How does a coal-fired power plant use the energy in coal to produce electricity?

Answer:

option B and C

Explanation:

Control rod are used to regulate the nuclear reactor.

When you insert control rod in the reactor it slows down the nuclear fission inside the reactor and the energy produced in the reactor will be less.

When you remove control road from the reactor the nuclear fission increase inside the reactor and the energy production is high.

Control rod consist of boron, boron absorb the neutron which help to control the nuclear fission.

Hence, the correct answer is option B and C

Answer:

Work function of the metal, [tex]W_o=3.38\times 10^{-17}\ J[/tex]

Explanation:

We are given that  

Frequency of the electromagnetic radiation,  [tex]f=5.16\times 10^{16}[/tex] Hz

The maximum kinetic energy of the emitted electron, [tex]K=4.04\times 10^{-19}\ J[/tex]

We need to find the work function of the metal.

We know that the maximum kinetic energy of ejected electron

[tex]K=h\nu-w_o[/tex]

Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]

[tex]\nu[/tex] =Frequency of light source

[tex]w_o[/tex]=Work function

Substitute the values in the given formula  

Then, the work function of the metal is given by :

[tex]W_o=h\nu -K[/tex]

[tex]W_o=6.63\times 10^{-34}\times 5.16\times 10^{16}-4.04\times 10^{-19}[/tex]

[tex]W_o=3.38\times 10^{-17}\ J[/tex]

So, the work function of the metal is [tex]3.38\times 10^{-17}\ J[/tex]. Hence, this is the required solution.

Answer:

A.) A particle of water in a wave moves in a circular motion when viewed in cross section

Explanation:

Water waves are transverse waves. Waves transport energy, not water. As a wave crest passes, the water particles move in circular paths

The water particles don't move, they still end up in the place in which they began, the water particle moves up and down while the wave that passes through them moves in a perpendicular direction to the direction of the water particles moving up and down with respect to the crests and troughs of the wave (since water is a transverse wave). This will cause the wave to transverse in a circular path

The answer cannot be C because wavebase is equal to one-half the wavelength measured from still water level

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × [tex]\frac{ \lambda }{2}[/tex]      ........1

so  λ = [tex]\frac{2L}{10}[/tex]  

and velocity is express as

V = [tex]\sqrt{\frac{T}{\mu } }[/tex]    .................2

so

frequency for string A = f1 = [tex]\frac{V1}{\lambda}[/tex]

f1 = [tex]\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}[/tex]

f1 = [tex]\frac{10}{2L} \sqrt{\frac{T1}{\mu } }[/tex]      

and

f2 = [tex]\frac{10}{2L} \sqrt{\frac{T2}{\mu } }[/tex]

so

beat frequency is = f2 - f1

put here value

beat frequency = [tex]\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}[/tex]  - [tex]\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }[/tex]

beat frequency = 13.87 Hz

Answer:

The answer to your question is 11.2 m/s

Explanation:

Data

Initial speed (vo) = 6.0 m/s

Acceleration (a) = 3.0 m/s²

time = 15 s

Final speed = ?

Formula

                d = vot + [tex]\frac{1}{2} at^{2}[/tex]

                vf² = vo² + 2ad

Process

                d = (6)(15) + [tex]\frac{1}{2} (3)(15)^{2}[/tex]

                d = 90 + 337.5

                d = 427.5 m

                vf² = (6)² + 2(3)(15)

                vf² = 36 + 90

                vf² = 126

                vf = 11.2 m/s

Answer:

Explanation:

We shall apply law of  conservation of mechanical energy for projectile being thrown .

Total energy on the surface = total energy at height h required

a ) At height h , velocity = .351 x ( 2 GM/R x h )

[tex]\frac{-GM}{R} + \frac{m\times(.351\times\sqrt{2GM})^2 }{2R } = \frac{-GMm}{R+h} + 0[/tex]

[tex]\frac{-GMm}{R} +\frac{1}{2}\times  \frac{-2GMm}{R} \times0.123=\frac{-GMm}{R+h}[/tex]

[tex]\frac{0.877GMm}{R} =\frac{-GMm}{R+h}[/tex]

h = .14 R

b )

[tex]\frac{-GM}{R} + \frac{m\times(.351\times2GM) }{2R } = \frac{-GMm}{R+h} + 0[/tex]

[tex]\frac{-0.649GMm}{R} = \frac{-GMm}{R+h}[/tex]

h = .54 R

c ) least initial mechanical energy required at launch if the projectile is to escape Earth

= GMm / R + 1/2 m (2GM/R)

= 0

Answer:

number of molecules=  2.83 x 10^19

Explanation:

0.04 % means 0.04 L CO_2 in 100 L atmosphere

so for 2.9 L atmosphere CO2 vol = ( 0.04/100) x 2.9 = 0.00116

T = 27°C = 300 K , P = 1 atm ,

n = PV/RT = ( 1 x 0.00116) ÷ ( 0.08206x300) = 4.71×10^{-5}

number of molecules = 6.023 x 10^23 x 4.71×10^{-5}   = 2.83 x 10^19

The temperature at which hydrogen molecules would have a root-mean-square velocity equal to the Moon's escape velocity is approximately 112 million Kelvin.

The root-mean-square speed of an ideal gas is given by the equation: vrms = sqrt(3RT/M), where R is the ideal gas constant (8.314 J mol-1 K-1), T is the temperature in Kelvin, and M is the molar mass in kg/mol. To find the temperature necessary for hydrogen molecules to achieve the Moon’s escape velocity (2.38 km/s), we first need to convert the escape velocity to m/s and the molar mass of hydrogen to kg/mol.

So, 2.38 km/s is equal to 2380 m/s, and the molar mass of hydrogen is equal 2.016 g/mol, which is 0.002016 kg/mol. Then, we solve the equation (2380 = sqrt (3RT/0.002016)) for T to get the temperature.

By resolving this equation we find that the necessary temperature would be approximately 112 million Kelvin. Therefore, for hydrogen molecules to have a root-mean-square velocity equal to the Moon's escape velocity, they must be at this extremely high temperature.

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Answer:

the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex] along downward direction

Explanation:

Here, the acceleration is constant and it is equal to acceleration due to gravity in moon. Therefore the question depicts a situation of uniformly accelerated motion in a straight line. So, let us refresh the three equations of uniformly accelerated straight line motion.

v = u + at

[tex]s = ut + \frac{1}{2}at^{2}[/tex]

[tex]v^{2} = u^{2} +2as[/tex]

where,

u = initial velocity

v = final velocity

s = displacement

a = acceleration

t = time

Since we are dealing with vectors (velocity, acceleration and displacement), we have to take their directions in to account. So we must adopt a coordinate system according to our convenience. Here, we are taking point of throwing as origin, vertically upward direction as positive y axis and vertically downward direction as negative y axis.

t = 1s

u = 0 (since the hammer is dropped)

v = -1.6m/s (since its direction is downward)

a = ?

The only equation that connects all the above quantities is

v = u + at

therefore,

a = [tex]\frac{v - u}{t}[/tex]

substituting the values

a = [tex]\frac{-1.6 - 0}{1}[/tex]

a = -1.6m/[tex]s^{2}[/tex]

Thus, the value of acceleration due to gravity in moon is 1.6m/[tex]s^{2}[/tex]. The negative sign indicates that it is along downward direction.

The purpose of system is to make system more reliable and efficient for desired work completion

Explanation:

In most of the times, the system analysts operate in a dynamic environment where change is  the necessary process . A business application, a business firm or a computer system may be the required environment. In order To rebuild a system, the key elements must be considered which we need to examine are as follows:

1. Outputs and inputs.

2. Security and Control.

3. processor

4. Environment.

5. Feedback.

6. Boundaries and interface.

Final answer:

The work required to move a satellite from an orbit of radius 2R to 3R around a planet is calculated using the gravitational potential energy formula and is found to be 3.897×1010 J.

Explanation:

To calculate the work required to move a satellite from one circular orbit to another around a planet, we must consider the gravitational potential energy differences in the two orbits.

The gravitational potential energy (U) of an object of mass m in orbit around a planet of mass M at a distance r is given by U = -GmM/r, where G is the gravitational constant (6.67×10-11 N m2/kg2).

For the initial orbit at radius 2R, the potential energy is U1 = -GmM/(2R), and for the final orbit at radius 3R, the potential energy is U2 = -GmM/(3R). The work done (W) in moving the satellite is the difference in gravitational potential energy, W = U2 - U1. Substituting the values, we get:

W = (-GmM/3R) - (-GmM/2R) = (GmM/6R)

Let's calculate the work required using the given values: G = 6.67×10-11 N m2/kg2, m = 571 kg, M = 7.76×1024 kg, R = 8.8×106 m.

W = (6.67×10-11 N m2/kg2 × 571 kg × 7.76×1024 kg) / (6 × 8.8×106 m)

W = 3.897×1010 J

Therefore, the work required to move the satellite from a circular orbit of radius 2R to one of radius 3R is 3.897×1010 J.

The initial velocity of the car is 36.6 m/s

Explanation:

The motion of the car is a uniformly accelerated motion (=constant acceleration), therefore we can apply suvat equations:

[tex]v^2-u^2=2as[/tex]

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem, we have:

v = 0 is the final velocity (the car comes to a stop)

[tex]a=-3.2 m/s^2[/tex] is the acceleration

s = 210 m is the displacement of the car

Solving for u, we find the initial velocity:

[tex]u=\sqrt{v^2-2as}=\sqrt{0-(2)(-3.2)(210)}=36.6 m/s[/tex]

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Answer:

(ii) 1s2 2s2 2p6 3s2

Explanation:

Electron Affinity is the energy change that occur when an atom gains an electron.

                                  X₍₉₎ + e⁻  →  X⁻           ΔE = Eea

ΔE is change in energy

Eea is electron affinity

Often, electron affinity has negative energy values. The more negative the electron affinity, the easier it is to add an electron to a particular atom. Electron affinity increases across the period in the periodic table. However, there are few exceptions:

1. The electron affinities of group 18 (8A) elements are greater than zero. This is because the atom has a filled valence shell, an addition of electron causes the electron to move to a higher energy shell.

2. The electron affinities of group 2 (2A) elements are more positive because addition of an electron requires it to reside in the previously unoccupied p sub-shell.

3. The electron affinities of group 15 (5A) elements are more positive because addition of an electron requires it to be put in an already occupied orbital.

Applying these consideration to the elements given in the question:

(i) The sum of the electron in 1s²2s² 2p⁶ 3s¹ = 2+2+6+1 = 11 Sodium (Na)

(ii) The sum of the electron in 1s² 2s² 2p⁶ 3s² = 2+2+6+2 = 12 Magnesium (Mg)

(iii) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p¹ = 2+2+6+2+1 = 13 Aluminium (Al)

(iv) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁴ = 2+2+6+2+4 = 16 Sulfur (S)

(v) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁵ = 2+2+6+2+5 = 17 Chlorine (Cl)

The atom that is expected to have a positive electron affinity is Magnesium which is a group 2A element with electron configuration of 1s² 2s² 2p⁶ 3s².

An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Typically, these atoms tend to have half-filled or filled subshell configurations. In the given examples, the atom with electron configuration 1s2 2s2 2p6 3s2 3p1 is likely to have a positive electron affinity.

The electron affinity of an atom is a measure of the energy change when an electron is added to a neutral atom to form a negative ion. An atom with a positive electron affinity attracts electrons more than one with negative electron affinity. Usually, the atom would tend to have a half-filled (e.g. 3p3) or filled (e.g. 3p6) subshell configuration.

If we consider the given electron configurations: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p1 (iv) 1s2 2s2 2p6 3s2 3p4 (v) 1s2 2s2 2p6 3s2 3p5, it appears that (iii) 1s2 2s2 2p6 3s2 3p1 should have a positive electron affinity. This atom is a Potassium atom (K). It has a positive electron affinity because it is just one electron away from obtaining a full orbit shell (in the 3s and 3p orbitals) which would provide a more stable electron configuration.

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Answer:

gdsz

Explanation:

dsgzz cxvzdgctfgdsvftgdsftrdsfdtsardtgasfd5t6sgftsfdrstfdtgsv6cr5vsd5rw5

Explanation:

We know the equation

            [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]

   where l is length of pendulum, g is acceleration due to gravity and T is period.

Rearranging

              [tex]g= \frac{4\pi^2l}{T^2}[/tex]

Length of pendulum in Tokyo = 0.9923 m

Length of pendulum in Cambridge = 0.9941 m

Period of pendulum in Tokyo = Period of pendulum in Cambridge = 2s

We have

                     [tex]\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}= \frac{\frac{4\pi^2 l_{\texttt{Tokyo}}}{ T_{\texttt{Tokyo}}^2}}{\frac{4\pi^2 l_{\texttt{Cambridge}}}{ T_{\texttt{Cambridge}}^2}}\\\\\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}=\frac{\frac{0.9923}{2^2}}{\frac{0.9941}{2^2}}=0.998[/tex]

Ratio of free fall acceleration of Tokyo to Cambridge = 0.998

Answer:

The enthalpy for dissolution is - 305.558 J/g

Solution:

Mass of the ionic compound, m = 10.00 g

Mass of water, m' = 75.0 g

Initial temperature, T = [tex]23.2^{\circ}C[/tex]

Final Temperature, T' = [tex]31.8^{\circ}C[/tex]

Now,

To calculate the change in enthalpy:

We know that the specific heat of water is 4.18 [tex]J/g^{\circ}C[/tex]

Total mass of the solution, M = m + m' = 10.00 + 75.0 = 85.0 g

Temperature, difference, [tex]\Delta T = T' - T = 31.8 - 23.2 = 8.6^{\circ}C[/tex]

Thus

The heat absorbed by the solution is given by:

[tex]Q = MC_{w}\Delta T = 85.0\times 4.18\times 8.6 = 3055.58\ J[/tex]

Enthalpy, [tex]\Delta H = -\frac{Q}{m} = - \frac{3055.58}{10} = - 305.558\ J/g[/tex]

Answer:

The speed of cart is 5.5 m/s.

Explanation:

Given that,

Mass of cart = 100 kg

Distance = 3 m

Speed = 6 m/s

We need to calculate the speed of cart

Using relation of centripetal force and normal force

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

[tex]\dfrac{m}{2}+mg=\dfrac{mv^2}{r}[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}+9.8=\dfrac{v^2}{3}[/tex]

[tex]v=\sqrt{3(\dfrac{1}{2}+9.8)}[/tex]

[tex]v=5.5\ m/s[/tex]

Hence, The speed of cart is 5.5 m/s.

Final answer:

The speed of the cart at the top of the loop is 4.24 m/s.

Explanation:

In order to find the speed of the cart at the top of the loop, we can use the principle of conservation of mechanical energy. The total mechanical energy of the cart is the sum of its gravitational potential energy and its kinetic energy. At the top of the loop, the gravitational potential energy is zero, so the total mechanical energy is equal to the kinetic energy. The kinetic energy of the cart is given by the formula KE = 1/2 * m * v^2, where m is the mass of the cart and v is its velocity.

Given that the mass of the cart is 100 kg and the radius of the loop is 3 m, we can first find the centripetal acceleration at the top of the loop using the formula a = v^2 / r. Plugging in the values, we get a = (6 m/s)^2 / 3 m = 12 m/s^2. Since the force exerted by the track on the cart at the top of the loop is half the cart's mass, we can use this force to find the net force acting on the cart. The net force is equal to the centripetal force, which is given by F = m * a. Plugging in the values, we get F = (1/2 * m) * a = (1/2 * 100 kg) * 12 m/s^2 = 600 N.

Since the net force equals the centripetal force, we can equate it to the formula F = m * v^2 / r. Plugging in the values, we get 600 N = 100 kg * v^2 / 3 m. Solving for v, we get v^2 = 600 N * 3 m / 100 kg = 18 m^2/s^2. Taking the square root of both sides, we find that v = sqrt(18 m^2/s^2) = 4.24 m/s.

Answer:

The south-component of its velocity is the same

Explanation:

You could say several things concerning the object.

These are some of them:

- The south-component of its velocity is the same.

- Its speed (module of the velocity) is greater now.

- The velocity has now a west-component.

- The object is now moving due west of south.

- The object did not move in a straight line.

Final answer:

After experiencing a westward net force, the object will change its velocity and once the force is removed, it will continue at a new constant velocity in a direction that combines the original southward and the newly added westward components.

Explanation:

According to Newton's first law of motion, an object at rest or moving with a constant velocity will continue to do so unless acted upon by a net force. In the scenario provided, the object initially moving due south at a constant velocity experiences a net force acting due west. This force will cause the object to accelerate in the direction of the force, which is westward, changing the object's velocity.

After the short time interval, when the net force becomes zero, the object will then move at a constant velocity in a new direction that is the resultant of the south and west velocity components.

Answer:

Coefficient of static friction = 0.37

Explanation:

At the point the the quoll slides, quoll attains its maximum velocity.

So Ne = (mv^2)/r ....equa 1

And N =mg....equ 2

Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s

Sub equ 2 into equ 1

Mge= (mv^2)/r ...equa3

Simplfy equ3

e = v^2/(gr)...equ 4

Sub figures above

e = 5.76/(9.8*1.6)

e = 0.37

The new electric force is equal to the original force

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

In this problem, at the beginning we have

[tex]q_1 = -q\\q_2 = -q\\r=r[/tex]

So the force is

[tex]F=k\frac{(-q)(-q)}{r^2}=\frac{kq^2}{r^2}[/tex]

Later, the amount of both charges is doubled, so:

[tex]q_1' = -2q\\q_2' = -2q[/tex]

and the separation is doubled as well:

[tex]r'=2r[/tex]

So the new force is:

[tex]F'=k\frac{(-2q)(-2q)}{(2r)^2}=\frac{kq^2}{r^2}=F[/tex]

This means that the new electric force is equal to the original force.

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Answer:

μk = 0.488

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the inclined plane and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

FN : Normal force : perpendicular to the inclined plane

fk : kinetic Friction force: parallel to the inclined plane

Calculated of the W

W= m*g

W= 4.9 kg* 9.8 m/s² = 48.02 N

x-y weight components

Wx = Wsin θ = 48.02*sin27° = 21.8 N

Wy = Wcos θ = 48.02*cos27° = 42.786 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay    ay = 0

FN - Wy = 0

FN = Wy

FN = 42.786 N

Calculated of the fk

fk = μk* FN=  μk*42.786 Equation (1)

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the acceleration of the block :

d = v₀*t+(1/2)*a*t² Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

t: time interval   (m/s)

a: acceleration ( m/s²)

Data:

d= 2.7 m

v₀ = 0

t= 5.4 s

We replace data in the formula (2)  

d = v₀*t+(1/2)*a*t²

2.7 = 0+(1/2)*a*( 5.4)²

2.7 = (14.58)*a

a = 2.7 / (14.58)

a= 0.185 m/s²

We apply the formula (1) to calculated μk:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-fk= m*a     , fk=μk*42.786 of the Equation (1)

21.8 - (42.786)*μk = (4.9)*(0.185)

21.8 -0.907= (42.786)*μk

20.89 = (42.786)*μk

μk = (20.89) / (42.786)

μk = 0.488

Answer:

B = 0.0307 T = 30.74 mT

Explanation:

Given

m = 0.150 kg

I = 15.0 A

d = 0.510 m

μk = 0.16

B = ?

Balancing the forces on the rod in the j direction

N - m*g = 0   ⇒   N = m*g

and in the i direction

I*d*B - μk*N = 0     ⇒      B = μk*N / (I*d)

⇒      B = μk*m*g / (I*d)

⇒      B = (0.16)*(0.150 kg)*(9.8 m/s²) / (15.0 A*0.510 m)

⇒      B = 0.0307 T = 30.74 mT

Answer: [tex]4.8(10)^{-5} m[/tex]

Explanation:

We can solve this problem by the following equation:

[tex]\eta=\frac{F.h}{A \Delta x}[/tex]

Where:

[tex]\eta=3.5(10)^{10}Pa[/tex] is the shear modulus for brass

[tex]F=4.2(10)^{4}N[/tex] is the applied force

[tex]h=2.5 cm=0.025 m[/tex] is the height of the cube

[tex]A=h^{2}=(0.025 m)^{2}=625(10)^{-6} m^{2}[/tex] is the area of each surface of the cube

[tex]\Delta x[/tex] is the shear displacement

Finding [tex]\Delta x[/tex]:

[tex]\Delta x=\frac{F.h}{A \eta}[/tex]

[tex]\Delta x=\frac{(4.2(10)^{4}N)(0.025 m)}{(625(10)^{-6} m^{2})(3.5(10)^{10}Pa)}[/tex]

Finally:

[tex]\Delta x= 4.8(10)^{-5} m[/tex]

Answer: The temperature increases on the stratosphere with the altitude, given that absorption of the ultraviolet rays by the ozone.

Explanation: On the stratosphere, the water vapor and the umidity are almost nonexistents and, in view of the absorption of ultraviolet rays by the ozon, the temperature increases, reaching 35,6º Fahrenheit.  

The ozone is a unusual type of oxygen molecule. In the stratosphere, the ozone appears on a large scale and warms it up by the absorption of the ultraviolet rays energy.

The temperature in the stratosphere increases with altitude due to the absorption of UV radiation by ozone, which leads to exothermic reactions that generate heat, especially at higher altitudes where ozone concentration is greater.

The temperature increases with height in the stratosphere primarily because of the absorption of ultraviolet (UV) radiation by ozone. Ozone, a form of oxygen molecule, is very good at absorbing UV radiation from the Sun. When UV radiation is absorbed by ozone, it leads to exothermic chemical reactions which generate heat, thus raising the temperature of the surrounding atmosphere.

At higher altitudes in the stratosphere, the concentration of ozone is greater and, consequently, more UV radiation is absorbed, leading to a higher temperature in this region compared to the lower part of the stratosphere. It is this temperature inversion that distinguishes the stratosphere from the troposphere below, where the temperature normally decreases with altitude. The absorption of UV radiation and its conversion into heat makes the top of the stratosphere hotter as it is closer to the source of shortwave radiation, and the amount of absorbed shortwave radiation diminishes towards the bottom of the stratosphere.

Answer:a)8.01 m/s

Explanation:

Given

mass of ball [tex]m=0.9 kg[/tex]

initial speed [tex]u=9 m/s[/tex]

launch angle [tex]\theta =27 [/tex]

As the projectile reaches its highest point its vertical velocity component becomes zero and there will only be horizontal component

velocity at highest Point [tex]v=u\cos \theta [/tex]

[tex]v=9\cos 27[/tex]

[tex]v=8.01 m/s[/tex]

(b)maximum height h

Maximum height is given by

[tex]H_{max}=\frac{u^2\sin^2 \theta }{2g}[/tex]

[tex]H_{max}=\frac{9^2\sin^2 (27)}{2\times 9.8}[/tex]

[tex]H_{max}=0.85 m[/tex]

Final answer:

The speed of the ball at its highest point is its horizontal velocity component, calculated using the cosine of the angle. The height reached by the ball is found using the kinematic equation for vertical motion, considering the initial vertical velocity and the acceleration due to gravity.

Explanation:

When addressing the question of a 0.90-kg ball thrown at an upward angle, we are dealing with a two-dimensional projectile motion problem in physics. This type of problem can be broken down into horizontal and vertical components. Since gravity acts only in the vertical direction, it does not affect the horizontal velocity of the projectile.

(A) Speed at the highest point: The speed of the ball at its highest point is solely the horizontal component of its initial velocity because the vertical component will be zero at that point. The horizontal velocity (
vx) can be calculated using
vx =
v cos(
θ). So,
vx = 9.0 m/s cos(27°).

(B) Maximum height: To find the maximum height, we need to consider the vertical component of the velocity (
vy) and the acceleration due to gravity (
g), which is -9.81 m/s² (negative sign indicates the direction is opposite to the velocity). The initial vertical velocity (
vy) is
v sin(
θ), thus
vy = 9.0 m/s sin(27°). We can then use the kinematic equation
v2 = u2 + 2as to find the maximum height (
h), where u is the initial velocity, s is the displacement, and a is the acceleration. At the highest point, the final vertical velocity (
v) is 0 m/s. Rearranging the equation to solve for
s, we get
s = v2 / (2g).

Answer: A Whip

Explanation:

Final answer:

Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane.

Explanation:

Facilitated transport, also known as facilitated diffusion, is the process by which a molecule moves down its concentration gradient using transport proteins in the plasma membrane. This process does not require the input of energy and allows substances to diffuse across the membrane more easily. For example, glucose is transported into cells using glucose transporters that utilize facilitated transport. This process is important for the movement of larger or charged molecules that cannot freely diffuse across the cell membrane.

Answer:

Explanation:

The particles are in x-y plane with coordinates of masses as follows

m₂ at (0,0 ) m₁ at ( 0,2 ), m₄ at ( 2,2 ) and m₃ at (2,0 )

Moment of inertia about z axis

I_z = 0 + 3 x 2² + 4 x (2√2)² + 3 x 2²

= 12 + 32 + 12

= 56 kgm²

Now let us find out moment of inertia about axis through CM

According to theorem of parallel axis

I_z = I_g + m x r²  

Here m is total mass that is 14 kg and r is distance between two axis which is √2 m

56 = I_g + 14 x (√2)²

I_g  = 56 - 28

= 28 kgm²

We can directly compute  I_g  as follows

I_g = 4 x (√2)² +3 x (√2)² +4 x (√2)²+3 x (√2)²

= 8 +6 +8 +6

= 28 kgm²

So the result obtained earlier is correct.

Answer:

The tangential acceleration of the pedal is 0.0301 m/s².

Explanation:

Given that,

Length = 19 cm

Diameter = 23 cm

Time = 10 sec

Initial angular velocity = 65 rpm

Final velocity = 90 rpm

Suppose we need to find the tangential acceleration of the pedal

We need to calculate the tangential acceleration of the pedal

Using formula of tangential acceleration

[tex]a_{t}=r\alpha[/tex]

[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{\omega_{2}-\omega_{1}}{t}[/tex]

[tex]a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{90\times\dfrac{2]pi}{60}-65\times\dfrac{2\pi}{60}}{10}[/tex]

[tex]a_{t}=0.0301\ m/s^2[/tex]

Hence, The tangential acceleration of the pedal is 0.0301 m/s².

Answer:

a)Time taken will be 2.783 s

b)Time taken will be 2.564 s

Explanation:

a)Since the pulley has mass ,

[tex]m_{1}g-T_{1}= m_{1}a//T_{2}-m_{2}g=m_{2}a[/tex] ------3

             αR = a ;                    ------1

[tex](T_{1} - T_{2})*R=m*R^{2}*(alpha)=5*R^{2}*\frac{a}{R}[/tex] ------2

solving these we get ,

[tex]a=\frac{4g}{33}[/tex]

∴

a)time taken :

[tex]s=ut+\frac{1}{2} at^{2}\\u=0\\a=\frac{4g}{33} \\s=4.6\\4.6=\frac{1}{2} * \frac{4*9.8}{33} *t^{2}\\t= 2.783 sec[/tex]

ANS : 2.783 sec

b)

In case of B, mass is zero.

Thus, there is no rotation of the pulley. this is equivalent to a normal 1 Dimension motion question

[tex]m_{1}g-T_{1}=m_{1}a\\T_{1}-m_{2}g=m_{2}a\\a= \frac{m_{1}-m_{2}}{m_{1}+m_{2}}g\\a=\frac{4g}{28} \\a=\frac{g}{7} \\a=1.4ms^{-2}[/tex]

Thus time t will be ,

[tex]s=\frac{1}{2} at^{2}\\4.6=\frac{1}{2}*1.4*t^{2}\\ t=2.564 sec[/tex]

ANS : 2.564 sec

The time it takes the block to reach the ground can be found by making

use of the law of conservation of energy.

Reasons:

Given parameters are;

Mass of block m₁ = 16.0 kg

Mass of block m₂ = 12.0 kg

Mass of the pulley, M = 5.00 kg

By conservation of energy, we have;

m₁g·h - m₂·g·h = 0.5×(m₁ + m₂)·v² + 0.5·I·ω²

[tex]\omega = \dfrac{v}{R}[/tex]

[tex]Moment \ of \ inertia\ of \ pulley, I = \dfrac{1}{2} \cdot M \cdot R^2[/tex]

Therefore;

[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2 + 0.5 \cdot I \cdot \dfrac{v}{R}[/tex]

Which gives;

(16 - 12)×9.81×4.6 = 0.5×(16+12)×v² + 0.5×(0.5×5×0.3²)× [tex]\left(\dfrac{v}{0.3} \right)^2[/tex]

Solving gives, v ≈ 21.93 m/s

We have;

v ≈ 3.544

v² = 2·a·h

[tex]a = \dfrac{v^2}{2 \times h}[/tex]

Which gives;

[tex]a = \dfrac{3.544^2}{2 \times 4.6} \approx 1.365[/tex]

v = a×t

[tex]t = \dfrac{v}{a} = \dfrac{3.544}{1.365} \approx 2.596[/tex]

The time it takes the the block m₁ to hit the floor, t ≈ 2.596 seconds

b. When the mass of the pulley is neglected, we have;

[tex]m_1 \cdot g \cdot h - m_2 \cdot g \cdot h = 0.5 \cdot (m_1 + m_2) \cdot v^2[/tex]

(16 - 12)×9.81×4.6 = 0.5×(16+12)×v²

180.504 = 14·v²

[tex]v = \sqrt{\dfrac{180.504}{14} } \approx 3.591[/tex]

[tex]a = \dfrac{3.591^2}{2 \times 4.6} \approx 1.401[/tex]

[tex]t = \dfrac{v}{a} = \dfrac{3.591}{1.401} \approx 2.562[/tex]

The time it takes the the block m₁ to hit the floor,  if the mass of the pulley were neglected, t ≈ 2.562 seconds.

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