If a wave hits a smooth surface at an angle of incidence of 40 degrees, the angle of reflection is

Answer:

The power of the quarry belt is, P = 2.8 kW

Explanation:

Given data,

The mass of the sand, m = 1800 kg

The displacement of the quarry belt, s = 9.5 m

The initial velocity of the sand, u = 0

The final velocity of the sand, v = 0.28 m/s

The acceleration of the belt, a = (v-u)/ t

                                                   = 0.28/10

                                                  = 0.028 m/s²

The net force acting on the quarry belt is,

                         F = mg + ma

                             = m (g +a)

                             = 1800 (9.8 + 0.028)

                            = 17690 N

The work done by the quarry belt,

                         W = F S

                              = 17690 x 9.5

                              = 168058 N

The power of the quarry belt,

                          P = W / t

                              = 168058 / 60

                              = 2800 W

                              = 2.8 kW

Hence, the power of the quarry belt is, P = 2.8 kW

Answer:

The time depends on the distance that they have to travel

[tex]x(t) = \frac{0.3846t^{2} }{2}[/tex]

Explanation:

The only horizontal force exerts over the car and you, it is the force that your friend is applied

Newton's Second Law of Motion defines the relationship between acceleration, force, and mass, thus

[tex]\sum{F} = ma[/tex]

        550 = 1430a

            a = 0.3846 m/s2

The car and you have a motion under constant acceleration, then theirs position to a time-based is:

[tex]x(t) = x_{0} + v_{0}t +\frac{at^{2} }{2}[/tex]

By the initial conditions

[tex]x(t) = \frac{at^{2} }{2}[/tex]

[tex]x(t) = \frac{0.3846t^{2} }{2}[/tex]

The time depends on the distance that they have to travel  

Explanation:

Here some of the pairs of terms are given. We need to find the incorrect relation. The relation are as follows :

1. amplitude of a sound: intensity of the sound

2. frequency of sound waves: number of wavelengths

3. quality of a sound: frequency of the sound

4. frequency of sound waves: loudness of the sound

The intensity of sound is directly proportional to the square of its amplitude. So, relation 1 is correct.

If f is the frequency and [tex]\lambda[/tex] is the wavelength. Speed of sound wave is,

[tex]v=f\lambda[/tex]

So, relation 2 is correct.  

The quality of sound wave depends on its amplitude. So, relation 3 is not correct.

The vibration depends on two factors i.e amplitude and the frequency. The loudness of sound depends on the amplitude of wave. So, relation 4 is incorrect.

The incorrectly related pair in the given set is "quality of a sound: frequency of the sound" because it is the unique combination of frequencies and intensities, and not only the frequency, that determines the quality (timbre) of a sound.

In the provided set of term pairs, the pairing "quality of a sound: frequency of the sound" is incorrectly related. While the frequency of a sound wave indeed influences the pitch of a sound, it doesn't determine the quality (or timbre) of the sound directly. The timbre of a sound is determined by the unique set of frequencies and intensities produced by a musical instrument or a voice, not by the frequency of the sound alone.

On the other hand, aspects like the "amplitude of a sound: intensity of the sound" and "frequency of sound waves: number of wavelengths" are correctly related. Larger-amplitude waves indicate greater pressure maxima and minima, thus higher intensity sounds. Similarly, the frequency is related to the number of wavelengths that pass in a given amount of time.

The pair "frequency of sound waves: loudness of the sound" can be both correct and incorrect depending on interpretation. Loudness is perceived volume and is influenced by both intensity and frequency. At the same frequency, greater intensity equates to more loudness. However, if the frequency changes without adjusting intensity, it can affect our perceived volume due to the varying sensitivity of human ears at different frequencies. Thus, frequency alone does not specify 'loudness'

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Answer:

Explanation:

During any collision whether elastic or inelastic, force of action and reaction are generated at the point of touch for a brief period during which they remain in touch with each other. These action and reaction forces are equal and opposite. Impulse is defined by force multiplied by time duration . Hence we can say that during any collision , the impulse generated are equal and opposite .

In inelastic collision , colliding objects coalesce with each other . Hence in the given case , collision is inelastic.

In elastic collision , when an object collides with a similar object at rest , there is exchange of velocity . In the given case also , similar objects are colliding and exchange of velocity is taking place.So it is an example of elastic collision.

In an inelastic collision , momentum is conserved. So final total momentum is equal to initial total momentum.

When a cannon fires a cannonball, the cannon will recoil backward because the total momentum is conserved.  

Answer:

36 hours

Explanation:

Transmission of Borrelia burgdorferi is most likely to occur after a tick feeds a minimum of 36 hours.

The infected ticks are most likely to transmit infection after approximately 2 or more days after feeding. If it remains unchecked, the B. Burgdorferi may trigger systemic infection by passing through the bloodstream and being rooted in different body tissues.

Answer:

A) Motion-sensor activated cameras

Explanation:

Motion sensor activated cameras are the ones which actuate a trigger when any sort of motion is detected by them. They can play alarming sound, send e-mails and zoom the frame of motion as a response to the detected motion.

Motion sensor cameras can be of two types:

1. Fixed view camera

2. Rotating view camera

Fixed view cameras are in fixed position and they always point in the same direction for the view while the rotating cameras are the ones which can change their direction of view.

For a small and high security space we can use the fixed type camera so that we do not miss any suspicious movements. For wide areas and distant vision we use rotating cameras.

Answer:

The charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] respectively.

Explanation:

It is given that,

Radius of the conducting sphere, r = 0.15 m

Potential, V = 200 V

Potential on the surface of sphere is given by :

[tex]V=\dfrac{kq}{r}[/tex]

q is the charge on the sphere

[tex]q=\dfrac{Vr}{k}[/tex]

[tex]q=\dfrac{200\times 0.15}{9\times 10^9}[/tex]

[tex]q=3.34\times 10^{-9}\ C[/tex]

Charge per unit area is called charge density on the surface. it is given by :

[tex]\sigma=\dfrac{q}{A}[/tex]

[tex]\sigma=\dfrac{q}{4\pi r^2}[/tex]

[tex]\sigma=\dfrac{3.34\times 10^{-9}}{4\pi (0.15)^2}[/tex]

[tex]\sigma=1.18\times 10^{-8}\ C/m^2[/tex]

So, the charge and the charge density on the surface of a conducting sphere is [tex]3.34\times 10^{-9}\ C[/tex] and [tex]1.18\times 10^{-8}\ C/m^2[/tex] respectively. Hence, this is the required solution.

Final answer:

The charge on the conducting sphere with a radius of 0.15 m and a potential of 200V is approximately 3.34 × 10⁻⁶ C, and the surface charge density is approximately 1.18 × 10⁻⁵ C/m².

Explanation:

To find the charge (q) and the surface charge density (σ) on the surface of a conducting sphere with a given potential (V), we can use the relationship between potential, charge, and radius of a sphere, along with the equation that relates charge to surface charge density.

The potential (V) of a conducting sphere is given by V = (k × q)/R, where k is Coulomb's constant (k = 8.99 × 10⁹ N.m²/C²), q is the charge on the sphere, and R is the radius of the sphere.

In this case, V = 200V and R = 0.15m. Using the formula, we can find the charge q on the sphere:

V = (k × q)/R
200V = ((8.99 × 109 N.m²/C²) × q)/0.15m
q = (200V × 0.15m)/(8.99 × 10⁹ N.m²/C²)
q ≈ 3.34 × 10⁻⁶ C

Once we have q, we can use the equation q = σ(4πR2) to find the surface charge density σ:

σ = q / (4πR²)
σ ≈ 3.34 × 10⁻⁶ C / (4π × (0.15m)²)
σ ≈ 1.18 × 10⁻⁵ C/m²

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

[tex]-h= vsinA\times t-gt^2/2[/tex]

putting values h=15 m, v=0.8

[tex]-15 = 0.8vt - 4.9t^2[/tex]  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

[tex]-15 = 0.8\times40/0.6 - 4.9t^2[/tex]

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = [tex]-H =v sinA t - gt^2/2[/tex]

⇒ [tex]4.9t^2 - 8.9\times0.8t - 100 = 0[/tex]

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

Answer:

The acceleration of the satellite is [tex]0.87 m/s^{2}[/tex]

Explanation:

The acceleration in a circular motion is defined as:

[tex]a = \frac{v^{2}}{r}[/tex]  (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

[tex]v = \frac{2 \pi r}{T}[/tex] (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ([tex]1.50x10^{7} m[/tex]) and the Earth radius ([tex]6.38x10^{6} m[/tex]) :

[tex]r = 1.50x10^{7} m+6.38x10^{6}m [/tex]

[tex]r = 21.38x10^{6}m [/tex]

Then, equation 2 can be used:

[tex]T = 8.65 hrs \cdot \frac{3600 s}{1hrs}[/tex] ⇒ [tex]31140 s[/tex]

[tex]v = \frac{2 \pi (21.38x10^{6}m)}{31140s}[/tex]

[tex]v = 4313 m/s[/tex]

Finally equation 1 can be used:

[tex]a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}[/tex]    

[tex]a = 0.87 m/s^{2}[/tex]

Hence, the acceleration of the satellite is [tex]0.87 m/s^{2}[/tex]

The acceleration of the satellite is 0.87 m/s²

The given parameters;

distance of the satellite, r₁ = 1.5 x 10⁷ m

radius of the earth, r₂ = 6.38 x 10⁶ m

time of motion, t = 8.65 hours

The radius of the satellite is calculated as;

R = r₁ + r₂

R = (1.5 x 10⁷) m + (6.38 x 10⁶) m

R = 2.138 x 10⁷ m

The time of motion of the satellite in seconds;

T = 8.65 hours x 3600 s

T = 31,140 s

The velocity of the satellite is  calculated as;

[tex]V = \frac{2\pi R}{T} \\\\V = \frac{2\pi \times (2.138\times 10^7)}{31, 140} \\\\V = 4314.45 \ m/s[/tex]

The acceleration of the satellite is  calculated as;

[tex]a_c = \frac{V^2}{R} \\\\a_c = \frac{4314.45^2}{2.138\times 10^{7}} \\\\a_c = 0.87 \ m/s^2[/tex]

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Answer:

468449163762.0812 W

Explanation:

m = Mass = [tex]\rhoV[/tex]

V = Volume =[tex]\dfrac{4}{3}\pi r^3[/tex]

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

[tex]\rho[/tex] = Density = 19 g/cm³

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

A = Area = [tex]\pi R^2[/tex]

I = Intensity = [tex]\dfrac{P}{4\pi r^2}[/tex]

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

[tex]F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}[/tex]

According to the question

[tex]F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W[/tex]

The power required of the light source is 468449163762.0812 W

Final answer:

To calculate the power required of the light source, you need to find the upward radiation force exerted on the sphere, which should match the downward gravitational force acting on the sphere. The power is given by P = I * A.

Explanation:

To calculate the power required of the light source, we first need to find the upward radiation force exerted on the sphere, which should match the downward gravitational force acting on the sphere. The radiation force is given by the formula:

Fr = (P / c)A

where Fr is the radiation force, P is the power of the light source, c is the speed of light, and A is the cross-sectional area of the sphere.

Since the sphere is totally absorbing, all the light incident on it is absorbed, so the intensity of the light is given by:

I = P / A

where I is the intensity of the light. Rearranging the equation, we get:

P = I * A

Substituting the value of A for the area of the sphere, we can find the power required.

Answer:

C. Quadruple

Explanation:

[tex]m[/tex] = mass

[tex]v_{i}[/tex] = initial speed = [tex]v[/tex]

[tex]K_{i}[/tex] = initial kinetic energy

Initial kinetic energy is given as

[tex]K_{i} = (0.5) m v_{i}^{2}\\K_{i} = (0.5) m v^{2}[/tex]

[tex]v_{f}[/tex] = Final speed = [tex]2 v[/tex]

Final kinetic energy is given as

[tex]K_{f} = (0.5) m v_{f}^{2}\\K_{f} = (0.5) m (2v)^{2}\\K_{f} = 4 (0.5) m v^{2}\\K_{f} = 4 K_{i}[/tex]

Hence the kinetic energy is quadruple

Answer:D

Explanation:

D. About half of first marriages end in divorce. - Gradpoint

Answer:

1) It will be potential to kinetic energy. 2) kinetic energy is partly transformed into heat and partly spent in generating elastic waves within the two colliding bodies.

Explanation:

We have to first understand the principle of energy conservation and that it is that energy is transformed from one form to another, from potential energy to kinetics or vice versa. When a body is immobile at a height relative to a reference point it is said that at that height the body has potential energy, as the reference point is taken as a point where all potential energy is zero and has been transformed into kinetic energy

1)

In the moment when the ball is in the highest point of elevation with respecto to the reference point the potential energy will be:

[tex]E_{p} =m*g*h\\where:\\E_{p}= potential energy [J]\\m= mass [kg]\\h=elevation [m], respect to the reference point or level[/tex]

When the ball is in the lowest point the energy is kinetic and it will be defined by:

[tex]E_{k} =E_{p}\\E_{k} =\frac{1}{2}*m*v^{2}  \\where:\\E_{k} =kinetic energy [J]\\v= velocity of the body [m/s][/tex]

2)

When the ball is rolling and moving with a velocity and hits a wedge it will be kinetic energy transformed into kinetic energy but with less amount of the initial energy because the lost kinetic energy is partly transformed into heat and partly spent in generating elastic waves within the two colliding bodies.

Answer:

Cis, Trans.

Explanation:

Rhodopsin also known as visual purple, pigment which contains sensory protein that helps to convert light into an electrical signal. Rhodopsin present in wide range of organisms from bacteria to vertebrates.

Rhodopsin is composed of opsin, and 11-cis-retinaldehyde which is derived from vitamin A. When the eye contact with light the 11-cis component converted to all trans-retinal, which results in the changes in configuration fundamental in the rhodopsin molecule.

The oldest star cluster is cluster 2, which has B type main-sequence turnoff stars. These stars are cooler and less massive, implying that they have been burning their hydrogen fuel longer than the A and G types in the other clusters.

The age of a star cluster can be determined by the spectral type of its main-sequence turnoff stars. In the main sequence, stars with higher masses tend to burn out faster because they deplete their hydrogen fuel more quickly. When these stars fuse all their hydrogen, they leave the main sequence, becoming red giants. This point is the 'turn-off' point.

Spectral type is related to a star's mass and temperature. Type A stars are hotter and more massive than type G, who are in turn hotter and more massive than type B. As a result, cluster 1 with the A type turn-off stars is the youngest, followed by the G type in cluster 3 and finally, the B type in cluster 2 is the oldest.

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The oldest star cluster can be determined by looking at the main-sequence turnoff point in an H-R diagram. In this case, the cluster with main-sequence turnoff stars of spectral type G is the oldest.

The main-sequence turnoff point in an H-R diagram is a key indicator of the age of a star cluster. The turnoff point represents the stage at which stars begin to leave the main sequence and evolve into red giants. In general, the lower the turnoff point, the older the cluster. Therefore, based on the given information, cluster 3 with main-sequence turnoff stars of spectral type G is the oldest.

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a) The force is repulsive

b) The force is attractive

Explanation:

Every magnet has a magnetic field around it. It is possible to distinguish two different poles in the magnet, according to the direction of the magnetic field: in particular, the lines of the field go out from the North pole and go into the South Pole. When a magnet is broken, two new magnets are formed, each of them having its own north and south pole.

The force between two magnets can be either attractive or repulsive, depending on which poles are facing each other. We have the following situation:

Therefore, we have the following situations in this problem:

a)

Here we are holding the two north poles together: since they are like poles, they repel each other, so the force in this case is repulsive

b)

Here we are holding a north pole and a south pole together: since they are opposite poles, they attract each other, so the force in this case is attractive

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Answer:

20.6 cm

Explanation:

charge per cm = 0.14 μC

number of electrons (e) =  [tex]1.8 x 10^{13}[/tex]

to get the length of tape pulled we can apply the formula below

length of tape = [tex]\frac{magnitude of charge of the electrons}{charge per cm}[/tex]

therefore we need to find the magnitude of the charge of the electrons

         [tex]1.8 x 10^{13} electrons = 1.8 x 10^{13} x 1.602 x 10^{-19} = 2.88μC}[/tex]

now that we have the magnitude of the charge, we can find the length of the tape

A length of 20.6 cm of tape must be pulled if a total of 1.8 × 10^13 electrons are transferred during the pulling, given that the tape acquires 0.14 microcoulombs of charge per centimeter.

To find out what length of tape must be pulled if a total of 1.8 × 10^13 electrons are transferred during the pulling, we first need to determine the total charge in coulombs transferred to the tape. Since each electron has a charge of approximately -1.602 × 10^-19 coulombs, we can multiply this value by the number of electrons to get the total charge:

Total charge = Number of electrons × Charge per electron = 1.8 × 10^13 × -1.602 × 10^-19 C = -2.8836 × 10^-6 C.

Since the charge is negative and we are looking for the positive charge transferred to the tape, we take the absolute value of this charge. Now, given that each centimeter of tape acquires 0.14 microcoulombs (or 0.14 × 10^-6 C), we can divide the total charge by the charge per centimeter to find the length of the tape:

Length of tape = Total charge / Charge per cm = 2.8836 × 10^-6 C / (0.14 × 10^-6 C/cm) = 20.6 cm

Therefore, a length of 20.6 cm of tape must be pulled to have 1.8 × 10^13 electrons transferred to it, assuming the tape was electrically neutral before being pulled.

Answer:

Plasma

Explanation:

For a fusion reaction to take place, there must be conditions in which the particles have extreme thermal kinetic energies, in this way the collisions that cause the nuclear fusion are generated. Therefore, it is necessary to reach very high temperatures, in which the state of matter will necessarily be plasma.

Answer:

3801.13 N

Explanation:

Pressure exerted on a surface is equivalent to applied force divided by the cross sectional area. Then, the applied force will be equal to the product of the pressure exerted and the cross sectional area.

Where given:

Atmospheric pressure (P1) = 1.013*10^5 Pa

T1 =  20+273.15 = 293.15 K

P2 = ?

T2 = 120+273.15 = 393.15 K

Using the gas equation: P1/T1 = P2/T2

Therefore, P2 = P1*T2/T1 = 1.013*10^5 *393.15/293.15 = 13.6*10^4 Pa

The net pressure = P2 - P1 = 13.6*10^4 - 1.013*10^5 = 34.6 kPa

The net force [tex]F_{120} = net pressure * area[/tex]

Area = 0.11 m^2

Thus:  

The net force [tex]F_{120} = 34555.7*0.11[/tex] = 3801.13 N

Explains the concept of pressure cookers and the force latches must withstand due to high pressure inside. Uses physics principles to calculate the force required.

Pressure cookers work by increasing the boiling temperature through high pressure inside the sealed vessel. In this scenario, to calculate the force the latches must withstand, you can utilize the formula for pressure, force, and area.

Answer:

Transform active margins are associated with which type of boundary?

Transform boundary

Explanation:

The transform boundary is a boundary where one plates(crust) slides past another plate horizontally. This kind of plate movement have been detected to exist between the interaction of the North pacific plates(continental plate) and the pacific plates(oceanic plates) .

At the transform margin the crust are usually broken. But overall crust are neither created nor destroyed . The transform margin region are active as it is marked by shallow-focus earthquakes .

Along the fractured zone where this transform movement occurs is known to create extensive transform faults .Notable transform fault that exist in this kind of boundary(transform) is the San Andrea fault and Alpine Fault.  

The motion of this plates can occur on a single fault or on a group of faults.

Answer:

Determine the blood glucose level

Explanation:

Determining the blood glucose level of the infant will uncover if the child is affected with hypoglycemia. The increase in the Apgar scores from 7 to 9 in five minutes shows that the newly born child is adapting very well to the outside life.

Explains the significance of Apgar scores in assessing a newborn's health after birth.

The Apgar score is a quick evaluation of a newborn's health status with scores ranging from 0 to 10 based on heart rate, respiration, muscle tone, reflex response, and color. A score of 7 at 1 minute and 9 at 5 minutes are generally positive signs, indicating improvements in the baby's condition after birth. Based on these scores, continuous monitoring and supportive care should be provided to ensure the baby's well-being.

Answer:

The space in seconds that will be kept = 2.27 seconds

Explanation:

S = d/t..................... Equation 1

making t the subject of formula in the equation above,

t = d/S.................... Equation 2

Where S = speed, d = distance, t = time.

Conversion: (i)if 1 mph = 0.44704 m/s,

                then, 30 mph = 30×0.44704    

               = 13.41 m/s

              (ii) If 1 foot = 0.3048 m

           then, 100 foot = 30.48 m.

Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m

Substituting these values into equation 2

t = 30.48/13.41

t = 2.27 seconds.

Therefore the space in seconds that will be kept = 2.27 seconds

Answer:

np

Explanation:

If

n= number of trials

p= probability of success

then the expected number of successes in a binomial experiment is given by

E(X) = n×p.

standard deviation

σ= √(npq)

where q is probability of failure.

The expected number of successes in a binomial experiment is calculated as μ = np, where 'n' is the number of trials and 'p' is the probability of success on each trial. The trials are independent and occur under identical conditions. The binomial distribution can approximate the normal when np and nq are both greater than five.

The formula for the expected number of successes in a binomial experiment with n trials and probability of success​ p is given by μ = np, where 'n' denotes the number of independent trials and 'p' represents the probability of success on one trial. For a binomial distribution, there are only two possible outcomes per trial, labeled success and failure. These trials are independent and possess identical conditions. Moreover, the outcomes of a binomial experiment fit a binomial probability distribution. The random variable X signifies the number of successes, with outcomes x = 0, 1, 2, 3, ..., n.

Further, it's crucial to note that the probability of failure, 'q', is represented by q=1-p. So, both np (number of trials times probability of success) and nq (number of trials times probability of failure) should be greater than five (np > 5 and nq > 5) for the binomial distribution of a sample to approximate the normal distribution.

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Answer:

If the mass is tripled, the kinetic energy is also tripled

Explanation:

Kinetic Energy: This is the energy possessed by a body due to motion.

The S.I unit of kinetic energy is Joules.

Mathematically it can be represented as,

Ek = 1/2mv²................ Equation 1

Where Ek = kinetic energy, m = mass, v = velocity of the body.

From the equation above,

(i) Kinetic Energy is directly proportional to mass

(ii) Kinetic energy is directly proportional to velocity squared.

When the mass is tripled,

Ek₁ = 3/2mv².................... Equation 2

Comparing equation 1 and 2,

Ek₁ = 3×Ek

Therefore if the mass is tripled, the kinetic energy is also tripled

Tanungin mo si mama mo Alam nya yan

Explanation:

Or search ka sa google

Answer:

Tire blowout

Explanation:

Tire Blowout

A tire experiences a blowout when it rapidly and explosively loses the inflation pressure. It can happen when some sharp object cuts or tears the surface and the air is suddenly released, making things worse.

The driver frequently loses control of the traveling direction and can be involved in serious accidents. It is recommended not to use brakes during or after the blowout because it could bring little help or make things worse.

Focusing on steering and let the vehicle softly brake.

Answer:

(a) The greater the frequency of the incident light is, the greater is the stopping potential.

Explanation:

The stopping potential is crucial to terminate the braking of the photoemitted electrons, stopping the current completely.

Since the kinetic energy of electrons depends on the light's incidence frequency (rather than the intensity), therefore, the stopping potential is proportional to the light's incidence frequency.

The cutoff frequency, in turn, is a limiting frequency below which no  photoelectric effect occurs. The cutoff frequency depends on the material from which it is made the emitting surface (and its work function).

Answer:

The time taken by the projectile to fall to the ground is 1 second.

Explanation:

It is given that,

Acceleration due to gravity on the planet Zorb, [tex]a=10\ m/s^2[/tex]

Angle of projectile is 30 degrees

Initial velocity of the projectile, u = 10 m/s

Let t is the time taken for the projectile to fall to the ground. Using second equation of motion to find it

[tex]h=u\ sin\theta t+\dfrac{1}{2}at^2[/tex]

When it fall to the ground, h = 0 and a = -g

[tex]10\ sin(30)t-\dfrac{1}{2}\times 10t^2=0[/tex]

On solving the above quadratic equation, we get the value of t as,

t = 1 second

So, the time taken for the projectile to fall to the ground is 1 second. Hence, this is the required solution.

Answer:

c

Explanation:

Answer:

p= 1.50289×10⁷ N/m²

Explanation:

Given

HA = (564 m)................(River Elevation)

HB = (2096 m).............(Village Elevation)

Area = A =(π/4){Diameter}² = (π/4){0.15 m}² = 0.017671 m²

ρ = (1 gram/cm³) = (1000 kg/m³)........(Water Density)

p(pressure)=?

Solution

p=PA - PB

p= ρ*g*HB - ρ*g*HA

p= (ρ*g)*(HB - HA)

p= (1000×9.81 )×{2096  - 564}  

p= 1.50289×10⁷ N/m²

Answer:

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