitivity analysis.

The shadow price represents the change in the objective function's value per unit increase in the availability of Resource 2 (Machine II in this case). We can determine it by evaluating the sensitivity of the objective function to changes in the constraint.

Since the constraint is x + 3y ≤ (300 + k), the shadow price associated with Resource 2 is the coefficient of the Machine II term, which is 3.

Therefore, the shadow price for Resource 2 (associated with constraint 2) is $3 per minute.

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the correct answer is probably 70km

Answer:

uranis

Explanation:

Answer:

Saturn

Explanation:

Saturn is most like earth because it has moons and the atmosphere is denser like Earth.

Answer:

I am not sure really.

Explanation:

I just don't know

The kinetic energy of the ball is 100j

100j of work would be required to stop the ball.

Kinetic energy is a form of energy that an particle has by reason of it's motion.if we accelarate a object we must apply a force.

Here the mass (m) is 0.5 kg

It's velocity (v) is 20 m/s

The kinectic energy is

KE = 1/2 mv^2 = 1/2(0.5)20.20= 100 J

Again the workdone to stop it is calculated by work energy theorem

Change in KE = workdone

Or 100 J - 0 J or 100 J needed to stop the ball.

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Generally, the coefficient of static friction should be at least 0.7 for a car traveling at this speed, but this value could be higher depending on the specific conditions.

The coefficient of static friction, or μs, is the ratio of the maximum static force that can be applied without causing slipping to the normal force between two surfaces. The coefficient is affected by the surfaces in contact and the material they are made of.

The tires of a car also have an effect on the coefficient of static friction. Tires with a greater tread depth can provide better grip on the road and therefore have a higher coefficient of static friction than tires with a shallow tread. Tire pressure also plays a role, as low pressure can reduce the coefficient of static friction.

The weight of the vehicle is also an important factor. Heavier vehicles tend to have a higher coefficient of static friction, as the greater mass increases the normal force between the car and the surface.

In conclusion, the coefficient of static friction for a car traveling at 115 km/h around a turn depends on several factors, including the road surface, the vehicle's tires, and the vehicle's weight.

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Answer:

3552.6N

Explanation:

F=mg

=3.72*955

=3552.6N

Answer:

3552.6N

Explanation:

Answer:

It would stretch about 6 inches, but it depends on the metal.

Explanation:

pls mark brainliest with the crown

Answer:

The shape of planetary orbits affects their orbital velocity because the speed of a planet in its orbit is not constant. According to Kepler's laws of planetary motion, the planets move in elliptical orbits with the sun at one of the two foci of the ellipse.

Kepler's second law, also known as the law of equal areas, states that a line that connects a planet to the sun sweeps out equal areas in equal times as the planet travels around the sun. This means that a planet's speed varies throughout its orbit.

When a planet is closer to the sun (at perihelion), it travels faster as it is subject to a stronger gravitational pull. Conversely, when a planet is farther from the sun (at aphelion), it travels slower due to the weaker gravitational pull.

Therefore, the shape of a planet's orbit determines its distance from the sun and, consequently, the strength of the gravitational force acting on it, which in turn affects its orbital velocity.

When a diffraction grating is substituted for the double slit slide, the resulting pattern on the screen will be different.

A diffraction grating is a device with many slits that diffracts light into its component colors, similar to a prism. When light passes through a diffraction grating, it is diffracted, or spread out, into many different directions. This diffraction produces a pattern of bright and dark bands on a screen placed behind the grating, similar to the pattern produced by a double slit.

However, the pattern produced by a diffraction grating is different from the pattern produced by a double slit. The diffraction pattern produced by a diffraction grating has bright and dark bands that are wider and further apart than the bands produced by a double slit. This is because the diffraction pattern produced by a diffraction grating is determined by the spacing of the slits in the grating and the wavelength of the light passing through it, whereas the pattern produced by a double slit is determined by the interference of the two slits.

To observe the pattern on the screen with a diffraction grating, the screen does not need to be brought closer to the grating. Instead, the grating is typically placed at a distance from the screen, and the screen is placed at a distance from the grating such that the bright and dark bands on the screen are clearly visible.

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Based on the circled elements in the periodic table’s location, we could infer that most of the circled elements are gases at room temperature. The correct answer is B.

There are 11 elements that are in gaseous state at room temperature. They are Hydrogen (H), Helium (He), Neon (N), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn), Fluorine (F), Chlorine (Cl), Nitrogen (N) and Oxygen (O).

Elemental hydrogen (H, element 1), nitrogen (N, element 7), oxygen (O, element 8), fluorine (F, element 9), and chlorine (Cl, element 17) are all gases at room temperature, and also are found as diatomic molecules (H2, N2, O2, F2, Cl2).

Although part of your question is missing, you might be referring to this full question: Consider the circled elements in the periodic table as seen on attached image. Based on their location, we could infer that most of them are:

A) very reactive gases.

B) gases at room temperature.

C) do not react readily with metals.

D) nonreactive solids at room temperature.

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In a single slit experiment, decreasing the width of the slit would increase the amount of diffraction, which would result in a broader diffraction pattern.

Diffraction is the bending of waves around obstacles, and in a single slit experiment, it occurs when light passes through a narrow slit and spreads out into a pattern of bright and dark fringes on a screen. As the width of the slit is decreased, the diffraction of light increases, resulting in a wider central maximum and more pronounced side maxima. This means that the intensity of the fringes decreases, and the fringes become broader and less sharp.

Therefore, in general, the narrower the slit, the wider and less intense the diffraction pattern will be, which is due to the increased diffraction of light caused by the smaller opening.

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Answer:

Water cannot be used in thermometer because of its higher freezing point and lower boiling point than other liquids .  If water is used in a thermometer , it will start phase change at 0\(degree\\\)C and 100\(degree\)C and will not measure temperature , out of this range . This range is very small as compared  to other liquids as mercury , having freezing point about −39\(degree\)C and boiling point 356\(degree\)C.

Explanation:

Note : Please discard your other answers as they are incorrect. I have personally solved this problem and have attached the appropriate image.

Have a blessed day, if you have any follow-up questions please reach out!

Stay safe!

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The values of gain K for which the closed-loop system is stable cannot be determined without plotting the Root Locus.

To obtain the Root Locus plot for the given open-loop system, we need to determine the poles and zeros of the system and then plot the loci of the roots as the gain K varies.

The given open-loop transfer function is:

G(s) = K(s + 3) / ((s + 5)(s + 2)(s - 1))

The poles of the system are the values of 's' that make the denominator of the transfer function equal to zero. So, we have poles at s = -5, s = -2, and s = 1.

The zeros of the system are the values of 's' that make the numerator of the transfer function equal to zero. In this case, there is a zero at s = -3.

To find the values of gain K for which the closed-loop system is stable, we need to determine the regions of the Root Locus plot that lie on the left-hand side of the complex plane. In other words, the regions where the number of poles to the right of a point is an odd number. From the given transfer function, we can see that there are three poles at s = -5, s = -2, and s = 1. Therefore, the Root Locus plot will start from these three poles and extend towards infinity. To find the breakaway and break-in points on the Root Locus plot, we can perform calculations and analysis using the characteristic equation. However, since the calculations are involved and require step-by-step analysis, it is best to refer to a graphical representation of the Root Locus plot. Please refer to a Root Locus plot software or tool to obtain the complete Root Locus plot for the given open-loop system. The plot will show the regions of stability and the values of gain K for which the closed-loop system is stable.

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Answer: 6 west

Explanation:

Based on the calculations, the electric sander (sanding) is the more expensive in terms of electricity used.

Mathematically, the quantity of energy consumed by an electrical device can be calculated by using this formula:

Energy consumption = power × time

For the electric saw, we have:

Substituting the given parameters into the formula, we have;

Energy consumption = 1,200 × 20

Energy consumption = 24,000 Joules.

For the electric sander, we have:

Substituting the given parameters into the formula, we have;

Energy consumption = 1,000 × 30

Energy consumption = 30,000 Joules.

Therefore, the electric sander (sanding) is the more expensive in terms of electricity used.

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Answer: sanding

Explanation:

Energy used by saw = power × time = 1,200W × 1,200s = 1,440,000J.

Energy used by sander = power × time = 1,000W × 1,800s = 1,800,000J.

The sander uses more energy.