A symmetric projectile spends a total of 6 seconds in the air. How long does it take to reach its peak height? The projectile has a starting velocity of 8m/s. What is its horizontal range? What is the projectile's horizontal acceleration? What is the projectile's vertical acceleration?

Both the objects travel equal distance before stopping.

Explanation:

Given-

Mass of object 1, m₁ = 4kg

Speed of object 1, v₁ = 2m/s

Mass of object 2, m₂ = 1kg

Speed of object 2, v₂ = 4m/s

Force₁ = Force₂ = F

Distance, s = ?

We know,

[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 distance

When the brake is applied, the object comes to rest and the final velocity, v becomes 0. So,

[tex]s = \frac{u^2}{2a}[/tex]

We know,

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

The stopping distance becomes,

[tex]s = \frac{u^2m}{2F}[/tex]

For object 1:

[tex]s = \frac{(2)^2 X 4 }{F}[/tex]

[tex]s = \frac{16}{F}[/tex]

For object 2:

[tex]s = \frac{(4)^2 X 1}{F}\\ \\s = \frac{16}{F}[/tex]

For both the objects the distance travelled is same.

Therefore, both the objects travel equal distance before stopping.

The maximum safe speed on a curved road can be calculated using the formula v = √(2.3r), where v represents speed in miles per hour and r is the radius of curvature in feet. By substituting the given radius of curvature into the formula, we can find the maximum safe speed. In this case, the maximum safe speed is approximately 29.18 miles per hour.

The formula v = √(2.3r) models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r, in feet. To find the maximum safe speed at a given radius of curvature, we can substitute the value of r into the formula and solve for v. In this case, the radius of curvature is given as 370 feet, so we substitute r = 370 into the formula:

v = √(2.3 * 370)

Simplifying the expression, we get:

v = √(851)

Taking the square root, we find that the maximum safe speed is approximately v ≈ 29.18 miles per hour.

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Final answer:

The force of gravity and air resistance do work on a simple pendulum, with gravity doing positive and negative work as the pendulum moves, and air resistance dissipating energy from the system. Tension in the string does not do work since it is always perpendicular to the pendulum's motion.

Explanation:

In the context of a simple pendulum, the forces capable of doing work are the force of gravity and air resistance. The force of gravity does work on the pendulum as it swings back and forth because it has a component along the direction of the pendulum's movement. Specifically, as the pendulum swings, gravity pulls it downward, causing it to accelerate towards the lowest point of its path. Upon reaching this lowest point, the pendulum has maximum kinetic energy because the potential energy due to its elevated position has been converted into kinetic energy. The pendulum then slows down as it climbs against gravity until it stops momentarily at the highest point of its swing, and gravity starts doing negative work, converting kinetic energy back into potential energy. The tension in the string, although a force present in the pendulum system, does no work because it acts perpendicular to the direction of the pendulum's motion at all times.

Air resistance does work, although typically very small, by opposing the motion of the pendulum and thus removing energy from the system, mostly in the form of thermal energy due to the friction between the air and the pendulum bob. In simple pendulum motion, air resistance is usually considered negligible, but it does contribute to the eventual stopping of the pendulum due to energy dissipation.

The force of gravity does work in both directions, the tension in the string does no work, and air resistance does work by dissipating the pendulum's energy.

Force of Gravity (Weight): The force of gravity does work on the pendulum bob. Gravity acts downward and provides the restoring force that accelerates the pendulum back toward its equilibrium position. As the pendulum swings, gravity causes the bob to move, thereby doing positive work as the bob descends and negative work (taking energy away from the pendulum's motion) as the bob ascends.

Tension in the String: The tension in the string does no work on the pendulum bob. This is because the tension force always acts perpendicular to the direction of the bob's instantaneous motion. Work is defined as the force component in the direction of displacement times the displacement itself. Since there is no displacement in the direction of the tension force, it does no work.

Air Resistance: Air resistance does work on the pendulum bob, but in the opposite sense to gravity. Air resistance acts against the direction of motion of the bob, causing it to lose energy over time. This dissipation of energy due to air resistance results in damping, gradually reducing the amplitude of the pendulum's oscillation.

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

   mass per unit length ([tex]\mu[/tex]) = [tex]\frac{M}{l}[/tex] = 1 kg/m

      Tension = 4 N

    Speed (v) = [tex]\sqrt{\frac{F}{\mu}}[/tex]

So, when F is doubled then change in value of F will be as follows.

               F = 4 + 4 = 8 N

Therefore, speed will be calculated as follows.

           v = [tex]\sqrt{\frac{8 N}{1 kg/m}}[/tex]

              = 2.8 m/s

Thus, we can conclude that the wave speed is 2.8 m/s.

Answer:

-7.8×[tex]10^{2}[/tex] N

Explanation:

see the attachment plz

Answer:

(a)

[tex]\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s[/tex]

(b)

1120 N

Explanation:

Change in velocity, [tex]\triangle v[/tex] is given by subtracting the initial velocity from the final velocity and expressed as [tex]\triangle v= v_f -v_i[/tex]

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

[tex]\triangle v=0-8=-8\ m/s[/tex]

To find [tex]m\triangle v[/tex] then we substitute 7 kg for m and -8 m/s for [tex]\triangle v[/tex] therefore [tex]\triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s[/tex]

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

[tex]a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}[/tex]

Substituting t with 0.05 s then [tex]a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}[/tex]

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

Answer:

Orbital period of satellite is 5.83 x 10³ s

Explanation:

The orbital period of satellite revolving around Earth is given by the equation :

[tex]T=\sqrt{\frac{4\pi ^{2} (R+h)^{3} }{GM} }[/tex]      .....(1)

Here R is radius of Earth, h is height of satellite from the Earth's surface, M is mass of Earth and G is gravitational constant.

In this problem,

Height of satellite, h = 500 km = 500 x 10³ m

Substitute 6378.1 x 10³ m for R, 500 x 10³ m for h, 5.972 x 10²⁴ kg for M and 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻² for G in equation (1).

 [tex]T=\sqrt{\frac{4\pi ^{2} [(6378.1+500)\times10^{3} ]^{3} }{6.67\times10^{-11} \times5.972\times10^{24} } }[/tex]

T = 5.83 x 10³ s

Answer:

lead

Explanation:

Lead was once widely used in the United States as a gasoline additive.

Addition of lead is in the form of tetra ethyl lead(II).

It helps to improve the octane rating of gasoline and to produce more useful energy via each combustion step.

Answer:

C

Explanation:

Both technicians are correct.

Cheers

Technician B is correct because the ECT and IAT should read similarly after a car has been sitting for a while, reflecting the same ambient temperature. TSBs can address a variety of issues, not necessarily involving a specific DTC as Technician A suggests.

Technician B is correct. Technical Service Bulletins (TSBs) typically address widespread problems or issues found in a particular model of a vehicle. While these bulletins may include information about specific Diagnostic Trouble Codes (DTCs), they do not always involve a specific stored DTC, as Technician A suggests. TSBs can cover various issues, including non-DTC-related performance, noise, and other iterative improvements that are not related to any stored codes.

Technician B is accurate in saying that the Engine Coolant Temperature (ECT) sensor and the Intake Air Temperature (IAT) sensor readings should be close to the same after the vehicle has been sitting for several hours, typically overnight. This is because both the coolant in the engine and the air in the intake manifold would have reached ambient temperature, reflecting similar temperatures if the vehicle's sensors are functioning correctly.

Answer choice B, which credits Technician B, is the correct option.

Answer:

0.28 m/s

Explanation:

From the equation of a moving wave,

V = λf.................. Equation 1

Where V = speed of the water wave, λ = wave length of the water wave, f = frequency of the water wave.

Given: λ = 7.0 cm = 0.07 m, f = 4 Hz.

Substitute into equation 1

V = 0.07(4)

V = 0.28 m/s.

Hence the speed of the water wave = 0.28 m/s

Final answer:

The force exerted by the toy locomotive on the caboose is 3.08 N, which is calculated by adding the force required to overcome friction and the force necessary for acceleration according to Newton's second law.

Explanation:

To determine the magnitude of the force exerted by the toy locomotive on the caboose, we can use Newton's second law, which states that the force is equal to the mass times the acceleration (F = ma).

However, we also need to account for the frictional force acting on the caboose. The total force required to accelerate the caboose includes the force to overcome friction and the force needed for acceleration:

Step 1: Calculate the force needed to overcome friction, which is already given as 0.48 N.

Step 2: Determine the force needed to accelerate the caboose using the formula F = ma, where m is the mass of the caboose (1.0 kg) and a is the acceleration (2.6 m/s2).

Step 3: Add the force to overcome friction to the force required for acceleration to get the total force exerted by the locomotive on the caboose.

Step 2 in detail: F = ma = 1.0 kg × 2.6 m/s2 = 2.6 N.

Step 3 in detail: The total force exerted by the locomotive on the caboose is 2.6 N (to accelerate) + 0.48 N (to overcome friction) = 3.08 N.

Therefore, the force exerted by the toy locomotive on the caboose is 3.08 N.

Answer:

The answer to the question is;

(a) The amount of work required to withdraw the mica sheet is 4.0×10⁻⁵ J

(b) The potential difference across the capacitor after the mica is withdrawn is 500 V.

Explanation:

To solve the question, we note that

The Capacitance of the capacitor is proportional to the dielectric constant, that is removal of the dielectric material will reduce the capacitance of the capacitor by a factor of the dielectric constant as follows

Energy stored in a capacitor is given by

[tex]E = \frac{1}{2} CV^{2}[/tex]

Where:

C = Capacitance of the capacitor = 2.00 nF

V = Voltage = 100 V

Therefore before the dielectric material is removed, we have

E = [tex]\frac{1}{2}[/tex]×2.00 × 10⁻⁹×100² = 1.0×10⁻⁵ J

Asfter the dielectric material is removed we have

E = 5× [tex]\frac{1}{2}[/tex]×2.00 × 10⁻⁹×100² = 5.0×10⁻⁵ J

Since energy is neither created nor destroyed, we have

Change in energy of system  = 5.0×10⁻⁵ J - 1.0×10⁻⁵ J = 4.0×10⁻⁵ J

Work done on the system to remove the mica sheet = energy change of the system = 4.0×10⁻⁵ J.

(b) From the equation [tex]C = \frac{Q}{V}[/tex]

Where

Q = Charge of the capacitor

V= Voltage

C = Capacitance of the capacitor, we have

Q, the charge of the capacitor is constant

and the final capacitance = [tex]\frac{Initial .capacitance}{5} = \frac{C_1}{5}[/tex]

Therefore we have

V₁ = [tex]\frac{Q}{C_1}[/tex]

V₂ = [tex]\frac{Q}{\frac{C_1}{5} }[/tex] = [tex]{5} *\frac{Q}{C_1}[/tex] = 5·V₁

The velocity increases by a factor of 5 and

V₂ = 5×V₁ =  5 × 100 V = 500 V.

Full Question:

What is the Lewis structure of the covalent compound that contains one nitrogen atom, one hydrogen atom, and one carbon atom?

Explanation:

This covalent compound is the hydrogen cyanide, HCN.

The following steps are used to obtain it;

Step 1. Draw a skeleton structure of the compound based on the elements

Put the least electronegative atom C in the middle with H and Cl on either side.

H-C-N

Step 2. Count the valence (outermost) electrons you can use

H + C + N = 1 + 4 + 5 = 10

Step 3. Add these electrons to give every atom an octet

You have to put a triple bond between C and N.

Note: Each bond is made up of two electrons.

The lewis structure is given in the image below;

Answer:

Magnitude of electric field = E = q/Aε0

Explanation:

Consider plates are placed at a distance of d. As given in the question the charge stored on the plates have magnitude q and given by:

                                          q = CV

And  

                                          V = q/C    ……. (i)

The capacitance is given by the following equation:  

                                         C = Aε0/d ……. (ii)

Put equation (ii) in (i) ,

                                          V = qd/ Aε0 …..(iii)    

The electric field is defined as:  

                                            E = V/d   …… (iv)

Put equation (iii) in (iv),

                                            E = qd/ Aε0d

                                            E = q/Aε0

Hence, the magnitude of electric field will be q/Aε0 .

The electric field between two plates with charges −q and +q is directed from the positive to the negative plate and can be calculated using the charge density and vacuum permittivity. For systems of parallel conducting plates, the electric field can be found using potential differences and distances, while the interaction of a charge with a dipole requires considering forces from each dipole component.

The magnitude of the electric field E between two plates with charges −q and +q can be found using the formula E = σ/ε₀, where σ is the surface charge density (σ = q/A, A is the area of the plate) and ε₀ is the vacuum permittivity. The direction of the electric field is from the positive to the negative plate. Considering a scenario with two parallel conducting plates 15 cm apart and each with a charge of −0.225 C and +0.225 C, you would first calculate the electric field strength using the aforementioned formula and then state that its direction is from the positive to the negative plate.

For the case involving three parallel conducting plates with specified potentials, you calculate the average electric field experienced by an electron by finding the potential difference between the plates and dividing it by the distance between the plates. The electric field between the plates is uniform if we assume the plates are infinitely large and parallel.

The net force on a charge −q placed equidistant from a dipole with charges +2q and −2q can be found using Coulomb's law. Since the forces due to the +2q and −2q charges will be equal in magnitude but opposite in direction, the result will be a null net force in the direction along the line connecting the two dipole charges. However, there will be a resultant force perpendicular to this line, attracting the −q charge towards the dipole's center.

Answer: 3.889N/m

Explanation:

f=(1/2)*√k/m

f=n/t

f= 21/3.3=6.4Hz

6.4*2=√k/m

12.73=√k/m

12.73^2=k/m

162.0529=k/m

Since m=24g

Convert g to kg

m=24/1000

m=0.024kg

K=162.0529*0.024

K=3.889N/m

Final answer:

The spring constant of the spring is approximately 24 N/m.

Explanation:

When a 24-gram mass is attached to a spring, causing it to make 21 complete vibrations in 3.3 seconds, we are looking at a problem that involves simple harmonic motion and can use the properties of a mass-spring system to determine the spring constant (k).

The formula to calculate the frequency (f) is:

f = number of vibrations / time
f = 21 / 3.3 s
f = 6.36363636 Hz

The frequency is related to the spring constant (k) and mass (m) as follows:

f = (1 / (2π)) * (√(k / m))

Given the mass (m) and frequency (f), we can solve for k:

k = (2πf)² * m
k = (2 * π * 6.36363636 Hz)² *0.024 kg
k ≈ 24 N/m

Therefore, the spring constant of the spring is approximately 24 N/m.

Answer:

NO

Explanation:

Acceleration is change in velocityΔv in respect to timeΔt

so if the velocity of the car is greater than the truck it does not mean that the car acceleration is greater than the truck.

Sometimes with constant velocity it means no accelaration ,but the truck may have accelaration

so, higher velocity of the car does not mean higher acceleration

Answer:Hydrogen lines will be weak if the star is too hot or too cold.

Explanation:

At higher temperatures, the hydrogen atom ionizes due to the atomic collisions. M spectral class stars are mainly the main sequence and red stars. They are in a temperature range of 3000 K which means that these stars have maximum ionized calcium lines. In this star, hydrogen atoms have electrons in the lower energy state and it is difficult to absorb photons. These stars do not have enough temperature for absorption and undergo fusion.

A negative test charge moves toward regions of higher electric potential and lower electric potential energy, which is the opposite direction of the electric field defined by positive charges.

A negative test charge will accelerate toward regions of higher electric potential and lower electric potential energy. This happens because a negative charge moves oppositely to the electric field direction, which is defined from high to low potential. When a negative charge, such as an electron, moves toward a higher potential, it is moving towards a region where it would have a lower potential energy if it were positive; however, since it is negative, its potential energy actually decreases as it moves in this direction.

Understanding electric fields and potentials, consider that a positive test charge is repelled by positive charges and attracted to negative charges. Since the field lines point away from positive charges and toward negative charges, a negative test charge would move in the direction opposite to the field lines, meaning it moves from lower to higher potential but in doing so, it lowers its electric potential energy.

Answer:

Check attachment for solution

Explanation:

Given that 12V battery

Answer:

Terminal voltage = 8.36 V

Current = 1.82 A

Explanation:

E.M.F of battery = 12V

Internal resistance of battery (r) = 2Ω

Resistance of resistor (R) = 4.6Ω

Now the formula for terminal voltage across the battery is;

V = ε - Ir

Where ε is EMF and I is electric current

Using ohms law, we know that V = IR and I = V/R.

Thus, let's put V/R for current in the potential difference equation;

V = ε - r(V/R)

Thus, lets make V the subject of the formula ;

V + (rV/R) = ε

V(1 + r/R) = ε

So, V = ε/(1 + r/R)

V = 12/(1 + (2/4.6))

V = 12/(1 + 0.4348)

V = 12/1.4348 = 8.36 V

Thus from V=IR, we can find current. So 8.36 = I(4.6)

I = 8.36/4.6 = 1.82 A

Answer:[tex]a_c=46.24\ m/s^2[/tex]

Explanation:

Given

radius of container [tex]r=13.1\ cm[/tex]

cylinder is rotating with [tex]N=2.99 rps[/tex]

Centripetal acceleration at the outer wall is given by

[tex]a_c=\omega ^2\times r[/tex]

where [tex]\omega [/tex]=Angular speed

[tex]\omega =2\pi N[/tex]

[tex]\omega =2\pi \times 2.99[/tex]

[tex]\omega =18.78\ rad/s[/tex]

[tex]a_c=(18.78)^2\times 0.131[/tex]

[tex]a_c=46.24\ m/s^2[/tex]

Answer:

7.85 m/s

Explanation:

We are given that

Mass of object=m=0.900 kg

[tex]F(x)=\alpha x-\beta x^2[/tex]

[tex]\alpha=60 N/m[/tex]

[tex]\beta=18N/m^2[/tex]

[tex]F(x)=-60x-18x^2[/tex]

U=0 when x=0

Potential energy=[tex]-\int F(x)dx[/tex]

Substitute the values

[tex]U(x)=-\int (-60x-18x^2)dx[/tex]

[tex]U(x)=60(\frac{x^2}{2})+18(\frac{x^3}{3})+C[/tex]

Using the formula

[tex]\int x^n dx=\frac{x^{n+1}}{n+1}+C[/tex]

Substitute x=0

[tex]U(0)=C\implies C=0[/tex]

[tex]U(x)=30x^2+6x^3[/tex]

[tex]x_1=0.5,x_2=1[/tex]

[tex]v_2=0[/tex]

Using law of conservation energy

[tex]\frac{1}{2}mv^2_1+U(x_1)=\frac{1}{2}mv^2_2+U(x_2)[/tex]

Substitute the values

[tex]\frac{1}{2}(0.9)v^2_1+30(0.5)^2+6(0.5)^3=0+30(1)^2+6(1)^3[/tex]

[tex]\frac{1}{2}(0.9)v^2_1+8.25=36[/tex]

[tex]\frac{1}{2}(0.9)v^2_1=36-8.25=27.75[/tex]

[tex]v^2_1=\frac{27.75\times 2}{0.9}[/tex]

[tex]v_1=\sqrt{\frac{27.75\times 2}{0.9}}[/tex]

[tex]v_1=7.85 m/s[/tex]

Answer:

x=61

y=35

Explanation:

vertical component= 70sin(30)

Horizontal component=70cos(30)

Answer:

Technician A

Explanation:

The stator is a STATIONARY component of a rotary system. The stator is a stationary component in the alternator, therefore, technician B saying that it is a rotating component is WRONG.

Also, slip rings allows the flow of electrical signals and electrical power from a stationary structure to a rrotating structure. Therefore, technician A is correct in his take that the slip rings allow current to flow from the stationary end frame to the moving rotor.

Final answer:

Technician A is correct about slip rings allowing current flow to the rotor, while Technician B is incorrect as the stator is the stationary part of an alternator.

Explanation:

Technician A is correct in saying that slip rings allow current to flow from the stationary part to the moving rotor. Slip rings are made of a ring and brushes; the ring is attached to the rotating part of a machine (like an alternator's rotor), and the brushes are attached to the stationary part (such as the end frame), facilitating electrical contact as the rotor spins.

Technician B is incorrect; the stator is the stationary part of an alternator or electric motor that contains conductors where the electric current is induced by the motion of the rotor. In an alternator, the rotor is the rotating component, not the stator.

Answer:

[tex]E_a=1124.83 J/mol[/tex]

Explanation:

Given that second order equation

K₁ = 0.0796 L/mol-s , T₁= 737⁰C

T₁ = 737 + 273 K = 1010 K

K₂ = 0.0815 L/mol-s , T₂=947°C

T₂=947+273 K= 1220 K

The activation energy given as follows

[tex]\ln\dfrac{K_2}{K_1}=\dfrac{E_a}{R}\left ( \dfrac{1}{T_1}-\dfrac{1}{T_2} \right )[/tex]

Now by putting the values we can get

[tex]\ln\dfrac{0.0815}{0.0796}=\dfrac{E_a}{8.314}\left ( \dfrac{1}{1010}-\dfrac{1}{1220} \right )[/tex]

[tex]0.023=0.00017\times \dfrac{E_a}{8.314}[/tex]

[tex]E_a=0.023\times \dfrac{8.314}{0.00017}[/tex]

[tex]E_a=1124.83 J/mol[/tex]

Therefore the activation energy will be 1124.83 J/mol

Answer:

Center of mass

Explanation:

The Doppler technique is a good method for discovering exoplanets. It uses the Doppler effect to analyze the motion and properties of the star and planet. Both the planet and the star are orbiting a common center of mass. This means that the star and the planet gravitationally attract one another, causing them to orbit around a point of mass central to both bodies.

If we use the Doppler method to measure the period with which a star alternately moves toward and away from us due to an orbiting planet, then we also know the planet's orbital period

When utilizing the Doppler method to measure the periodic shifts in a star's spectral lines caused by an orbiting planet, not only do we discern the star's radial velocity variations, but crucially, we also ascertain the orbital period of the planet. This period signifies how long it takes for the planet to complete one orbit around its host star.

The Doppler effect causes a star's light to shift towards the blue end of the spectrum as it moves closer to Earth and towards the red end as it moves away. By analyzing these shifts, astronomers can deduce the star's motion induced by the gravitational pull of its companion planet. The periodicity of these shifts corresponds directly to the planet's orbital period. This information is fundamental in understanding the planet's characteristics, including its distance from the star and its mass, contributing significantly to our knowledge of exoplanetary systems in the vast cosmos.

To know more about orbital period

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Answer:

a) τ = 4.47746 * 10^25 N-m

b) E = 2.06301 * 10^13 J

c) P = 3.25511*10^21 W

Explanation:

Given that,

The radius of earth r = 6.3781×10^6 m

The angular speed of earth w = 7.27*10^-5 rad/s

The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s

The mass of earth m = 5.972 × 10^24 kg

The inertia of sphere I = 2/5 * m* r^2

Solution:

angular acceleration of the earth from rest to w is given by α:

                               α = w / t

                               α = (7.27*10^-5) / (1.57784760 * 10^8)

                               α = 4.60754*10^-13 rad/s^2

The required torque τ is given by:

                               τ = I*α

                               τ = 2/5 * m* r^2 * α

 τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)

 τ = 4.47746 * 10^25 N-m

Power required P to turn the earth to the speed w is:

                          P = τ*w

                          P = (4.47746 * 10^25)*(7.27*10^-5)

                          P = 3.25511*10^21 W

Energy E required is :

                          E = P / t

                          E = (3.25511*10^21) / (1.57784760 * 10^8)

                          E = 2.06301 * 10^13 J

Answer: yes.

Explanation: The light that will be incidented on that metal is visible light.

It depends on 3 factors:

1. The temperature

2. The specific heat capacity of the metal

3. The thermal conductivity of the metal.

The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.

When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer. 

The photoelectric effect causes the metals to get warm when light is shined on them.

Photoelectric effect:

It states that the metals emit electrons when by absorbing electromagnetic radiation.

When light sine on the metal, electrons present on the metals absorb some energy of the light or radiations and leave their atoms. This causes the metal to heat up when light sine on it.

The formula is,

[tex]h \nu= W + E[/tex]

Where,

[tex]h[/tex]- Plank's constant

[tex]\nu[/tex] -  Threshold frequency

[tex]W[/tex] - Energy needed to remove the electron

[tex]E[/tex]-  Kinetic energy of the emitted electron

Therefore, the metal gets warm due to the Photoelectric effect.

To know more about the Photoelectric effect:

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Answer:

Ranked from left to right based on their abundance in the solar nebula:

Hydrogen and Helium-Hydrogen compounds-Rocks-Metals

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The materials in the solar nebula can be ranked from highest to lowest abundance as hydrogen and helium gas, icy and rocky planetesimals, dust and vapor, and meteorites, comets, and asteroids.

The materials that made up the solar nebula can be categorized into four general types based on their abundance. These types, ranked from highest to lowest abundance, are:

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Answer:

[tex]v=87.46m/s[/tex]

Explanation:

Objects moving in circular path would be have either centripetal or centrifugal  force.The force is either to center or away from center. When the object is moving along the circular path the centripetal force is

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

Here m is mass, v is velocity and r is radius of circular path

The acceleration is given by:

[tex]a_{r}=\frac{v^{2}}{r}[/tex]

The point of interest is lowest point on circle.The acceleration of plane at  this position point up.The speed of plane from radial acceleration equation is:

[tex]v=\sqrt{a.r}\\[/tex]

Substitute 17 m/s² for a and 450m for r

So

[tex]v=\sqrt{17m/s^{2}*450m }\\ v=87.46m/s[/tex]

Answer:

1600 cal

Explanation:

The formula

q=mΔ[tex]_{fus}[/tex][tex]H^{0}[/tex]

is used to calculate the heat required to melt a solid where q=amount of heat, m=mass and Δ[tex]_{fus}[/tex][tex]H^{0}[/tex] is the enthalphy of fusion

now we substitute, m=20g, Δ[tex]_{fus}[/tex][tex]H^{0}[/tex]=80cal/g

Therefore, q=20g x 80cal/g =1600 cal

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The heat released when 20.0 g of water at 0°C is frozen to ice is 1600 calories.

When water freezes, it releases heat energy. The quantity of heat energy needed to convert a substance from a solid to a liquid or vice versa without affecting its temperature is known as the heat of fusion.

In this case, we are asked to find the number of calories of heat released when moment 20 g of water freezes to ice at zero degrees.

To calculate this, we can use the formula:

Heat released = mass of water * heat of fusion

Substituting the given values:

Heat released = 20.0 g * 80. cal/g = 1600 cal

Therefore, 1600 calories of heat are released when 20.0 g of water freezes to ice at 0 °C.

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