A Punnett square is used to determine the *

According to Newton's law, action and reaction are equal and opposite.

According to the Newton's third law of motion, action and reaction are equal and opposite. As such, when the arrow hits the vertical wood board, the board exerts some force on the arrow.

We can not determine the magnitude of this force numerically because the question is incomplete.

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Answer:

newtons law action and reaction are equal and opposite

Explanation:

The merry-go-round takes approximately 29.41 seconds to complete 2.00 rotations.

Given data:

Acceleration (α) = 0.135 rad/\(s^2\)

Number of rotations (θ) = 2.00

To find the time taken (t) for 2.00 rotations, we need to use the formula:

θ = 0.5 * α * \(t^2\)

Rearranging the formula, we get:

\(t^2\) = (2 * θ) / α

Plugging in the given values, we have:

\(t^2\) = (2 * 2.00) / 0.135

\(t^2\) = 29.63

Taking the square root of both sides, we find:

t ≈ √29.63

t ≈ 5.439

Therefore, the time taken for the merry-go-round to complete 2.00 rotations is approximately 5.439 seconds.

Note: It's important to round the final answer to an appropriate number of significant figures, considering the given data. In this case, we have used four significant figures in the final answer.

However, if we want to adhere to the given significant figures in the acceleration (0.135 rad/\(s^2\)), the answer should be rounded to three significant figures. In that case, the final answer would be approximately 5.44 seconds.

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Answer:

The fireworks give off heat.

Explanation:

The difference in power for two people carrying the boxes up the ramp is 30 W.

Given the following data:

W = 40.0 N is the weight of a box.

The ramp's length is L = 2.00 m.

The platform height is h = 1.0 m.

t = 2.0 s is the time interval for the first person.

t' = 4.0 s is the time interval for the other person.

Power is the rate at which energy is used. The expression for the Power is given in the given question as,

P = W×(L+h)/t

Assume that you are solving for the first person.

P₁ = W(L+h)/t₁.................................................................. (1)

Substitute the following values into equation (1):

P₁ = 40(2+1)/2

P₁ = 20(3) (3)

P₁ = 60 W.

Regarding the second person,

P₂ = W(L+h)/t₂..................................................................... (2)

Fill in the blanks in equation (2) as follows:

P₂ = 40(2+1)/4

P₂ = 10*(3) *3)

P₂ = 30 W

Obtaining the difference in power as

P = P₁ - P₂

P = 60-30

P = 30 W

As a result, we can conclude that the difference in power for two people carrying the boxes up the ramp is 30 W.

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a) The potential difference between the plates after a sheet of Teflon is inserted between them is 20.0V

b) The potential difference between the plates after a sheet of Teflon is inserted between them is  9.52Volts.

The amount of potential energy that exists between two points in a circuit is what we can refer to as the voltage. While the other points have potential, one has a higher potential. The term "voltage" or "potential difference" refers to the charge difference between higher and lower potentials.

Part a) Capacitor plates have a certain voltage when they are fully charged. ΔV = 20.0V

Considering that the battery is currently disconnected, the potential difference between the plates is given as

ΔV = Q/C

As long as the capacitance remains constant and the charge is conserved on the plates, the potential difference will remain unchanged.

b) Teflon sheet is now placed between the capacitor's plates.

Consequently, the capacitor's capacitance will alter by a factor of the dielectric constant.

Therefore, we will

C' = KC

Kteflon = 2.1

Consequently, the new capacitance given here is

C' = 2.1C

ΔV = Q/C

ΔV = 20/2.1

ΔV = 9.52Volts.

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a. The speed Se of the spacecraft when it crashes into the earth is approximately 11.2 km/s.

b. The spacecraft's speed when its distance from the center of the earth is R = αRe is Se / √(α).

a. To find the speed Se of the spacecraft when it crashes into the earth, we can use the law of conservation of energy. At R = infinity, the spacecraft has zero kinetic energy and potential energy, so its total mechanical energy is zero. As it falls towards the earth, the potential energy decreases while the kinetic energy increases. At the moment of impact, all of the potential energy has been converted to kinetic energy.

Using the law of conservation of energy, we have:

\(0 = 1/2 mv^2 - GM_em/r\)

where m is the mass of the spacecraft, v is its speed at impact, G is the universal gravitational constant, and r is the distance from the center of the earth to the spacecraft at impact.

We can rearrange this equation to solve for v:

\(v = \sqrt{(2GM_e/r)}\)

Substituting the values for G, M_e, and r, we get:

\(v = \sqrt{(2 \times 6.6743 \times 10^{-11} m^3/kg s^2 \times 5.97 \times 10^{24} kg / 6.38 \times 10^6 m)}\)

\(v = 11.2 km/s\)

Therefore, the speed Se of the spacecraft when it crashes into the earth is approximately 11.2 km/s.

b. To find the spacecraft's speed when its distance from the center of the earth is R = αRe, we can use conservation of energy again. The spacecraft still has zero kinetic energy and potential energy at R = infinity, so we can use the same equation as before:

\(0 = 1/2 mv^2 - GM_em/r\)

But now r = αRe, so we can solve for v in terms of Se and α:

v = √(2GM_e/αRe)

Substituting the value of \(GM_e\) from before and simplifying, we get:

v = Se / √(α)

Therefore, the spacecraft's speed when its distance from the center of the earth is R = αRe is Se / √(α).

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Answer:

Explanation:

Let the angle required be θ with direction opposite to current .

cos θ = 3 / 5

θ = 53° .

Resultant velocity towards the opposite side

= √ ( 5 ² - 3²)

= 4 km /h

distance covered = 1.2 km

time taken = 1.2 / 4 = .3 h

= 18 min .

Newton's first and second laws of motion both do, but I think the one you're looking for is: The First Law of Motion.  That description is a little more direct.

It says that if an object is not acted on by a net external force, then it continues in "constant, uniform motion".

Answer:

2.15 m/s²

Explanation:

We'll begin by calculating the force of attraction between two charges. This can be obtained as follow:

Charge of 1st object (q₁) = +11.5 μC = +11.5×10¯⁶ C

Charge of 2nd object (q₂) = –7.55 μC = –7.55×10¯⁶ C

Electrical constant (K) = 9×10⁹ Nm²/C²

Distance apart (r) = 0.925 m

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 11.5×10¯⁶ × 7.55×10¯⁶/ 0.925²

F = 0.781425 / 0.855625

F = 0.91 N

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass of object (m) = 0.423 Kg

Force (F) = 0.91 N

Acceleration (a) =?

F = ma

0.91 = 0.423 × a

Divide both side by 0.423

a = 0.91 / 0.423

a = 2.15 m/s²

Thus, the magnitude of the object's acceleration is 2.15 m/s²

Answer:

2.15

Explanation:

Acellus

Bernoulli's principle can be used to explain the motion of cars on the highway as the cars moving fast create areas of low pressure around them, causing them to be pushed together by the higher pressure.

Bernoulli's principle can be formulated by Daniel Bernoulli. The principle states that as the speed of a moving fluid increases such as liquid or gas, the pressure within the fluid decreases through it. Although, Bernoulli deduced this law, it was Leonhard Euler who derived the Bernoulli's equation in its usual form in the year 1752.

When the two cars passes one another in opposite directions on the highway, then they tend to be forced together. Use the Bernoulli's principle to explain why this happens with the cars. The cars moving fast on the highway create areas of low pressure around them, causing them to be pushed together by the higher pressure.

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Answer:

Incomplete question, but let analyse generally a torque acting on a rotating body

Explanation

Toque is given as

τ = r × F

Where

τ is the torque (Nm)

r is the position vector (m)

F is the force (N).

Let assume we have the following data

F= x•i + y•j + z•k

Also let assume the position vector is

r'= a•i + b•j + c•k

a. The first question, torque at the origin

The origin will have a position vector of r1(0,0,0)

Then, the displacement is

r=AB=AO+OB=-OA+OB

r = r'-r1 = (a, b, c)-(0,0,0)

r=(a, b, c)

Then, the torque is given as

τ = r×F

τ = (a, b, c) × (x, y, z)

Note that

i×i=j×j=k×k=0

i×j=k, j×i=-k

j×k=i, k×j=-i

k×i=j, i×k=-j

τ = (a•i + b•j + c•k) × (x•i + y•j + z•k)

τ = a•i×(x•i + y•j + z•k) + b•j×(x•i + y•j + z•k) + c•k×(x•i + y•j + z•k)

τ = (a•i × x•i)+ (a•i × y•j) + (a•i × z•k) + (b•j × x•i) + (b•j × y•j) + (b•j × z•k) + (c•k × x•i) + (c•k × y•j) + (c•k × z•k)

Now, using the above law.

τ = 0 + ay•k - az•j - bx•k + 0 + bz•i + cx•j - cy•i + 0

τ= ay•k - az•j - bx•k + bz•i + cx•j -cy•i

Then, rearranging

τ = (bz - cy)•i + (cx - az)•j + (ay - bx)•k

This is general formula for the torque at the origin, so you can apply it to the given values in your question because the question given is not complete.

b. Now let, analyse at the position vector (2.61 m, 0, -8.03 m)

Using the same analysis

position vector of r1(2.61, 0, -8.03)m

Then, the displacement is

r=AB=AO+OB=-OA+OB

r = r'-r1 = (a, b, c)-(2.61,0,-8.03)

r=(a-2.61, b, c+8.03)

Then, the torque is given as

τ = r×F

τ = (a-2.61, b, c+8.03) × (x, y, z)

Note that

i×i=j×j=k×k=0

i×j=k, j×i=-k

j×k=i, k×j=-i

k×i=j, i×k=-j

τ =([a-2.61]•i, b•j, [c+8.03]•k) × (x•i + y•j + z•k)

τ = (a-2.61)•i×(x•i + y•j + z•k) + b•j×(x•i + y•j + z•k) + (c+8.03)•k×(x•i + y•j + z•k)

τ = ((a-2.61)•i × x•i)+ ((a-2.61)•i × y•j) + ((a-2.61)•i × z•k) + (b•j × x•i) + (b•j × y•j) + (b•j × z•k) + ((c+8.03)•k × x•i) + ((c+8.03)•k × y•j) + ((c+8.03)•k × z•k)

Now, using the above law.

τ = 0 + (a-2.61)y•k - (a-2.61)z•j - bx•k + 0 + bz•i + (c+8.03)x•j - (c+8.03)y•i + 0

τ= (a-2.61)y•k - (a-2.61)z•j - bx•k + bz•i + (c+8.03)x•j -(c+8.03)y•i

Then, rearranging

τ = (bz - cy - 8.03y)•i + (cx + 8.03x - az - 2.61z)•j + (ay - 2.61y - bx)•k

This is general formula for the torque at the point (2.61 m, 0, -8.03 m), so you can apply it to the given values in your question because the question given is not complete.

(a) Torque acting on the pebble about the origin is (-1.5i - 4j - 1k) Nm.

(b) Torque acting on the pebble about the given coordinate is 20j Nm.

Torque is defined as the rotational analogue of force. It is the cross product of force and the perpendicular distance.

Here,

The position coordinates of the pebble, r = (0.5 m)j - (2 m)k

Force acting on the pebble, F =  (2 N)i - (3 N)k

(a) About the origin

 Torque acting on the pebble, τ = r x F

   τ = (0.5j - 2k) x (2i - 3k)

   τ = (0.5j x 2i) + (0.5j x -3k) + (-2k x 2i) + (-2k x -3k)

   τ = -1k - 1.5i - 4j + 0

   τ = (-1.5i - 4j - 1k) Nm

(b) At a point with coordinates (8.62m, 0, -2.93m)

    Torque acting on the pebble,

   τ = r x F

   τ = (8.62i - 2.93k) x (2i - 3k)

   τ = (8.62i x 2i) + (8.62i x -3k) + (-2.93k x 2i) + (-2.93k x -3k)

   τ = 25.86j - 5.86j

   τ = 20j Nm

Hence,

(a) Torque acting on the pebble about the origin is (-1.5i - 4j - 1k) Nm.

(b) Torque acting on the pebble about the given coordinate is 20j Nm.

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Your question was incomplete, but most probably your question was:

Force (2 N)i - (3 N)k acts on a pebble with position vector (0.5 m)j - (2 m)k, relative to the origin. What is the resulting torque acting on the pebble about (a) the origin and (b) a point with coordinates (8.62 m, 0, -2.93 m)

The speed of the electron when it is 10.0 cm from the 3.00-nC charge is approximately \(2.19x10^6 m/s.\)

To answer this question, we need to use Coulomb's law and the principle of conservation of energy.

Coulomb's law states that the force between two point charges is given by:

\(F = kq1q2/r^2\)

where F is the force, q1 and q2 are the charges, r is the distance between the charges, and k is the Coulomb constant.

In this case, the force between the 3.00-nC charge and the electron is:

\(F =\)\(kq1q2/r^2 = (9x10^9 N m^2/C^2)(3.00x10^-9 C)(1.60x10^-19 C)/(0.100 m)^2 =\) \(2.88x10^-10 N\)

The force on the electron is directed towards the 3.00-nC charge, so it will accelerate towards it. The work done by the electric force is converted into kinetic energy, so we can use conservation of energy to relate the speed of the electron to the distance from the charge.

At a distance of 10.0 cm, the potential energy of the electron is:

\(U = kq1q2/r = (9x10^9 N m^2/C^2)(3.00x10^-9 C)(1.60x10^-19 C)/(0.100 m) = 4.32x10^-18 J\)

At a distance r from the charge, the kinetic energy of the electron is:

\(K = (1/2)mv^2\)

where m is the mass of the electron and v is its speed. At a distance of infinity, the electron is at rest, so its kinetic energy is zero. Therefore, the total energy of the electron is conserved:

U = K

or

\((1/2)mv^2 = kq1q2/r\)

Solving for v, we get:

\(v = sqrt(2kq1*q2/mr)\)

Substituting the values we obtained earlier, we get:

\(v = sqrt[(2*9x10^9 N m^2/C^2 * 3.00x10^-9 C * 1.60x10^-19 C) / (9.11x10^-31 kg * 0.100 m)]v = 2.19x10^6 m/s\)

Therefore, the speed of the electron when it is 10.0 cm from the 3.00-nC charge is approximately 2.19x10^6 m/s.

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Answer:

a = - 11.53[m/s^2]

Explanation:

The airplane slows down as its speed decreases from the initial value of 610 [m/s] to zero.

To calculate the acceleration value we use the following kinematics equation:

\(v_{f} = v_{i}+(a*t)\\\)

where:

Vf = final velocity = 0

Vi = initial velocity = 610 [m/s]

a = acceleration [m/s2]

t = time = 53 [s]

Now replacing:

0 = 610 + (a*53)

-610 = 53*a

a = - 11.53[m/s^2]

The negative sign means that the aircraft is losing speed, i.e. slowing down

In the illustration with two charged particles of the same magnitude, one positive and the other negative, the equipotential line of the two charged particles would be B) The vertical line that is located halfway between the charges.

An equipotential line is a line in space where all points on the line have the same electric potential, and in this case, the line halfway between the charges fulfills this condition.

Option B) The vertical line that is located halfway between the charges could be an equipotential line of the two charged particles. An equipotential line is a line in space where all points on the line have the same electric potential. Since the vertical line is equidistant from both charged particles, the electric potential at each point on the line would be the same.

Option A) the circle around one charged particle and Option C) the horizontal line that runs directly through the center of each cannot be equipotential lines for both charges as the distance from each point on the line to the charged particles is not the same. Option D) None of the three lines could be an equipotential line for both charges is not correct as the vertical line satisfies the condition for being an equipotential line.

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From the description of the problem, the total velocity is obtained as 3.7 yards/ second.

The term velocity is the ratio  of the distance to the time taken. Let us note that in this case we are being asked for the average velocity of the ball and that is what we shall set to do after a consideration of the problem.

We are told that the ball is snapped to him on the 20 yard line and he jumps back 7 yards to find a receiver before running to the 50 yard line before being tackled, all during a 10 second play.

We can see that the total distance that has been covered by the ball is 7 yards + 30 yards = 37 yards and the time that has been taken is 10 seconds.

Thus velocity = distance / time = 37 yards /10 seconds

= 3.7 yards/ second

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A 5.26 kg object falls freely (ignore air resistance), after being dropped from rest. Determine the initial kinetic energy (in J), the final kinetic energy (in J), and the change in kinetic energy (in J) for the following. KE_f = 52.8 J.KE

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. The body exerts the same amount of effort when slowing down from its current pace to a condition of rest. Formally, a kinetic energy is any term that includes a derivative with respect to time in the Lagrangian of a system.

In classical physics, an object of mass m moving at a speed of v has kinetic energy.

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State-jail felonies, like possession of 4 ounces to 1 pound of marijuana, can result in penalties such as incarceration in a state jail facility, fines, and potential probation or community service.

The severity of the punishment typically depends on the specific circumstances and any prior convictions the offender may have.

In many jurisdictions, possession of a certain quantity of marijuana within the range you mentioned can be classified as a state-jail felony. This means that it is considered a more serious offense than a misdemeanor but less severe than a felony.

State-jail felonies typically carry potential penalties that include incarceration in a state jail facility, fines, and the possibility of probation or community service.

In terms of incarceration, individuals convicted of a state-jail felony may be sentenced to serve time in a state jail facility. State jails are correctional institutions that are designed to house individuals convicted of state-jail felonies.

The length of incarceration can vary depending on the jurisdiction and the specific circumstances of the case. It is important to note that state jails differ from prisons, which typically house individuals convicted of more serious offenses.

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Answer:

M.E = 0.241 Joules.

Explanation:

Given the following data;

Spring constant, k = 33.5 N/m

Extension, e = 0.12m

Mass = 1.20kg

To find the mechanical energy (M.E);

Mathematically, mechanical energy is given by this formula;

M.E = ½ke²

Where;

Substituting into the equation, we have;

M.E = ½*33.5*0.12²

M.E = ½*33.5*0.0144

M.E = 0.4824/2

M.E = 0.241 Joules.

B. Lead is a conductor....

Density of liquids and gases.

Data:

d = ?

m = 40.0 g

v = 80.0 cm³

The formula for the density of liquids and gases

                d = m/v

Since it asks us to calculate the density, it is not necessary to clear the formula. We just substitute data and solve.

                d = 40.0 g/80.0 cm³

                d = 0.5 g/cm³

The speed of the rock when it reaches the ground is 19.32 m/s.

The speed of the rock when it reaches the ground is calculated by applying the law of conservation of energy.

Kinetic energy of the rock at the bottom = potential energy of the rock at the top

K.E = P.E

¹/₂mv² = mgh

v² = 2gh

v = √ ( 2gh )

where;

The speed of the rock when it reaches the ground is calculated as;

v = √ ( 2gh )

v = √ ( 2 x 9.8 x 19.04 )

v = 19.32 m/s

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Answer:

a.25J

Explanation:

m=0.5kg

g=10

h=5

potential energy=m×g×h

=0.5×10×5

=25J

Answer:Speed is the time rate at which an object is moving along a path

Velocity is the rate and direction of an object's movement.

Explanation:

Hope it helps

While quantum mechanics predicts the existence of nodes and regions of very low probability, the wave function never truly reaches zero at any point along the spring.

The wave function represents the probability distribution of locating the particle (mass) at various places along the spring in the setting of a quantum harmonic oscillator.

The points where there is no chance of identifying the particle are known as the wave function's nodes.

For the quantum harmonic oscillator, the wave function is given by:

\(\[ \psi(x) = A \cdot H_n \left(\frac{x}{\sqrt{2} l}\right) e^{-\frac{x^2}{4l^2}} \]\)

The nodes of the wave function occur where \(\( \psi(x) = 0 \)\). Since the Hermite polynomials do not become zero, the nodes are determined by the exponential term:

\(\[ e^{-\frac{x^2}{4l^2}} = 0 \]\)

This demonstrates that the wave function never actually approaches zero along the spring in quantum mechanics.

The nodes relate to locations where there is a very little chance of detecting the particle. Away from the centre, the exponential term rapidly decays, creating areas with extremely low probability. These regions aren't precisely zero, though.

Consequently, the wave function never fully reaches zero at any point along the spring, despite the fact that quantum physics predicts the occurrence of nodes and areas with extremely low probability.

Thus, fundamental feature of quantum systems is this.

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Your question seems incomplete, the probable complete question is:

Consider a harmonic oscillator with mass m=0.100 kg and k = 50 N/m. You may have worked similar problems before, as a mass on a spring using classical mechanics, but this time you will use the solution to the Schrödinger equation for the harmonic oscillator. Keep in mind that this system would be enormous by quantum standards, and in practice you would never expect to use quantum mechanics to describe a mass on a spring. Nonetheless, it is interesting to see what quantum mechanics predicts here.  

Nodes are the points where the wave function (and hence the probability of finding the particle) is zero. What is the separation between nodes of the wave function for the mass on a spring described in this problem? Assume that all of the nodes occur in the classically allowed region. Since the diameter of an atomic nucleus is on the order of 10-15 m, the separation that you've calculated is far too small to be measureable in any experiment. Just as for a classical harmonic oscillator, the position of this mass would appear to be able to take all values.

At constant velocity, the crate would be in equilibrium, so that Newton's second law tells us

p + (-f ) = 0

where p and f denote the magnitudes of the added pushing force and friction force, respectively.

The friction force is proportional to the normal force by a factor of µ, the coefficient of friction. There's no vertical movement going on, so Newton's second law says

n + (-w) = 0

where n and w are the magnitude of the normal force and the weight of the crate, respectively.

Compute n :

n = w = (25 kg) (9.80 m/s²) = 245 N

Use this and µ = 0.45 to compute f :

f = µ n = 0.45 (245 N) = 110.25 N

Solve for p :

p + (-110.25 N) = 0

p = 110.25 N ≈ 110 N

The force i.e required to move the crate at a constant velocity across the floor is 110 N.

At the same velocity, the crate should be in equilibrium, so as per Newton's second law, the following formula should be used:

p + (-f ) = 0

p means the magnitudes of the added pushing force

f means the friction force

n = w = (25 kg) (9.80 m/s²) = 245 N

Now

force should be

= µ n

= 0.45 (245 N)

= 110.25 N

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The initial acceleration of the balloon in the horizontal direction is 3.85 m/s^2. The initial acceleration of the balloon can be calculated using the formula for drag force, Fd = 0.5*Cp*rho*A*V^2, where rho is the density of air, A is the cross-sectional area of the balloon, V is the velocity of the wind, and Cp is the drag coefficient.

The weight of the balloon, W = mg, where m is the mass of the balloon and g is the acceleration due to gravity. Since the balloon is standing still, the weight is balanced by the buoyant force, Fb = rhoVg, where V is the volume of the balloon.

Once the balloon is subjected to wind, the net force in the horizontal direction is Fnet = Fd. The initial acceleration of the balloon is then given by a = Fnet/m. Substituting the given values, we get:

A = pi*(7/2)^2 = 38.5 m^2
Fd = 0.5*0.2*1.20*38.5*(40/3.6)^2 = 1233 N
W = 320*9.81 = 3139 N
Fnet = Fd = 1233 N
a = Fnet/m = 1233/320 = 3.85 m/s^2

Therefore, the initial acceleration of the balloon in the horizontal direction is 3.85 m/s^2.

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Answer:

H = 0.06 kWh

Explanation:

Given that,

Power of an electric lamp, P = 60 W

Voltage, V = 240 V

It is operated for 1 hour

We need to find the heat generated by the lamp. Heat generated is given by :

\(H=P\times t\\\\H=60\ W\times 1\ h\\\\H=60\ Wh\\\\H=0.06\ kWh\)

So, 0.06 kWh of the heat is generated by the lamp.

Evidence doesn't support theories.

A theory is an explanation that's offered to support evidence.

The big bang theory is a proposed mechanism that can explain the observed abundance of the natural elements, the observed distribution of the cosmic microwave background, and the observed expansion of the universe (which itself is a theory that can explain the observed red shift of the light from distant galaxies).