An engine does 25 J of work and exhausts 20 J of waste heat during each cycle. If the cold-reservoir temperature is 30 ∘C, what is the minimum possible temperature in ∘C of the hot reservoir?

Answer:

The second ball

Explanation:

Both balls are under the effect of gravity, accelerating with exactly the same value. The first ball is dropped, therefore its initial velocity is zero. Since the second ball has horizontal and vertical velocity components, its initial velocity is given by:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

The vertical component is zero, however, it has a horizontal velocity, so its initial speed is not zero, therefore the secong ball has the greater speed at ground level.

The second ball will reach a greater total speed at ground level due to its initial horizontal speed added to its vertical speed. But, both balls will hit the ground at the same time if the horizontal speed of the second ball is not too high.

In the described situation, gravity is the only vertical force acting on both balls, hence they will reach the ground at the same speed in the vertical direction. But, the second ball has additional horizontal speed. Therefore, with this horizontal component added to the vertical detachment speed due to gravity, the second ball will have a greater total speed by the Pythagorean theorem: (total speed)^2 = (vertical speed)^2 + (horizontal speed)^2. However, the question of which hits the ground first depends on the initial horizontal speed of the second ball. If the horizontal speed is not too high, both balls should hit the ground at the same time.

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Answer:

(a) t = 2.97s

(b) h = 43.3 m

Explanation:

Let t be the time it takes to fall a distance h, then t - 1 (s) is the time it takes to fall a distance of h - 0.56h = 0.44 h

For the ball to fall from rest a distance of h after time t

[tex]h = gt^2/2[/tex]

Also for the ball to fall from rest a distance of 0.44h after time (t-1)

[tex]0.44h = g(t-1)^2/2[/tex]

We can substitute the 1st equation into the 2nd one

[tex]0.44gt^2/2 = g(t-1)^2/2[/tex]

and divide both sides by g/2

[tex]0.44t^2 = (t-1)^2[/tex]

[tex]0.44t^2 = t^2 - 2t + 1[/tex]

[tex]0.56t^2 - 2t + 1 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{2\pm \sqrt{(-2)^2 - 4*(0.56)*(1)}}{2*(0.56)}[/tex]

[tex]t= \frac{2\pm1.33}{1.12}[/tex]

t = 2.97 or t = 0.6

Since t can only be > 1 s we will pick t = 2.97 s

(b) [tex]h = gt^2/2 = 43.3 m[/tex]

Answer:

(B) Vertically upward

Explanation:

r = Side of triangle = 14 cm

q = Charge = -2.5 nC

Electrical force is given by

[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times (2.5\times 10^{-9})^2}{0.14^2}\\\Rightarrow F=2.86671\times 10^{-6}\ N[/tex]

For the top charge

Net force on both charges is given by

[tex]F_n=2Fcos\theta\\\Rightarrow F_n=2\times 2.86671\times 10^{-6}\times cos30\\\Rightarrow F_n=4.96529\times 10^{-6}\ N[/tex]

The net force acting on the top charge is [tex]4.96529\times 10^{-6}\ N[/tex]

Here the forces are symmetrical hence, the net force is along +y axis i.e., upward

The direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle is vertically upward.

The direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle can be determined using Coulomb's Law. Since the objects at the base of the triangle have a negative charge and the object at the top corner has a negative charge as well, the force between them will be repulsive. As a result, the direction of the electrical force on the top object will be vertically upward.

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm[/tex]

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

[tex]\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s[/tex]

The block's speed is 31.422 cm/s

Final answer:

This physics problem involves calculating the amplitude of oscillations and the speed of a block at a specific displacement by applying conservation of mechanical energy in harmonic motion. The amplitude is found to be approximately 6.37 cm, and the block's speed at 0.70 A (amplitude) is approximately 28.4 cm/s.

Explanation:

The problem involves a 0.600 kg block attached to a spring with a spring constant of 15 N/m that is initially at rest and is then given a speed of 44 cm/s. To find the amplitude of the subsequent oscillations and the block's speed when x = 0.70 A, where A is the amplitude, we first use the principle of conservation of mechanical energy in harmonic motion. The initial kinetic energy given to the block will be equal to the potential energy of the spring at maximum displacement, which allows us to calculate the amplitude. Next, we determine the speed of the block at a displacement of 0.70 A using the relationship between kinetic and potential energy at any point during the oscillation.

The amplitude can be found using: KE = 1/2 k A^2, where KE is the kinetic energy of the block. Converting 44 cm/s to m/s gives 0.44 m/s. The kinetic energy of the block is KE = 1/2 mv^2 = 1/2 (0.600 kg)(0.44 m/s)^2, and solving for A gives the amplitude in meters, which can then be converted back to centimeters. To find the block's speed at x = 0.70 A, we use the conversion of potential energy at this displacement back to kinetic energy, considering the total mechanical energy of the system remains constant.

Using these principles, calculations show the amplitude of the oscillations to be approximately 6.37 cm, and the block's speed at x = 0.70 A is about 28.4 cm/s.

Answer:

Explanation:

Given mass of bucket is [tex]m=2\ kg[/tex]

mass of cylinder [tex]M=4\ kg[/tex]

Suppose T is the tension in the rope

For bucket [tex]mg-T=ma[/tex]

where a=acceleration

For cylinder with Radius R

[tex]I\times \alpha =T\cdot R[/tex]

[tex]\frac{MR^2}{2}\times \frac{a}{R}=T\times R[/tex]

[tex]T=\frac{Ma}{2}[/tex]

[tex]a=\frac{mg}{m+0.5M}[/tex]

[tex]a=4.9\ m/s^2[/tex]

Using [tex]v^2-u^2=2a s[/tex] for bucket

v=final velocity

u=initial velocity

s=displacement

[tex]v^2-0=2\times 4.9\times 1.5[/tex]

[tex]v=\sqrt{14.7}[/tex]

[tex]v=3.83\ m/s[/tex]              

Final answer:

The speed of the bucket after falling a distance of 1.50 m is 3.83 m/s.

Explanation:

To find the speed of the bucket after falling a distance of 1.50 m, we can use the principle of conservation of energy. The initial potential energy of the bucket is given by the equation PE = mgh, where m is the mass of the bucket, g is the acceleration due to gravity, and h is the height from which the bucket is released. In this case, the initial potential energy is PE = (2.0 kg)(9.8 m/s^2)(1.50 m).

The final kinetic energy of the bucket is given by the equation KE = 0.5mv^2, where m is the mass of the bucket and v is the speed of the bucket. Since the bucket starts from rest, the initial kinetic energy is zero. Therefore, the final kinetic energy is equal to the initial potential energy. We can set up the equation (2.0 kg)(9.8 m/s^2)(1.50 m) = 0.5(2.0 kg)v^2 and solve for v.

Simplifying the equation, we get ([tex]29.4 kg*m^2/s^2[/tex]. Dividing both sides by (2.0 kg), we find v^2 = 14.7 m^2/s^2. Taking the square root of both sides, we find v = 3.83 m/s.

A very useful way to graphically schematize a field is to draw lines that go in the same direction as that field at several points. This is done through the electric field lines, which are imaginary lines that describe, if any, changes in the direction of the electric field when passing from one point to another, so that these lines are tangent, in each point of the space where the electric field is defined, to the direction of the electric field at that point.

According to Newton's first law, the force acting on a particle produces a change in its velocity; therefore, the movement of a charged particle in a region will depend on the forces acting on it at each point in that region.

Therefore the electric field lines in space surrounding a charge distribution will show tangents to the directions in which either static or moving charges would accelerate when passing through points on those lines.

The correct answer is D.

Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s  (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

Answer with explanation :

The negative sign means that the potential energy decreases by the movement of the electron.

negative charge at rest in an electric field moves toward the region of an electric field , so that its potential energy will diminish and change into the kinetic energy of motion. The total energy remains constant.

Positive charges will move downhill because of convention. It is to stay in accordance with other potential theories, particularly gravity, where the "charge" is mass, that moves downwards in the gravitational potential field expressed by ϕ(r)=−GM|r|ϕ(r)=−GM|r|. In an electronic system, howbeit, positive charges are fixed in their position within a component (e.g., a wire), therefore instead of the mobile,the negative charges, electrons, move uphill.

Final answer:

A negative charge at rest in an electric field moves toward a region of higher potential, thus decreasing its electric potential energy. A positive charge moves toward a region of lower potential, similarly decreasing its electric potential energy. The strength of the electric field relates to potential difference and is zero where electric potential is constant.

Explanation:

If a negative charge is initially at rest in an electric field, it will move toward a region of higher potential. This is because negative charges are attracted to areas of positive potential and repelled from areas of negative potential. Conversely, a positive charge will move toward a region of lower potential, much like how fluid moves from a region of high pressure to low pressure.

The potential energy of a charge in an electric field depends on both the electric potential and the sign and magnitude of the charge. For a negative charge, moving to a higher potential region decreases its electric potential energy, while a positive charge moving to a lower potential region also decreases its electric potential energy. The electric field is defined as the force per unit charge, and its strength is directly related to the potential difference. If the electric potential is constant in a region, the electric field strength is zero there.

When discussing the potential difference and electric field strength, a good example is a parallel plate capacitor which has a uniform electric field. If an electron, being negatively charged, is released at rest between the plates, it will accelerate toward the positively charged plate due to the attractive electrostatic force, thereby moving to a region of higher potential.

Final answer:

When the charge on each of two small spheres is halved, the force of attraction between them, according to Coulomb's Law, will be quartered (Option D). This is because the force is proportional to the product of the charges, which would be reduced to a quarter.

Explanation:

The question is related to Coulomb's Law, which describes the electrostatic force between two charges. According to Coulomb's Law, the force F between two charges is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r^2) between them: F = k * (q1*q2) / r^2, where k is Coulomb's constant.

If the charge on each of the two small spheres is halved, we have new charges q1/2 and q2/2. Substituting these into the formula, we get the new force F' as F' = k * ((q1/2)*(q2/2)) / r^2. Simplifying this, F' = (1/4) * k * (q1*q2) / r^2, which is one quarter of the original force F. Therefore, the force of attraction between the spheres will be quartered when the charge on each sphere is halved.

Final answer:

To calculate the pressure of the water in the hose at the spigot, you can use Bernoulli's equation, which relates pressure, density, velocity, and height of a fluid. By assuming the height at the spigot is the same as the nozzle and plugging in the given values, the pressure can be calculated.

Explanation:

The pressure of the water in the hose right after it comes out of the spigot can be calculated using Bernoulli's equation. Bernoulli's equation relates the pressure, density, velocity, and height of a fluid. In this case, the pressure can be found using the equation:

P1 + 1/2 ρv1² + ρgh1 = P2 + 1/2 ρv2²+ ρgh2

Where P1 is the pressure at the spigot, P2 is the pressure at the nozzle, ρ is the density of water, v1 is the initial velocity of water at the spigot, v2 is the velocity at the nozzle, h1 is the height of the spigot, and h2 is the height of the nozzle.

Since the height of the spigot is not given, we can assume it is at the same level as the nozzle, which means h1 = h2 = 7.25 m. The density of water, ρ, is 1000 kg/m³. The velocity at the nozzle, v2, is given as 11.2 m/s. Given these values, we can solve for the pressure at the spigot, P1.

Final answer:

The pressure of the water in the hose right after it comes out of the spigot can be calculated using Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. Assuming the flow is laminar and neglecting viscosity, the pressure is found to be 229000 N/m².

Explanation:

The pressure of the water in the hose right after it comes out of the spigot can be calculated using Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. Assuming the flow is laminar and neglecting viscosity, we can use the equation:

P + 1/2ρv² + ρgh = constant

Where P is the pressure, ρ is the density of water, v is the velocity of the water, g is the acceleration due to gravity, and h is the height difference between two points. Since the hose and nozzle are connected, we can assume that the pressures at these two points are the same. Also, the velocity of the water inside the hose can be considered negligible compared to the velocity at the nozzle. Therefore, we can simplify the equation to:

1/2ρv²+ 0 + ρgh = constant

The pressure at the spigot is atmospheric pressure, which is given as 1.013x10⁵ N/m². Rearranging the equation and solving for P, we have:

P = 1.013x10⁵ N/m² + 1/2ρv² + ρgh

Using the given values, we can substitute them into the equation and calculate the pressure

P = 1.013x10⁵ N/m² + 0.5x1000 kg/m³x(11.2 m/s)² + 1000 kg/m³x9.8 m/s²x7.25 m

P = 1.013x10⁵ N/m²+ 62656.96 N/m² + 68300 N/m²

P = 228957.96 N/m²

Rounding to three significant digits, the pressure of the water in the hose right after it comes out of the spigot is 229000 N/m².

Answer:

Layer of glass rod rubbed with silk.

Explanation:

Some of the atoms in the surface layer of a glass rod positively charged by rubbing it with a silk cloth have lost electrons, leaving a net positive charge because of the unneutralized protons of their nuclei. A negatively charged object has an excess of electrons on its surface.

Final answer:

Conducting materials like metals have freely moving electrons that enable the flow of charge, supported by experimental evidence like studies on conductivity and direct experiments confirming the presence of electrons as charge carriers.

Explanation:

Conducting materials such as metals have freely moving electrically charged particles within them. These free electrons can move through the material allowing the flow of charge. Experimental evidence supporting this idea includes studies on conductivity, Ohm's law for electromagnetics, and direct experiments like the one by Tolman and Stewart confirming that charge carriers in metals are indeed electrons.

Answer:

Albedo Alto: Snow

Albedo bass: ocean

Explanation:

Albedo is the fraction of solar radiation reflected by a surface. The term has its origin from the Latin word albus, which means "white." It is quantified as the proportion or percentage of solar radiation of all wavelengths reflected by a body or surface to the amount incident on it.

The color of the soil certainly affects the reflectivity, the lighter colors that have greater albedo than the dark colors and, therefore, present greater albedo. Soil texture is also a factor that affects albedo. Some studies have shown that snow-covered soils, being a light color, reflect most of the light and therefore do not absorb it and present the greatest amount of terrestrial surfaces. While with the ocean and seas most of the radiation is absorbed it has a dark color increasing with the depth of the place and therefore has the lowest surface albedo. In the end of the year the snow has 86% of reflected light while the oceans 8% of reflected light.

Answer:

A) t₁ = 0.56 s t₂ =3.26 s

B) vh₁=vh₂ = 25.1 m/s

C) v₁ = 13.2 m/s  v₂ = -13.2 m/s

D) v = 31. 3 m/s

E) 36.7º below horizontal.  

Explanation:

A) As the only acceleration of the baseball is due to gravity, as it is constant, we can apply the kinematic equations in order to get times.

First, we can get the horizontal and vertical components of the velocity, as the movements  along these directions are independent each other.

v₀ₓ = v* cos 36.7º = 31.3 m/s * cos 36.7º = 25.1 m/s

v₀y = v* sin 36.7º =  31.3 m/s * sin 36.7º = 18.7 m/s

As in the horizontal direction, movement is at constant speed, the time, at any point of the trajectory, is defined by the vertical direction.

We can apply to this direction the kinematic equation that relates the displacement, the initial velocity and time, as follows:

Δy = v₀y*t -1/2*g*t²

We can replace Δy, v₀y and g for the values given, solving a quadratic equation for t, as follows:

4.9*t²-18.7t + 9 = 0

The two solutions for  t, are just the two times at which the baseball is at a height of 9.00 m above the point at which it left the bat:

t = 1.91 sec +/- 1.35 sec.

⇒ t₁ = 0.56 sec   t₂= 3.26 sec.

B) As we have already told, in the horizontal direction (as gravity is always downward) the movement is along a straight path, at a constant speed, equal to the x component of the initial velocity.

⇒ vₓ = v₀ₓ = 25.1 m/s

C) In order to get the value of  the vertical components at the two times that we have just found, we can apply the definition of acceleration (g in this case), solving for vfy, as follows:

vf1 = v₀y - g*t₁ = 18.7 m/s - (9.8m/s²*0.56 sec) = 13.2 m/s

vf₂ = v₀y -g*t₂ = 18.7 m/s - (9.8 m/s²*3.26 sec) = -13.2 m/s

D) In order to get the magnitude of  the baseball's velocity when it returns to the level at which it left the bat, we need to know the value of the vertical component at this time.

We could do in different ways, but the easiest way is using the following kinematic equation:

vfy² - v₀y² = 2*g*Δh

If we take the upward path, we know that at the highest point, the baseball will come momentarily to an stop, so at this point, vfy = 0

We can solve for Δh, as follows:

Δh = v₀y² / (2*g) = (18.7m/s)² / 2*9.8 m/s² = 17.8 m

Now, we can use the same equation, for the downward part, knowing that after reaching to the highest point, the baseball will start to fall, starting from rest:

vfy² = 2*g*(-Δh) ⇒ vfy = -√2*g*Δh = -√348.9 = -18. 7 m/s

The horizontal component is the same horizontal component of the initial velocity:

vx = 25.1 m/s

We can get the magnitude of the baseball's velocity when it returns to the level at which it left the bat, just applying Pythagorean Theorem, as follows:

v = √(vx)² +(vfy)² = 31.3 m/s

E) The direction below horizontal of the velocity vector, is given by the tangent of the angle with the horizontal, that can be obtained as follows:

tg Ф = vfy/ vx = -18.7 / 25. 1 =- 0.745

⇒ Ф = tg⁻¹ (-0.745) = -36.7º

The minus sign tell us that the velocity vector is at a 36.7º angle below the horizontal.

A major leaguer hits a baseball and the question asks for the time, velocity components, velocity magnitude, and direction at different points of the ball's trajectory when it reaches a certain height.

A) Let's consider the vertical motion of the baseball. The equation to determine the time it takes for an object to reach a certain height is:

h = v0t + 0.5gt2

Where:

By rearranging the equation, we can solve for t:

t2 + (2v0/g)t - (2h/g) = 0

Using the quadratic formula, we can plug in the values and solve for t. The two possible values of t will correspond to the two times when the ball is at a height of 9.00 m above the point at which it left the bat.

B) The horizontal component of velocity remains constant throughout the motion. Therefore, the horizontal component of velocity at each of the two times found in part (a) will be the same as the initial horizontal velocity, which is 31.3 m/s * cos(36.7º).

C) The vertical component of velocity changes due to the acceleration of gravity. At each of the two times found in part (a), the vertical component of velocity can be found using the equation:

vv1,2 = v0 + gt

Where:

D) When the baseball returns to the level at which it left the bat, its vertical velocity will be -v0, which means it will be moving downward with the same magnitude as the initial velocity but in the opposite direction. The horizontal component of velocity remains unchanged. Therefore, the magnitude of the baseball's velocity when it returns to the level at which it left the bat will be the square root of the sum of the squares of the horizontal and vertical components, which can be calculated using the formula:

v = √(vh2 + vv2)

Where:

E) The direction of the baseball's velocity when it returns to the level at which it left the bat can be determined using trigonometry. The angle below the horizontal can be found as:

θ = tan-1(vv/vh)

Where:

Answer:

    The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275  J

W= - 125 J

W' = 50 J

W(net)= -125  + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the  internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

The internal energy of the gas is the energy contained by the gas. The increase in the internal energy of the gas in the given container is 350 J.

The internal energy of the gas can be calculated by the formula

Q= ΔU + W(net)

Where,

Q - energy absorbed by the system = 275  J

ΔU - internal energy of the gas= ?

W(net)=  W - W' = -125  + 50 = -75 J

Put the values in the formula,

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

Therefore, the increase in the internal energy of the gas in the given container is 350 J.

To know more about internal energy,

https://brainly.com/question/11278589

Answer:

A. [tex]KE_h=\frac{1}{16}\times KE_l[/tex]

B. the lighter block travels at a speed 4 times faster than the heavier block.

C. b The heavy block must be pushed 4 times farther than the light block.

Explanation:

Given that the blocks are  acted upon by equal force.

A.

Then the kinetic energy of the blocks depends on their individual velocity.

And velocity is related to momentum through Newton's second law of motion:

[tex]\frac{d}{dt} .p=F[/tex]

[tex]\frac{d}{dt} (m.v_l)=F[/tex]

considering that the time for which the force acts on each mass is equal.

[tex]dv_l=\frac{F.dt}{m}[/tex]

For the heavier block:

[tex]dv_{_h} =\frac{F.dt}{4m}[/tex]

Therefore:

Kinetic energy of lighter block:

[tex]KE_l=\frac{1}{2}\times m.(\frac{F.dt}{m} )^2[/tex]

[tex]KE_l=\frac{1}{2m} \times (F.dt)^2[/tex]

Kinetic energy of heavier block:

[tex]KE_h=\frac{1}{2} \times m.(\frac{F.dt}{4m} )^2[/tex]

[tex]KE_h=\frac{1}{16}\times (\frac{1}{2m} \times (F.dt)^2)[/tex]

[tex]KE_h=\frac{1}{16}\times KE_l[/tex]

B.

From the above calculations and assumptions we observe that the lighter block travels at a speed 4 times faster than the heavier block.

C.

Since the lighter block is having the speed 4 times more than the heavier block so it must be pushed 4 times farther because the speed is directly proportional to the  distance.

A) The statement that is true from the attached link about the kinetic energy of the heavier block after the push is;

B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block

B) Compared to the speed of the heavierblock, the speed of the light block travel is; twice as fast.

C) If we assume that both blocks have the same speed after being pushed with the same force F_vec, what we can say about the distances the two blocks are pushed is;

The heavyblock must be pushed 4 times farther than the lightblock.

A) The formula for the work done on each block is gotten from the formula;

W = F × d

We are told that the two blocks were pushed the same distance and with the same force F_vec being exerted on them. Thus, the work done on each of the blocks will be the same.

From work-energy theorem, we recall that the work done on an object is equal to its change in kinetic energy. Thus;

W = ΔKE

Where;

ΔKE = Final KE - Initial KE

Thus;

W = Final KE - Initial KE

Since the blocks were initially at rest, then it means that;

Initial KE = 0

Thus;

W = Final KE - 0

W = Final KE

The work done on each block is the same and as a result their final kinetic energies will be the same.

B) We are told that one of the blocks is ¼ times as heavy as the other block.

Thus;

½(m)v_h² = ½(¼mv_l²)

v_h² = ¼v_l²

Where;

v_h is speed of heavy block

v_l is speed of light block

Thus, taking square root of both sides gives;

v_h = ½v_l

Thus, the speed of the light block is twice as fast.

C) We assume the blocks have the same speed after being pushed with the same force. Thus;

It means that heavier block must be pushed four times farther than lighter block.

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Answer:

F = 0.0034 N

Explanation:

Given:

[tex]B = 5.2*10^(-5) T\\Q = 57 degrees\\I_{wire} = 12 A\\L_{wire} = 10 m[/tex]

The angle between B and wire = 90 - 57 = 33 degrees

Using formula:

[tex]F = B*I*L*sin (90-Q)\\F = (5.2*10^(-5)*(12)*(10)*sin (33)\\F = 0.0034 N[/tex]

(A) The magnetic force exerted on the wire is 3.4×10⁻³N

(B) The direction of the force is to the west.

Given that the magnetic field B = [tex]5.2\times10^{-5}T[/tex] which points in the direction  57° below a horizontal line pointing north.

Length of the wire L = 11m

current in the wire I = 11A

The angle between the wire and the magnetic field is θ = (90-57) = 33°

(A) The magnetic force on a finite wire of length L carrying a current I is given by:

[tex]F=BILsin\theta\\\\F=5.2\times10^{-5}\times11\times11\sin33\\\\F=3.4\times10^{-3}N[/tex]

(B) The direction of the force is given by dl×B, now B is at 57° with the north direction and the wire is verticle, so the direction of the field will be to the west.

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Answer:

a) Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Explanation:

a) Statement A : 2.567km to two significant figures.

2.567km 2. S.F = 2.6km

Statement B : 2.567km to three significant figures.

2.567km 3 S.F = 2.57km

Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) statement A: (2.567 km + 3.146km) to 2 S.F

(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km

Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).

2.567km to 2 S.F = 2.6km

3.146km to 2 S.F = 3.1km

2.6km + 3.1km = 5.7km

Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Considering significant figures, statement A: 2.567 km is less than Statement B: 2.567 km, since it rounds to 2.6 km. For the addition, Statement A (5.713 km rounded to 5.7 km) is equal to Statement B (adding 2.6 km and 3.1 km).

When we consider significant figures, statement A: 2.567 km round off to two significant figures would be 2.6 km and Statement B: 2.567 km is the same to three significant figures. Therefore, statement A is less than statement B because 2.6 km is less than 2.567 km. Similarly, for the next case, if we add 2.567 km and 3.146 km, it gives us 5.713 km, rounded off to two significant figures it becomes 5.7 km for statement A. Whereas statement B, 2.567 km rounded off becomes 2.6 km and 3.146 km becomes 3.1 km. Adding both gives 5.7 km. Hence, statement A is equal to statement B.

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Explanation:

Given that,

Charge acting on the object, [tex]q=-4\ nC=-4\times 10^{-9}\ C[/tex]

Force acting on the object, [tex]F=19\ nC=19\times 10^{-9}\ C[/tex] (in downward direction)

(a) The electric force acting in the electric field is given by :

[tex]F=qE[/tex]

E is the electric field

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}[/tex]

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, [tex]q=1.6\times 10^{-19}\ C[/tex]

The force acting on the proton is :

[tex]F=qE[/tex]

[tex]F=1.6\times 10^{-19}\times 4.75[/tex]

[tex]F=7.6\times 10^{-19}\ N[/tex]

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

Answer:

Explanation:

The abundance of the chemical elements is a measure of the occurrence of the chemical elements relative to all other elements in a given environment. Abundance is measured in one of three ways: by the mass-fraction (the same as weight fraction); by the mole-fraction (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the volume-fraction. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and ideal gas mixtures. Most abundance values in this article are given as mass-fractions.

For example, the abundance of oxygen in pure water can be measured in two ways: the mass fraction is about 89%, because that is the fraction of water's mass which is oxygen. However, the mole-fraction is about 33% because only 1 atom of 3 in water, H2O, is oxygen. As another example, looking at the mass-fraction abundance of hydrogen and helium in both the Universe as a whole and in the atmospheres of gas-giant planets such as Jupiter, it is 74% for hydrogen and 23–25% for helium; while the (atomic) mole-fraction for hydrogen is 92%, and for helium is 8%, in these environments. Changing the given environment to Jupiter's outer atmosphere, where hydrogen is diatomic while helium is not, changes the molecular mole-fraction (fraction of total gas molecules), as well as the fraction of atmosphere by volume, of hydrogen to about 86%, and of helium to 13%.[Note 1]

The abundance of chemical elements in the universe is dominated by the large amounts of hydrogen and helium which were produced in the Big Bang. Remaining elements, making up only about 2% of the universe, were largely produced by supernovae and certain red giant stars. Lithium, beryllium and boron are rare because although they are produced by nuclear fusion, they are then destroyed by other reactions in the stars.[1][2] The elements from carbon to iron are relatively more abundant in the universe because of the ease of making them in supernova nucleosynthesis. Elements of higher atomic number than iron (element 26) become progressively rarer in the universe, because they increasingly absorb stellar energy in their production. Also, elements with even atomic numbers are generally more common than their neighbors in the periodic table, due to favorable energetics of formation.

The abundance of elements in the Sun and outer planets is similar to that in the universe. Due to solar heating, the elements of Earth and the inner rocky planets of the Solar System have undergone an additional depletion of volatile hydrogen, helium, neon, nitrogen, and carbon (which volatilizes as methane). The crust, mantle, and core of the Earth show evidence of chemical segregation plus some sequestration by density. Lighter silicates of aluminum are found in the crust, with more magnesium silicate in the mantle, while metallic iron and nickel compose the core. The abundance of elements in specialized environments, such as atmospheres, or oceans, or the human body, are primarily a product of chemical interactions with the medium in which they reside.

Answer:

A)Qa=80 J

B)Qr= 100 J

Explanation:

Given that

W= 20 J

COP = 4

Heat rejected from cold reservoir = Qa

Heat exhausted to hot reservoir = Qr

The COP of refrigerator is given as

[tex]COP=\dfrac{Qa}{W}[/tex]

[tex]4=\dfrac{Qa}{20}[/tex]

Qa= 4 x 20 J

Qa=80 J

By using first law of refrigerator

Qr= Qa + W

Qr= 80 + 20 J

Qr= 100 J

A)Qa=80 J

B)Qr= 100 J

The heat extracted from the cold reservoir per cycle is 80 J and the heat exhausted to the hot reservoir per cycle is 100 J.

The equation for the coefficient of performance (COP) of a refrigerator is COP = Q_c / W, where Q_c is the heat extracted from the cold reservoir and W is the work done. Given the COP and the work done, we can rearrange this equation to find Q_c: Q_c = COP * W. Substituting the given values: Q_c = 4.0 * 20 J = 80 J.

The heat exhausted to the hot reservoir (Q_h) for a refrigerator can be found using the equation: Q_h = W + Q_c. Substituting the values we found: Q_h = 20 J + 80 J = 100 J.

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Answer:

v = 26.7 mph

Explanation:

During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:

Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi

Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:

Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi

So, the total displacement during all the trip wil be:

Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi

So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:

v = Δx / Δt = 160 mi / 6 h = 26.7 mph

Answer:

Heat required to raise the temperature will be 563.625 J

Explanation:

We have given number of moles n = 2.70 mole

Temperature is raises from 22°C to 32°C

So increase in temperature [tex]\Delta T=32-22=10^{\circ}C[/tex]

It is given that air is diatomic so [tex]c_v=\frac{5}{2}R=2.5\times 8.314=20.875[/tex]

We know that heat is given by [tex]Q=nc_v\Delta T[/tex]

So heat will be equal to [tex]Q=2.70\times 20.875\times 10=563.625J[/tex]

So heat required to raise the temperature will be 563.625 J

Answer:

The distance of the bee from the hive is 740 m.

Explanation:

Given that,

Bee starts fly 500 m due east, 430 m west and 670 m east.

The direction of the bee

500 m in positive direction

430 m in negative direction

670 m in positive direction

We need to calculate the net distance

Using formula of distance

[tex]D=500-430+670[/tex]

[tex]D=740\ m[/tex]

Hence, The distance of the bee from the hive is 740 m.

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I is 1360 W/m²

P sun =  2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars =  3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

Answer:

K E = 300 J

Explanation:

given,

Force of pushing crate, F = 100 N

distance of push = 10 m

frictional force = 70 N

kinetic energy is gained by the crate = ?

using work energy theorem

K E = work done

K E  = F . s

K E  = (100 - 70) x 10

K E = 30 x 10

K E = 300 J

kinetic energy is gained by the crate is equal to 300 J.

Final answer:

The kinetic energy gained by the crate is 300 J, calculated as the net work done on the crate, which is the work done by the applied force (1000 J) minus the work done by friction (700 J).

Explanation:

To determine how much kinetic energy is gained by the crate, we can use the work-energy theorem which states that the work done on an object is equal to the change in kinetic energy of the object. In this scenario, the net work done on the crate is the work done by the applied force minus the work done by friction. The work done by the applied force is force times distance, which is 100 N x 10 m = 1000 J. The work done by friction is 70 N x 10 m = 700 J. Therefore, the net work done on the crate, which is also the kinetic energy gained by the crate, is 1000 J - 700 J = 300 J.

Answer:

U= 23.544 KJ

Explanation:

Given that

mass ,m = 120 kg

Height above the datum ,h = 20 m

Take g = 9.81 m/s²

The potential energy U is given as

U= m g h

m=mass

h=Height above the datum

g=Gravitational constant

Now by putting the value in the above values we get

U= 120 x 9.81 x 20 J

U=23544 J

Potential energy in KJ

U= 23.544 KJ

Therefore the answer will be 23.544 KJ.

Answer:

d) Both a and b are correct

Explanation:

Displacement: It is defined as the distance between initial and final position during motion.

Distance: It is defined as the total path length traveled by object

Or

It is the distance of one place from other place.

Student said that the displacement between my dorm and the lecture hall is 1km.

It is not displacement .It is distance or we say path length.

Therefore, he is using  incorrect physical quantity for the information provided.

He should have called distance 1 km or path length 1 km.

Option d is true.

The student is accurately using the term 'displacement' to describe the 1 kilometer between their dorm and the lecture hall. Technically, 'distance' or 'path length' could also have been used assuming a straight-line path and ignoring direction.

In the context of this specific question, the student is accurately using the term displacement as a physical quantity. Displacement refers to the shortest distance between two points in a particular direction. In this case, the 1 kilometer refers to the displacement between the dorm and the lecture hall, assuming a straight-line path. So, technically, both a) Distance and b) Path length could also have been used to describe the 1 kilometer, if we don't consider the direction.

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Answer:

option B

Explanation:

given,

Length of field = 360 ft

                        = 360/3.281 = 109.72 m    ∵ 1 m = 3.281 ft

width of field = 160 ft

                      = 160/ 3.281 = 48.76 m

width of mower = 0.5 m

number of rounds required

   = [tex]\dfrac{48.76}{0.5}[/tex]

   = 97.5

we know,  1 km/h = 0.278 m/s

time taken for each round is equal to

[tex]t = \dfrac{length}{speed}[/tex]

[tex]t = \dfrac{109.72}{0.278}[/tex]

     t = 394.67 s

total time,

T = 394.67 x 94.5 = 37296.91 s

[tex]T = \dfrac{37296.91}{3600}\ hours[/tex]

[tex]T =11\ hours[/tex]

Hence. the correct answer is option B

Answer

given,

initial speed of the rocket A = 100 m/s

height of explode = 300 m

acceleration due to gravity = 9.8 m/s²

rocket b is launched after t₁ time

now, using equation of motion to calculate time

[tex]s = ut + \dfrac{1}{2}gt^2[/tex]

[tex]300 = 100t + \dfrac{1}{2}(-9.8)t^2[/tex]

 4.9 t² - 100 t + 300 = 0

using quadratic equation

[tex]t = \dfrac{-(-100)\pm \sqrt{100^2-4\times 4.9 \times 300}}{2\times 4.9}[/tex]

t₁ = 3.65 s   and  t₂ = 16.75 s

now, rocket A reaches 300 m on return at 16.75 s

rocket B reaches 300 m after 3.65 s

time difference of launch:

t = 16.75 - 3.65

t = 13.1 s

velocity of rocket A

v_a = u + g t

v_a =100 - 9.8 x 16.75

v_a = -64.15 m/s

velocity of rocket B

v_b = u + g t

v_b =100 - 9.8 x 3.65

v_b =+64.23 m/s

relative velocity of B relative to A at the time of the explosion

[tex]V_{BA} = v_b - V_a[/tex]

[tex]V_{BA} = 64.23 -(-64.15)[/tex]

[tex]V_{BA} = 128.38\ m/s[/tex]

relative velocity is equal to 128.38 m/s