A metal ball at 30°C is dropped from a height of 6.2 km. The ball is heated due to the air resistance and it completely melts just before reaching the ground. The molten substance falls slowly on the ground. If the specific heat of lead = 126 Jkg−1^o C^{-1} and melting point of lead = 130^oC and suppose that any mechanical energy lost is used to heat the ball, then the latent heat of fusion of lead is:__________.

Answer:

Planet Z will have the following properties;

Active Volcanoes

Active Tectonics

An Atmosphere produced by outgassing

Explanation:

The little terrestrial worlds have heat shorter than the much bigger terrestrial worlds, so the bigger worlds tend to have active volcanism and tectonics. These active volcanism and tectonics are likely to erase ancient craters. The active volcanism and tectonics would create an atmosphere by producing gases.

It is know that the Terrestrial worlds that are not far from the star have higher surface temperature.

Fast rate of rotation can cause winds and strong storms but here it is slower compared to earth. Also, a tilt of axis causes seasons.

The properties the star have are active volcanoes, active tectonics and an atmosphere produced by outgassing.

Final answer:

Planet Z's characteristics such as geological activity, seasons, and atmosphere can be inferred from its size, orbital distance, and rotation rate. A planetary mass similar to Earth's suggests active geology and an atmosphere from outgassing, while a proper distance from the sun allows for liquid water and polar ice caps. The planet's rotation influences the presence of seasons and potential for strong winds and violent storms.

Explanation:

Based on Planet Z's size, orbital distance, and rotation rate, it is possible to infer several characteristics that this planet might have. The level of geological activity on a planet is often proportional to its mass, suggesting that planets similar in size to Earth and Venus are more likely to exhibit geological activity such as active volcanoes or tectonics. Similarly, a planet's distance from its sun can influence the presence of liquid water, with those at optimum distances having the potential for erosion due to liquid water and possibly polar ice caps. A slower rotation might lead to more extreme temperature differences between day and night, which could impact atmospheric conditions and lead to strong winds and violent storms due to the larger temperature gradient.

Planetary rotation also contributes significantly to the development of seasons; hence, how Planet Z rotates will affect whether it experiences seasons. A planet that has active geology and volcanism will likely have an atmosphere that is at least partially produced by outgassing, as seen on Earth, and could also support active tectonics. Lastly, if the planet is not geologically active, it may have a surface crowded with impact craters, similar to the Moon and Mercury, which have less geological activity to renew their surfaces.

Answer:

Small sports car.

Explanation:

Lets take

mass of the small car = m

mass of the truck = M

As we know that when car collide with the massive truck then due to change in the moment of the car both car as well as truck will feel force.We also know that from Third law of Newton's ,it states that every action have it reaction with same magnitude but in the opposite direction.

Therefore

F = m a

a=Acceleration of the car

[tex]a=\dfrac{F}{m}[/tex]

F= M a'

a'=Acceleration of the massive truck

[tex]a'=\dfrac{F}{M}[/tex]

Here given that M > m that is why a > a'

Therefore car will experiences more acceleration.

Answer

The answer for the first one I think is false.

The second one would be true i think. I hope i got it right and have a wonderful day

Answer:

True

False

Explanation:

From 0 to E, the train moves a distance of 55 m.

From F to J, the train moves a distance of 59 m.

The total distance is 55 + 59 = 114 m.

The displacement is the difference between the final position and initial position.  Here, the distance between J and 0 is -4 m.

The mass of the mountain is 3.002 × 10^16 kg and its fraction of Earth's mass is approximately 4.92 × 10^-8. These results are unreasonable due to the large mass of the mountain compared to Earth and the assumptions made in the question.

To calculate the mass of the mountain, we can use Newton's Law of Universal Gravitation. The gravitational force exerted by the mountain is equal to 2.00% of the person's weight. Since weight is equal to mass multiplied by acceleration due to gravity, we can set up the equation:

0.02mg = GMm / d^2

where G is the gravitational constant, M is the mass of the mountain, m is the mass of the person, and d is the distance between them. Since the person's weight is equal to mg, we can rewrite the equation as:

0.02mg = (GMm / d^2)

Dividing both sides by mg gives us:

0.02 = (GM / d^2)

Now we can solve for the mass of the mountain (M):

M = (0.02d^2 / G)

Substituting the given values (d = 10.0 km, G = 6.673 × 10^-11 Nm²/kg²),

M = (0.02 × 10000^2) / (6.673 × 10^-11)

M = 3.002 × 10^16 kg

The mass of the mountain is 3.002 × 10^16 kg.

Comparing the mass of the mountain with that of Earth, we can use the equation:

Mountain's Mass / Earth's Mass = 3.002 × 10^16 / (6 × 10^24)

Mountain's Mass / Earth's Mass = 4.92 × 10^-8

The mass of the mountain is approximately 4.92 × 10^-8 of Earth's mass.

These results are unreasonable because the mass of the mountain and its fraction of Earth's mass are too large. It is unlikely for a mountain to have such a massive mass compared to the entire Earth.

The premises that are unreasonable or inconsistent in this scenario are the assumption that the gravitational force exerted by the mountain is 2.00% of the person's weight and the assumption that the distance between the mountain and the person is 10.0 km.

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The calculated mass of the mountain is 2.937 × 10¹⁷ kg, which is 4.91 × 10⁻⁸ of Earth's mass. This result is unreasonable because the mass and fraction are too large. The assumption about the gravitational force exerted by the mountain is inconsistent.

Let's start by calculating the mass of the mountain, given the gravitational force it exerts.

(a)

Given data:

The weight of the person is W = mg, where m is the mass of the person and g is the gravitational acceleration (9.8 m/s²).

The gravitational force F is given by Newton's law of gravitation: F = G * (m_p * M) / r², where:

Given that F = 0.02 * mg, we can equate:

0.02 * mg = G * (m * M) / (10,000 m)²

Cancel out m and solve for M:

M = (0.02 * g * (10,000 m)²) / G

M = 2.937 × 10¹⁷ kg

(b)

Mass of Earth ME = 5.972 × 10²⁴ kg

Fraction of Earth's mass = M / ME

Fraction = 2.937 × 10¹⁷ kg / 5.972 × 10²⁴ kg = 4.91 × 10⁻⁸

(c)

The mass of the mountain and its fraction of the Earth's mass are excessively large. Such a massive mountain would be geologically and physically improbable.

(d)

The gravitational force assumed to be exerted by the mountain is too large. In reality, a mountain would exert a much smaller percentage of gravitational force on a person.

Answer:

With an  Environmental  Engineering  and a broadcasting minor

You can work as an On Air  personality that host  programs that provide your audience with  documentaries about the environments and project carried out by Environmental Engineer

and also you can work as a journalist that explore the world making research that will preserve the environment  and leveraging the media as a broadcaster to provide this research findings as a video for you audience  

Explanation:

In order to get a better understanding let define some terms

Environmental Engineer :

Environmental engineers resolve and help prevent environmental problems. They work in many areas, including air pollution control, industrial hygiene, toxic materials control, and land management. The duties of an environmental engineer range from planning and designing an effective waste treatment plant to studying the effects of acid rain on a particular area. An environmental engineer is sometimes required to work outdoors, though most of her work is done in a laboratory or office setting. Career opportunities for environmental engineers exist in consulting, research, corporate, and government positions.

Broadcasting:

Broadcasting is the distribution of audio or video content to a dispersed audience via any electronic mass communications medium, but typically one using the electromagnetic spectrum (radio waves), in a one-to-many model.

Answer:2.13 m/s

Explanation:

Given

mass of ball [tex]m=24 gm[/tex]

Length of cord [tex]L=0.463 m[/tex]

acceleration due to gravity [tex]g=9.8 m/s^2[/tex]

minimum velocity after which cord will slack

[tex]\frac{mv^2}{r}=mg[/tex]

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

[tex]v=\sqrt{0.463\times 9.8}[/tex]

[tex]v=2.13 m/s[/tex]

Answer:

675 Joules

Explanation:

Considering that work can be calculated with the following formula:

W = Fx D

Where:

W = work

F = force

D = distance

We can directly use this formula in case the applied force remains constant

In case the force does not remain constant, we can calculate the work as a change, this is :

ΔW = ΔFxΔD

For this scenario, we have:

W₁ = F₁xD₁

F₁ = 210 N, D = 0 m

W₁ = 210Nx0m = 0 Joules

W₂ = F₂xD₂

F₂ = 45 N , D₂ = 15 m

W₂ = 45Nx15m = 675 Joules

Finally: ΔW = Total work performed when moving the car 15 m

ΔW = W₂ - W₁ = 675 Joules - 0 Joules = 675 Joules

Final answer:

The amount of work done on the car while pushing it is 1575 Joules.

Explanation:

Work is defined as the product of force and displacement. In this case, the force required to keep the car moving decreases as the car gets going. Work can be calculated using the formula:

Work = Force × Distance

Given that the force decreased from 210 N to 45 N over a distance of 15 m, we can calculate the work done as follows:

Work = (210 N + 45 N) / 2 × 15 m = 1575 J

Therefore, the amount of work done on the car while pushing it is 1575 Joules.

Answer:

F= 15.6 N,   a= 33.2 m/s^2

Explanation:

mass= m = 0.47 kg

spring constant= k = 130N/m

spring compression = x = 0.12 m

a).

force on the mass= F = k*x

F = 130 * 0.12 N

F= 15.6 N

b).

Acceleration of mass= a=?

F= ma

a=F / m

a= 15.6/ 0.47 m/s^2

a= 33.2 m/s^2

Final answer:

The force on the mass when the spring is released is 15.6 N, and the acceleration of the mass at that instant is approximately 33.19 m/s^2, calculated using Hooke's Law and Newton's second law, respectively.

Explanation:

Understanding Spring Force and Acceleration

To find the force on the mass at the instant the spring is released (in part A), we use Hooke's Law, which states that the force exerted by a spring (F) is equal to the negative spring constant (k) times the displacement from equilibrium (x), so F = -kx. Here, k = 130 N/m and x = 0.12 m, so the force is F = 130 N/m * 0.12 m = 15.6 N. The negative sign indicates that the force is in the opposite direction of the displacement.

To find the acceleration of the mass at the instant the spring is released (in part B), we apply Newton's second law, which relates force (F), mass (m), and acceleration (a) as F = ma. Rearranging for acceleration, we get a = F/m. Substituting the values, we have a = 15.6 N / 0.47 kg = approximately 33.19 m/s2.

Answer:

Because of heavy mass

Explanation:

When force acts on a body it tends to accelerate the body. The acceleration produced in the body depends on two things:

1). Magnitude of force

2). Mass of the body

F= ma

⇒ a = F/m  

As the force exerted on earth and another object are the equal in magnitude but opposite in direction. This forces will accelerate the object toward the earth but can't accelerate the earth as earth has very high mass.

a = F/m

This force tends to accelerate the earth but but due to earth's inertia the earth does not accelerate.

Answer:

27.5 days

0.92 month

Explanation:

[tex]r[/tex] = radius of the orbit of moon around the earth = [tex]3.85\times10^{8} m[/tex]

[tex]M[/tex] = Mass of earth = [tex]5.98\times10^{24} m[/tex]

[tex]T[/tex] = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

[tex]T^{2} = \frac{4\pi ^{2} r^{3}  }{GM}[/tex]

inserting the values, we get

[tex]T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3}  }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec[/tex]

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

[tex]T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days[/tex]

1 month = 30 days

[tex]T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month[/tex]

Final answer:

The period for the moon's motion around the earth is approximately 0.59 days, which is much shorter than the length of a month.

Explanation:

To find the period for the moon's motion around the earth, we can use Kepler's third law. According to Kepler's third law, the square of the period of a planet's orbit is directly proportional to the cube of its average distance from the center of the orbit.

We are given that the moon orbits the earth at a distance of 3.85 x 10^8 m. We can use this information to calculate the period as follows:

Hence, the period for the moon's motion around the earth is approximately 0.59 days. This is much shorter than the length of a month, which is about 30 days. Therefore, the moon completes multiple orbits around the earth in one month.

Answer:

a)

6.33 x 10⁸ N/C

Direction : Towards negative charge.

b)

1.11125 x 10²⁰ m/s²

Direction : Towards positive charge.

Explanation:

a)

[tex]Q_{1}[/tex] = magnitude of negative charge = 25 x 10⁻⁶ C

[tex]Q_{2}[/tex] = magnitude of positive charge = 50 x 10⁻⁶ C

[tex]r_{1}[/tex] = distance of negative charge from point P = 0.02 m

[tex]r_{2}[/tex] = distance of positive charge from point P = 0.08 m

Magnitude of electric field at P due to negative charge is given as

[tex]E_{1} = \frac{kQ_{1}}{r_{1}^{2} } = \frac{(9\times10^{9})(25\times10^{-6})}{0.02^{2} } = 5.625\times10^{8} N/C[/tex]

Magnitude of electric field at P due to positive charge is given as

[tex]E_{2} = \frac{kQ_{2}}{r_{2}^{2} } = \frac{(9\times10^{9})(50\times10^{-6})}{0.08^{2} } = 0.703125\times10^{8} N/C[/tex]

Net electric field at P is given as

[tex]E = E_{1} + E_{2}\\E = 5.625\times10^{8} + 0.703125\times10^{8} \\E = 6.33\times10^{8} N/C[/tex]

Direction:

Towards the negative charge.

b)

[tex]m[/tex] = mass of the electron placed at P = 9.31 x 10⁻³¹ C

[tex]Q_{1}[/tex] = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

Acceleration of the electron due to the electric field at P is given as

[tex]a = \frac{qE}{m}\\ a = \frac{(1.6\times10^{-19})(6.33\times10^{8})}{(9.11\times10^{-31})}\\a = 1.11125\times10^{20} ms^{-2}[/tex]

Direction: Towards the positive charge Since a negative charge experience electric force in opposite direction of the electric field.

Answer:

Explanation:

qA = - 25 x 10^-6 C

qB = 50 x 10^-6 C

AP = 2 cm

BP = 8 cm

(a)

Electric field at P due to the charge at A

[tex]E_{A}=\frac{kq_{A}}{AP^{2}}[/tex]

[tex]E_{A}=\frac{9\times 10^{9}\times 25\times 10^{-6}}{0.02^{2}}[/tex]

EA = 5.625 x 10^8 N/C

Electric field at P due to the charge at B

[tex]E_{B}=\frac{kq_{B}}{BP^{2}}[/tex]

[tex]E_{A}=\frac{9\times 10^{9}\times 50\times 10^{-6}}{0.08^{2}}[/tex]

EB = 0.70 x 10^8 N/C

The resultant electric field at P due to both the charges is

E = EA+ EB = (5.625 + 0.7) x 10^8

E = 6.325 x 10^8 N/C towards left

(b)  mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of electron

Force on electron, F = charge of electron x electric field

F = q x E

[tex]a = \frac{qE}{m}[/tex]

[tex]a = \frac{1.6\times 10^{-19}\times 6.325\times 10^{8}}{9.1\times 10^{-31}}[/tex]

a = 1.11 x 10^20 m/s^2

Answer:

Δ v =  125 m/s

Explanation:

given,

mass of space craft = 435 Kg

thrust = 0.09 N

time = 1 week

       = 7 x 24  x 60 x 60 s

change in speed of craft = ?

Assuming no external force is exerted on the space craft

now,

[tex]T= m_s a[/tex]

[tex]a=\dfrac{T}{m_s}[/tex]

[tex]a =\dfrac{0.09}{435}[/tex]

a = 2.068 x 10⁻⁴ m/s²

using equation of motion

Δ v = a t

Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60

Δ v =  125 m/s

Answer:

its a A. Transverse wave

Answer:There are three types of radiation

Alpha, Beta and Gamma radiation

Explanation: In an electric field produced by two parallel charged plates alpha particle would be deflected toward a - plate following a parabolic path, beta rays toward a +plate following a parabolic path and gamma radiation either - or + source.

Answer:

In computing the volume of a cube,

Maximum possible error = +/-1350cm³

Relative error = 0.05

Percentage error = 5%

In computing the surface area of a cube,

Maximum possible error = +/-180cm²

Relative error = 0.0333

Percentage error = 3.33%

Explanation:

A cube is a three dimensional solid object with six (6) faces, twelve (12) edges and eight(8) vertices.

The volume of a cube = x³

Where x= length of the edge of a cube

X = 30cm +/- 0.5cm

Differentiate V with respect to x (V = Volume of a cube)

dV/dx = 3 x²

dV = 3 x² . dx

dV= 3 × 30² × (+/-0.5)

= 2700(+/-0.5)

= +/-1350cm³

Maximum possible error =

+/- 1350cm³

Relative error = Maximum error /surface area

= ΔV/V

Recall that V = x³

V= (30)³

A = 27000cm³

Substitute the values for and V into the formula for Relative error

Relative error = 1350 / 270000

Relative error = 0.05

% error = Relative error × 100

= 0.05× 100

= 5%

Surface Area of a cube = 6x²

A = 6x²

Differentiate A with respect to x

dA/dx= 12x

dA = 12x . dx

dA= 12 × 30 (0.5)

= +/- 180cm²

Maximum possible error =

+/- 180cm²

Relative error = Maximum error / total area

= dA/dx

Recall that A = 6x²

A = 6(30)²

A = 5400cm²

Substitute the values for and A into the formula for Relative error

Relative error = 180/ 5400

Relative error = 0.0333(4 decimal place)

% error = Relative error × 100

= 0.0333 × 100

= 3.33%

Final answer:

Use of differentials to estimate maximum and relative errors in volume and surface area calculations for a cube.

Explanation:

Differentials for Cube:

Surface Area:

The force exerted on the bumper is 1397 N

Explanation:

We can solve this problem by using the equilibrium of the torques: in fact, the torque exerted on one side of the jack must be equal to the torque exerted on the other side of the jack.

Therefore, we can write:

[tex]F_h d_h = F_b d_b[/tex]

where

[tex]F_h = 163 N[/tex] is the force applied to the end of the jack handle

[tex]d_h = 15 cm[/tex] is the distance between the force applied on the handle and the pivot

[tex]F_b[/tex] is the force exerted by the jack on the car bumper

[tex]d_b = 1.75 cm[/tex] is the distance between this force and the car bumper

And solving for [tex]F_b[/tex], we find:

[tex]F_b = \frac{F_h d_h}{d_b}=\frac{(163)(15)}{1.75}=1397 N[/tex]

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Answer:

E=147898.01J

Explanation:

A 5.0-kg cannonball is fired over level ground with a velocity of 2.00 ⨯ 102 m/s at an angle of 25° above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip, find the change in the thermal energy of the cannonball and air

firstly , we look for the time of flight it takes to make the projectile path

T=2Usin∅/g

take g= 9.81m/s

T=2*200sin25/(9.81)

T=17.23Secs

energy is force *distance

E=f*d

f=m*g

f=5*9.81

f=49.05N

s=distance

s=(v+u)T/2

s=(150+200)17.23/2

s=3015.25m

49.05N*3015.25m

E=147898.01J

Answer:

[tex]0.54^{\circ}[/tex]

3.99322032 mm

Explanation:

n = Lines per mm = 125

Seperation between slits is given by

[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{125}\\\Rightarrow d=0.008\ mm[/tex]

[tex]\lambda_1[/tex] = 498 nm

[tex]\lambda_2[/tex] = 569 nm

We have the expression

[tex]dsin\theta_1=m\lambda_1[/tex]

For first maximum m = 1

[tex]\theta_1=sin^{-1}\dfrac{m\lambda_1}{d}\\\Rightarrow \theta_1=sin^{-1}\dfrac{1\times 498\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_1=3.57^{\circ}[/tex]

[tex]\theta_2=sin^{-1}\dfrac{m\lambda_2}{d}\\\Rightarrow \theta_2=sin^{-1}\dfrac{1\times 569\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_2=4.08^{\circ}[/tex]

Angular separation is given by

[tex]\Delta \theta=\theta_2-\theta_1\\\Rightarrow \Delta \theta=4.08-3.57\\\Rightarrow \Delta \theta=0.54^{\circ}[/tex]

Angular separation is [tex]0.54^{\circ}[/tex]

Now

[tex]\lambda_1[/tex] = 589 nm

[tex]\lambda_2[/tex] = 589.59 nm

[tex]\Delta \lambda=\lambda_2-\lambda_1\\\Rightarrow \Delta \lambda=589.59-589\\\Rightarrow \Delta \lambda=0.59]\ nm[/tex]

We have the relation

[tex]\dfrac{\lambda}{\Delta \lambda}=mN\\\Rightarrow N=\dfrac{\lambda}{m\Delta \lambda}\\\Rightarrow N=\dfrac{589}{2\times 0.59}\\\Rightarrow N=499.15254[/tex]

Width is given by

[tex]w=\dfrac{N}{n}\\\Rightarrow w=\dfrac{499.15254}{125}\\\Rightarrow w=3.99322032\ mm[/tex]

The width is 3.99322032 mm

The angular separation  \theta of the first maxima of these spectral lines generated by this diffraction grating is 0.54°

The width which this grating needs to be to allow you to resolve the two lines 589.00 and 589.59 nanometers is 3.99322032 mm

n = Lines per mm

= 125

The Separation between slits is given by:

d= 1/n

d= 1/125

= 0.008mm.

Where

line 1 = 498nm

line 2 = 569nm

The first maximum m= 1 will be:

θ1=  3.57°

θ2= 4.08°

The angular separation would be:

θ2- θ1= 0.54°.

Now, to find the width is:

w= N/n

= 499.15254/125

= 3.99322032 mm.

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Answer:

Explanation:

(a)mass [tex]m= 2.5 kg [/tex]

height [tex]h=6 m[/tex]

work required to raise

[tex]W=mgh[/tex]

[tex]W=2.5\times 9.8\times 6=147 J[/tex]

(b)Force [tex]F=20.8 N[/tex]

mass of cart [tex]m=3 kg[/tex]

length of track [tex]s=0.636 m[/tex]

[tex]Work\ done=F\cdot s[/tex]

Work done[tex]=20.8\cdot 0.636=13.22 J[/tex]

(c)mass of eddy [tex]m_e=65 kg[/tex]

height climbed [tex]h=1.6 m[/tex]

time [tex]t=1.2 s[/tex]

Energy required [tex]E=mgh=65\times 9.8\times 1.6=1019.2 J[/tex]

[tex]power=\frac{E}{t}=\frac{1019.2}{1.2}=849.33 W[/tex]

The work done in lifting the 2.5 kg object is 147 J, in moving the 3-kg cart is 13.2 J, and the power used by Eddy in climbing stairs is 851.7 W.

To solve these problems, we need to apply principles of physics, specifically related to work, energy, and power. For the first question, we use the concept of gravitational potential energy, which is calculated by multiplying together the object's mass, the acceleration due to gravity (~9.81 m/s² on Earth), and the height to which the object is lifted. Thus for the 2.5-kg object, the work done or energy required to lift it to a height of 6.0 meters is: W = m * g * h = 2.5 kg * 9.8 m/s² * 6.0 m = 147 J.

Next, for the 3-kg cart, since the cart moves at constant speed, we can say the work done is equal to the change in potential energy. Therefore, the work done is W = m * g * h = 3.0 kg * 9.8 m/s² * 0.450 m = 13.2 J.

Fianlly, for Eddy's case, power is defined as the work done per unit time. If Eddy lifts his own mass to the height of 1.6 m, the work done (again considering as change in gravitational potential energy) is W = m * g * h = 65 kg * 9.8 m/s² * 1.6 m = 1022 J. Given that he does this work in 1.2 seconds, the power expended would be P = W / t = 1022 J / 1.2 s = 851.7 W.

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Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

         y = y₀  L / (L-d)

The distance from the screen (d), we look for it with kinematics

         v = d / t

        d = v t

we replace

         y = y₀ L / (L - v t)

         y = 5.2 22 / (22 - 3 t)

         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

Answer:

option D

Explanation:

given,

coefficient of friction between wall and tire = µ

speed of motorcycle = s

friction force = f = μ N

where normal force will be equal to centripetal force

[tex]N = \dfrac{mv^2}{r}[/tex]

for motorcycle to not to slip weight should equal to the centripetal force

 now,

[tex]m g =\mu \dfrac{mv^2}{r}[/tex]

[tex]\mu =\dfrac{rg}{s^2}[/tex]

where "rg" is constant

[tex]\mu\ \alpha \ \dfrac{1}{s^2}[/tex]

[tex]\mu\ \alpha \ s^{-2}[/tex]

Hence, the correct answer is option D

radiation

convection

conduction

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The method of thermal energy transfer in this three cases:

A: Radiation

B: Convection

C: Conduction

Heat transfer is defined by thermodynamic systems as "The transfer of heat over the system boundary caused by a temperature difference between the system and its surroundings."

There are several ways for heat to go from one place to another. Conduction, convection, and radiation are some of the several ways that heat is transferred.

Conduction is the process of energy being transferred from one medium particle to another while the particles are in close proximity to one another.

The flow of fluid molecules from higher temperature regions to lower temperature regions is referred to as convection.

Radiant heat is the name for thermal radiations. Emission of electromagnetic waves results in the production of thermal radiation. These waves remove the energy from the body that is releasing them.

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Answer:

The time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours

Explanation:

Electric Charge: This is defined as the product of electric current to time in a an electric circuit. The S.I unit of charge is Coulombs (C).

Mathematically, it is represented as

Q = it.......................... Equation 1

Where Q =quantity of charge, I = current, t = time

making t the subject of formula in equation 1,

t = Q/I ....................... Equation 2

Given: Q = 3.5 C, I = 5 mA.

Conversion: We convert from mA to A

I.e 5 mA = (5 × 10⁻³) A = 0.005 A.

Substituting these values into equation 2

t = 3.5/0.005

t = 700 seconds  

or

(700/60) minutes = 11.67 minutes

or

(700/3600) hours = 0.194 hours.

Therefore the time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours

700 second will take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes.

What information do we have?

Charges of partical = 3.5 C

Current = 5 mA = (5 × 10⁻³) A

Charge = (Curernt)(time)

Q = iT

T = Q / i

Time taken = 3.5 / (5 × 10⁻³)

Time taken = 700 seconds or 11.67 minutes

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Answer:

20 metres

Explanation:

Speed = distance ÷ time

[tex]s = \frac{d}{t} [/tex]

If we substitute the values:

[tex]4 = \frac{d}{5} [/tex]

[tex]20 = d[/tex]

Answer: 20m

Explanation:

Speed = Distance/time taken

Speed = 4.00m/s

Time taken = 5.00s

Distance = D = ?

We insert the values in the formula

4.00m/s = D/5.00s

Multiply through by 5

D = 20m

Answer:

C. all above is true.

Explanation:

Energy releasing reactions are exothermic. Total energy of products ( [tex]  N_2 H_2[/tex]  ) is less than the total energy of reactants ( [tex]  H_2 + N_2 [/tex] ) gives negative enthalpy change.

hint: prefix exo- means "outside, external".

Answer:

V(h) = Eh

Explanation:

I will assume that the capacitor is a parallel-plate capacitor.

By Gauss' Law, electric field inside the capacitor is

[tex]E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}[/tex]

The relation between electric field and potential is

[tex]V_{ab} = -\int\limits^b_a {\vec{E}(h)} \, d\vec{h}  = \int\limits^h_0 {\frac{Q}{\epsilon_0 A}} \, dh \\V(h) - V(0) = V(h) - 0 = Eh\\V(h) = Eh[/tex]

The important thing in this question is that the electric field inside the parallel plate is constant. So, the potential is also constant and proportional to the distance, h.

The electric potential V(h) inside a capacitor as a function of height h, with zero potential at the bottom plate, is V(h) = Eh, using the relationship between the electric field E and potential V where E is constant.

To calculate the electric potential V(h) inside a capacitor as a function of height h, with the potential at the bottom plate taken to be zero, you can use a relation between the electric field E and the potential V. Given that E = V/d, where d is the separation between the plates, and that the electric field E is uniform, we have the relationship E = -dV/dh (the negative sign indicates the direction of the potential decrease). Integrating this equation from 0 to h, where V(0) = 0, gives us V(h) = -Eh. However, we can ignore the negative sign because we are interested in magnitude.

So, the final expression for V(h) inside the capacitor, in terms of the electric field E and the height h above the bottom plate, is:

V(h) = Eh

Answer:

The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.

Explanation:

Given that,

Acceleration = 9.5 m/s²

Diameter = 283 m

We need to calculate the frequency of the rotational motion for the wheel to produce this effect

Using formula of rotational frequency

[tex]a= r\omega^2[/tex]

[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]

Where, r = radius

a = acceleration

[tex]\omega[/tex] = rotational frequency

Put the value into the formula

[tex]\omega=\sqrt{\dfrac{9.5\times2}{283}}[/tex]

[tex]\omega=0.259\ rad/s[/tex]

The frequency in rev/min

[tex]\omega=0.259\times\dfrac{60}{2\pi}[/tex]

[tex]\omega=2.473\ rev/min[/tex]

Hence, The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.

Answer:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

Answer:

Option 1

Explanation:

The correct answer is option 1

The gold foil experiment was conducted by Rutherford. This experiment was conducted to study the Atom.

In the experiment Alpha rays from the emitter are passed through gold foil and there was a receiver that was present there to intercept the alpha rays.

The outcome of the result was that most of the alpha particle pass through foil undeflected and very few rays revert back on the original path from the heavy mass present at the center.

Later this heavy mass was known Nucleus.

Hence, most alpha particles passed through the foil.

Explanation:

a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.

1300 m east = 1300 i

2400 m north = 2400 j

Drops the penny from a cliff 640 m high = -640 k

Displacement of penny = 1300 i + 2400 j - 640 k

b) Displacement of man for return trip = -1300 i - 2400 j

    [tex]\texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m[/tex]

    Magnitude of his displacement = 2729.47 m

Answer:

Explanation:

d1 = 1300 m east

d2 = 2400 m north

d3 = 640 m downward

(a)

The displacement of penny is given by

[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]

[tex]\overrightarrow{d}=1300\widehat{i}+2400 \widehat{j}-640\widehat{k}[/tex]

(b) For the return journey of man, the displacement is given by

[tex]\overrightarrow{d}=-\overrightarrow{d_{1}}-\overrightarrow{d_{2}}[/tex]

[tex]\overrightarrow{d}=-1300\widehat{i}-2400 \widehat{j}[/tex]

The magnitude of the displacement is given by

[tex]d=\sqrt{1300^{2}+2400^{2}}=2729.47 m[/tex]

Answer:

1.457881265 kgm²

0.02842 Nm/rad

0.13962 rad/s

Explanation:

M = Mass = 3.97 kg

R = Radius = 85.7 cm

[tex]\tau[/tex] = Torque = 0.0688 Nm

[tex]\theta[/tex] = Angle of rotation = 2.42 rad

Moment of inertia about the center of the disk is given by

[tex]I=\dfrac{1}{2}MR^2\\\Rightarrow I=\dfrac{1}{2}\times 3.97\times 0.857^2\\\Rightarrow I=1.457881265\ kgm^2[/tex]

The rotational inertia of the disk about the wire is 1.457881265 kgm²

Torque is given by

[tex]\tau=\kappa \theta\\\Rightarrow \kappa=\dfrac{\tau}{\theta}\\\Rightarrow \kappa=\dfrac{0.0688}{2.42}\\\Rightarrow \kappa=0.02842\ Nm/rad[/tex]

The torsion constant is 0.02842 Nm/rad

Time period is given by

[tex]T=2\pi\sqrt{\dfrac{I}{\kappa}}[/tex]

Angular frequency is given by

[tex]\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\sqrt{\dfrac{\kappa}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{0.02842}{1.457881265}}\\\Rightarrow \omega=0.13962\ rad/s[/tex]

The angular frequency of this torsion pendulum when it is set oscillating is 0.13962 rad/s