On earth, a block is placed on a frictionless table on earth. When a horizontal force of 10 N is applied to the block, it accelerates at 5.3 m/s2. Suppose the block and table are set up on the moon. When a horizontal force of 5 N is applied to the block, what is the acceleration?
a.3.2m/s^2
b. 2.7m/s^2
c. 3.4m/s^2
d. 2.4m/s^2

Answer:

Part a)

[tex]v_f = 4 m/s[/tex]

Part b)

[tex]t = 0.001 s[/tex]

Part c)

[tex]d = 0.815 m[/tex]

Explanation:

Part a)

As we know that initially the grass hopper is at rest at the ground position

Now the acceleration is given as

[tex]a = 4000 m/s^2[/tex]

distance of the legs that it stretched is given as

[tex]s = 2.0 mm[/tex]

so we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 0 = 2(4000)(0.002)[/tex]

[tex]v_f = 4 m/s[/tex]

Part b)

time taken to reach this speed is given as

[tex]v_f - v_i = at[/tex]

[tex]4 - 0 = 4000 t[/tex]

[tex]t = 0.001 s[/tex]

Part c)

as the grass hopper reach the maximum height its final speed would be zero

so we will have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 4^2 = 2(-9.81) d[/tex]

[tex]d = 0.815 m[/tex]

Final answer:

The froghopper Philaenus spumarius takes off at an upward velocity of 4 m/s and reaches this velocity in 0.001 seconds. In the absence of air resistance, it can potentially jump to a height of 0.815 m, while air resistance reduces its actual jump to about 70 cm.

Explanation:

The froghopper Philaenus spumarius, known for its powerful jumping ability, can accelerate at 4.00 km/s² over a distance of 2.0 mm. To find the upward velocity (a) with which the insect takes off, the following kinematic equation can be used:v² = u² + 2as

where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration, and s is the distance.

Firstly, convert the given units to make them consistent. The acceleration becomes 4000 m/s² and the distance becomes 0.002 m. Plugging the values into the equation gives:

v² = 0 + 2(4000)(0.002) = 16

The takeoff velocity, v, is √16, which equals 4 m/s.

For the time interval (b), we use the equation:

v = u + at

So, t = v / a = 4 / 4000 = 0.001 s.

To calculate how high the insect would jump if air resistance were negligible (c), we can use the equation:

s = ut + ½at²

Since initial velocity u is 0 when coming back down, and final velocity v is 0 at the peak of the jump, we set v = 0 and solve for s:

0 = (4)2 - 2(9.81)s

s = 16 / 19.62 = 0.815 m (or 81.5 cm).

Comparing this to the actual jump height of 70 cm indicates a significant effect of air resistance on the insect's jump.

Answer:

Artherosclerosis develops in the nervous system.

Explanation:

Artherosclerosis  is a condition that develops when plaque or fatty deposits block the arteries. This restricts the blood flow through the arteries partially or completely. Artherosclerosis can develop in the arteries of heart, brain or legs.

When the arteries are blocked by the thickening of the blood vessels the blood flow to the corresponding organs reduces. Thus the supply of oxygen and other vital nutrients to these parts reduces.

Based on the location of development of plaque and the type of artery affected various conditions arise.

Examples are coronary heart disease, angina, carotid artery disease, peripheral artery disease, chronic kidney disease etc.

Answer:

Cardiovascular system

Explanation:

This is because Atherosclerosis is the build up of fats and cholesterol in your Artery walls, this can restrict blood flow in your body.Your Artery walls are apart of your Cardiovascular system.

Answer:

[tex]f_o = 132.91 Hz[/tex]

Explanation:

Frequency of the train whistle which is approaching him is given as

[tex]f_1 = (\frac{v}{v - v_s}) f_o[/tex]

[tex]f_1 = (\frac{343}{343 - 8.40}) f_o[/tex]

now similarly the frequency of the train whistle which is moving away from him is given as

[tex]f_2 = (\frac{v}{v + v_s})f_o[/tex]

[tex]f_2 = (\frac{343}{343 + 8.40})f_o[/tex]

now we know that

[tex]f_{beat} = f_1 - f_2[/tex]

so we have

[tex]6.50 = (\frac{343}{343 - 8.40}) f_o - (\frac{343}{343 + 8.40}) f_o[/tex]

[tex]6.50 = 1.025 f_o - 0.976 f_o[/tex]

[tex]f_o = 132.91 Hz[/tex]

Final answer:

The Doppler effect can be used to determine the frequency emitted by the two train whistles. By applying the formula for the Doppler effect, the equation can be derived and solved to find the frequency of each whistle. In this case, the frequency emitted by the whistle of each train is approximately 155.46 Hz.

Explanation:

The situation described in the question is an example of the Doppler effect, which explains the change in frequency of a sound wave when the source of the sound is moving relative to the observer. In this case, the boy is standing at a railroad crossing and is hearing two trains approaching and moving away from him.

When a sound source approaches the observer, the frequency of the sound waves appears higher, resulting in a higher pitch. Conversely, when a sound source moves away, the frequency appears lower, resulting in a lower pitch. The difference between the frequencies heard when the two trains are approaching and moving away from the boy is called the beat frequency, which is given as 6.50 Hz in the question.

To determine the frequency emitted by the two whistles, we can use the formula for the Doppler effect:

f' = fs * (v + vo) / (v + vs)

Where:

f' is the observed frequency

fs is the source frequency

v is the speed of sound

vo is the velocity of the observer

vs is the velocity of the source

In this case, both trains have the same speed of 8.40 m/s, so the velocity of the observer is zero. The speed of sound is given as 343 m/s. Given these values, we can substitute them into the formula.

Let's assume that the frequency emitted by the whistle of the approaching train (Train A) is fA and the frequency emitted by the whistle of the receding train (Train B) is fB.

For the approaching train:

f' = fA * (v + vo) / (v + vs)

Since the observer is not moving, the equation simplifies to:

f' = fA * v / (v + vs) .......... (1)

And for the receding train:

f' = fB * (v + vo) / (v - vs)

f' = fB * v / (v - vs) .......... (2)

We are given that the beat frequency is 6.50 Hz. This means that the difference in frequencies of the two trains is 6.50 Hz.

Substituting the values into equations (1) and (2), we get:

fA * v / (v + vs) - fB * v / (v - vs) = 6.50 Hz

fA / (v + vs) - fB / (v - vs) = 6.50 / v Hz

Since both trains have the same source frequency (fA = fB), we can simplify the equation:

2fA * vs / (v^2 - vs^2) = 6.50 / v Hz

Solving for fA, we can rewrite the equation as:

fA = (6.50 / v) * ((v^2 - vs^2) / (2vs)) Hz

Substituting the given values, we have:

fA = (6.50 / 343) * ((343^2 - 8.4^2) / (2 * 8.4)) ≈ 155.46 Hz

Therefore, the frequency emitted by the whistle of each train is approximately 155.46 Hz.

A basketball player shoots toward a basket 5.6 m away and 3.0 m above the floor, so the required initial speed is approximately 4.31 m/s.

To find the initial speed of the basketball, we can break the motion into horizontal and vertical components.

The horizontal component of the motion remains constant, as the player maintains his horizontal velocity.

The vertical component, however, is affected by gravity.

The ball is released at a height of 1.8 m above the floor and needs to go through the basket 3.0 m above the floor, which means it needs to rise by 1.2 m.

The required initial speed can be found using the equation:

vf² = vi² + 2 * a * d

where

vf is the final vertical velocity (0 m/s at the highest point)

vi is the initial vertical velocity

a is the acceleration due to gravity (-9.8 m/s²)

d is the vertical distance traveled (1.2 m).

Rearranging the equation, we get:

vi = sqrt(2 * a * d)

Plugging in the values, we have:

vi = sqrt(2 * (-9.8) * 1.2) = 4.31 m/s

Therefore, the initial speed of the basketball needs to be approximately 4.31 m/s in order to go through the basket.

Explanation:

3000 to 1,000,000 nm for IR-C

700 to 1400 nm for IR-A

Now, we are working with light, so we can use that the energy of a photon with frequency f is : E(f) = h*f

where h is the plank constant h = 6.62607015×10−34 j*s

and f is the frequency that f= c/λ, where λ is the wavelenght.

so, for IR-C the energy lies between h*c/3000 and h*c/1000000

where c is the light speed.

For IR-A the energy lies between h*c/700 and h*c/1400.

so here you can se that IR-A has a lot more energy than IR-C, you can se that the minimal energy of IR-A is h*c/1400 and the maximal of IR-C is h*c/3000 so the minimal energy of IR-A is almost twice times the maximum energy of IR-C

Answer:

ρm=4.14g/ml  :metal density

Explanation:

Conceptual analysis

We apply the formula to calculate the density:

ρ= m /V Formula (1)

Where:

ρ:density in g/mL

m: mass in g (grams)

V= volume in ml (milliliters)

Known data

[tex]V_{il}[/tex]: final volume of liquid=55.25 ml

[tex]V_{fl}[/tex]: final volume of liquid=61.00 ml

ms: metal mass=23.795 g

Problem development

The volume of the metal (vm) is equal to the change in volume of the liquid.

[tex]V_{m} = V_{fl}-V_{il}  =61.00 ml - 55.25 ml =5.75ml[/tex]

We replace the data in formula (1 )to calculate the density of the metal(ρm):

ρm= mm /Vm=23.795 g/5.75ml

ρm=4.14g/ml

Answer:

Speed of plane is 300.5 m/s at angle of 6.22 degree South of West

Explanation:

Air speed of the plane is given as

v = 264 m/s in direction 5 degree South of West

So we have

[tex]v_1 = 264 cos5 \hat i + 264 sin5 \hat j[/tex]

[tex]v_1 = 263 \hat i + 23 \hat j[/tex]

Also we have speed of air is given as

v = 37 m/s at 15 degree South of West

so it is

[tex]v_2 = 37 cos15\hat i + 37 sin15 \hat j[/tex]

[tex]v_2 = 35.74 \hat i + 9.58 \hat j[/tex]

So the net speed of plane with respect to ground is given as

[tex]v_p = v_1 + v_2[/tex]

[tex]v_p = (263 \hat i + 23 \hat j) + (35.74 \hat i + 9.58 \hat j)[/tex]

[tex]v_p = 298.74 \hat i + 32.58\hat j[/tex]

so it is

[tex]v_p = \sqrt{298.74^2 + 32.58^2}[/tex]

[tex]v_p = 300.5 m/s[/tex]

direction is given as

[tex]\theta =tan^{-1} \frac{v_y}{v_x}[/tex]

[tex]\theta = tan^{-1} \frac{32.58}{298.74}[/tex]

[tex]\theta = 6.22 degree[/tex]

Answer:

Explanation:

There are various examples around us that states " internal hold back property " and is different from frictional force or the force of Earth's gravity.

The plaster which is the mixture of cement and water has an adhesive nature and molecules inside them has a internal hold back property. They stick to the walls and ceiling and do not fall.

Though gravity is applied on it still its molecules tends to hold the ceiling firmly and do not fall upon.

Answer:

Given the equation of the particle, we know that:

[tex]x=(2.9\frac{m}{s^{3} } )*t^{3} -(2\frac{m}{s^{6} } )*t^{6}[/tex]

a) In [tex]t=0.0s[/tex] we evaluate [tex]t[/tex] in the former equation:

[tex]x_{1} =(2.9\frac{m}{s^{3} } )*(0)^{3} -(2\frac{m}{s^{6} } )*(0)^{6}=0m[/tex]

In [tex]t=1.3s[/tex]

[tex]x_{2} =(2.9\frac{m}{s^{3} } )*(1.3s)^{3} -(2\frac{m}{s^{6} } )*(1.3s)^{6}=-3,28232m[/tex]

a) So, we know that the displacement of te particle is given by:

[tex]x=x_{2}-x_{1} =-3.28232m-0m=-3.28232m[/tex]

To find it's velocity, we need to derivate the equation of position by the formula:

[tex]v_{x} =\frac{dx}{dt} =3ct^{2}-6bt^{5} =(8.7\frac{m}{s^{3} })t^{2} -(12\frac{m}{s^{6} } )t^{5}[/tex]

And evaluate this expression at each specified t:

b) [tex]v(1s)_{x} =(8.7\frac{m}{s^{3} })(1s)^{2} -(12\frac{m}{s^{6} } )(1s)^{5}=-3.3\frac{m}{s}[/tex]

c) [tex]v(2s)_{x} =(8.7\frac{m}{s^{3} })(2s)^{2} -(12\frac{m}{s^{6} } )(2s)^{5}=-349.2\frac{m}{s}[/tex]

d) [tex]v(3s)_{x} =(8.7\frac{m}{s^{3} })(3s)^{2} -(12\frac{m}{s^{6} } )(3s)^{5}=-2837.7\frac{m}{s}[/tex]

e) [tex]v(4s)_{x} =(8.7\frac{m}{s^{3} })(4s)^{2} -(12\frac{m}{s^{6} } )(4s)^{5}=-12148.8\frac{m}{s}[/tex]

To find it's acceleration, we need to derivate the equation of velocity by the formula:

[tex]a_{x} =\frac{dv}{dt} =(17.4\frac{m}{s^{3} } )t-(60\frac{m}{s^{6} } )t^{4}[/tex]

And evaluate this expression at each specified t:

f) [tex]a(1s)_{x} =(17.4\frac{m}{s^{3} } )(1s)-(60\frac{m}{s^{6} } )(1s)^{4}=-42.6\frac{m}{s^{2} }[/tex]

g) [tex]a(2s)_{x} =(17.4\frac{m}{s^{3} } )(2s)-(60\frac{m}{s^{6} } )(2s)^{4}=-925.2\frac{m}{s^{2} }[/tex]

h) [tex]a(3s)_{x} =(17.4\frac{m}{s^{3} } )(3s)-(60\frac{m}{s^{6} } )(3s)^{4}=-4807.8\frac{m}{s^{2} }[/tex]

i) [tex]a(4s)_{x} =(17.4\frac{m}{s^{3} } )(4s)-(60\frac{m}{s^{6} } )(4s)^{4}=-15290.4\frac{m}{s^{2} }[/tex]

To find the tensions in the ropes, we can use torque and set up an equation where the clockwise torque is equal to the counterclockwise torque. By solving this equation, we can determine the tensions. The tension in the rope closer to the person is 22.6 N, while the tension in the other rope is 16.6 N.

Tension in the ropes can be found by considering the forces acting on the beam. Since the beam is in equilibrium, the sum of the forces in the vertical direction is zero.

Using torque, we can determine the tension in each rope. We can set up an equation where the clockwise torque is equal to the counterclockwise torque:

Then, solve the equation to find the tensions in the ropes. The tension in the rope closer to the person will be 22.6 N, while the tension in the other rope will be 16.6 N.

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Answer:

a) -31.36 m/s

b) 50.176 m

Explanation:

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

[tex]V_{f}=V_{o} +a.t[/tex]  (1)

Where:

[tex]V_{f}[/tex] is the final velocity of the supply bag

[tex]V_{o}=0[/tex] is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)

[tex]a=g=-9.8m/s^{2}[/tex] is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

[tex]t=3.2s[/tex] is the time

Knowing this, let's solve (1):

[tex]V_{f}=0+(-9.8m/s^{2})(3.2s)[/tex]  (2)

Hence:

[tex]V_{f}=-31.36m/s[/tex]  Note the negative sign is because the direction of the bag is downwards as well.

In this case we will use the following equation:

[tex]y=V_{o}t-\frac{1}{2}gt^{2}[/tex] (3)

Where:

[tex]y[/tex] is the distance the bag has fallen

[tex]V_{o}=0[/tex] remembering the bag was dropped

[tex]g=-9.8m/s^{2}[/tex] is the acceleration due gravity (downwards)

[tex]t=3.2 s[/tex] is the time

Then:

[tex]y=-\frac{1}{2}gt^{2}[/tex] (3)

[tex]y=-\frac{1}{2}(-9.8m/s^{2})(3.2)^{2}[/tex] (4)

Finally:

[tex]y=50.176 m[/tex]

Answer:

second medium is less optical denser than first medium

Explanation:

As per Snell's law we know that

[tex]\mu_1 sin\theta_1 = \mu_2 sin\theta_2[/tex]

so here we know that

[tex]\theta_1 [/tex] = angle of incidence

[tex]\theta_2[/tex] = angle of refraction

As we know that light ray goes away from the normal when it move from medium 1 to medium 2

so here angle of refraction will me more than the angle of incidence

so we have

[tex]\frac{sin\theta_1}{sin\theta_2} = \frac{\mu_2}{\mu_1}[/tex]

since we know that

[tex]\theta_1 < \theta_2[/tex]

so we have

[tex]\frac{sin\theta_1}{sin\theta_2} < 1[/tex]

[tex]\frac{\mu_2}{\mu_1} <1[/tex]

so we can say that

[tex]\mu_2 < \mu_1[/tex]

so we can say that second medium is less optical denser than first medium

Hydrogen, nitrogen, and oxygen are examples of diatomic molecules naturally.

The atom consists of matter that may be split without releasing electrical charges.

In the nucleus proton and the neutron is exists. The condition of the atom to be electrically neutral is that the number of the proton and electron should be the same.

Two or more two atoms combine to form the molecules. A diatomic molecule is one that is made up of atoms from two distinct elements. Hydrogen, nitrogen, and oxygen are examples of diatomic molecules.

Hence, hydrogen, nitrogen, and oxygen are examples of diatomic molecules naturally.

To learn more about the atom refer to the link;

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Answer:

(a) [tex]5.88\times 10^{12}mi[/tex]

Explanation:

We have given speed of light [tex]c=3\times 10^8m/sec[/tex]

Given time = 1 year =365 days

We know that in 1 minute = 60 sec

1 hour = 60×60 = 3600 sec

In one day = 24 hour = 24×60×60=86400 sec

So in 365 days = 365×86400[tex]=3.1536\times 10^7sec[/tex]

We know that distance = speed ×time[tex]=3.1536\times 10^7\times 3\times 10^8=9.46\times 10^{15}m[/tex]

We have given that 1 mi = 1609 m

So [tex]9.46\times 10^{15}m=\frac{9.46\times 10^{15}}{1609}=5.88\times 10^{12}mi[/tex]So option (a) is correct

Answer:

C.

Explanation:

A meter is 8.56 centimeters longer than a yard. Something to keep in mind is that a meter is about 10% longer than a yard.

Hope this helps :)

Answer:

C. 8.56cm

Explanation:

Since we are to get the difference of cord length in centimeter, there fore we will have to convert the one yard and one metre length of wires to centimeters.

For 1m length of cord

1m = 100cm

For 1yard length of cord

1yard = 91.44cm

Difference in cord length = 100cm - 91.44cm

= 8.56cm

Answer:

7,14545 mph and 3,1936 m/s

Explanation:

The average speed is calculated by dividing the displacement over time, then it is 26,2 miles/(3 2/3 hours), here 3 (2/3) hours is a mixed number, that represents 11/3 hours or 3,66 hours. Then the average speed is 7,14545 mph, now to turn this into meters per second, we notice as mentioned that 1 mile =1609 meters and 1 hour=3600 seconds. Then 7,14545 miles/hour* (1 hour/3600 seconds) * (1609 meters/1 mile)=3,1936 m/s

Answer:

a)(2×40+2×80)/(2+2)=240/4=60

b)(240)/(120/40+120/80)=240/(3+1.5)

240/4.5=53.33333

no

Answer:

340 meters

Explanation:

For this we have to use the formula of constant velocity motion:

[tex]d=v*t\\where:\\d=distance\\v=velocity\\t=time[/tex]

Because you hear the blows exactly 1 per second. And you hear only one blow after she finished, the sound takes one second to get to you.

[tex]d=340(m/s)*(1(s))=340m[/tex]

Color: Apples are red or green or yellow. Oranges are normally orange.

Malleability: Apples are hard and brittle. Oranges are soft and squishy.

Compared to oranges, apples can be red, yellow or green, and they have a round shape with flattened top and bottom and a smooth, shiny skin. In contrast, oranges are typically orange, they're nearly spherical in shape, and their peel is thicker and rougher.

In comparing and contrasting the properties of apples and oranges, we look into their physical attributes such as their color, shape, and texture. Apples usually come in shades of red, green or yellow. On the other hand, oranges are typically orange in color. Regarding shape, apples tend to have a round shape with a flattened top and bottom, whereas oranges are nearly spherical. For the texture, apples have a smooth, shiny skin while oranges have a rough, textured external appearance with a slightly thicker rind.

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Answer:

B) Na has a lower first ionization energy than Ne.  

Explanation:

The atomic number¹ for Na has a value of 11 while in the case of Ne this value is 10. That means that Sodium (Na) has a total number of 11 protons, 11 neutrons and 11 electrons (since it is electrically neutral²). For the case of Neon (Ne) it has 10 protons, 10 neutrons and 10 electrons.

As the atomic number increases, the atomic radius³ shrinks (the orbitals are closer to the nucleus) as a consequence of the electric force. For the case of sodium (Na) the electron in the outermost orbital will experience a lower electric force than the electron placed in the outermost orbital in the atom of Neon (Ne).

Although, the sodium’s atom has more protons and therefore electrons, these eleven electrons will be organized according with the electronic configuration⁴ in the different shells (orbitals) of probabilities of their positions around the atom.

The electronic configuration for Na is:

1s²2s²2p⁶3s¹

The electronic configuration for Ne is:

1s²2s²2p⁶

Since Na needs another orbital to placed its outermost electron, the atomic radius will have a greater value than Ne. The electric force is inversely proportional to the square of the distance between two charged particles, as is established in Coulomb’s law:

[tex]F = \kappa_{0} \frac{q1q2}{r^{2}}[/tex]    (1)

Where q1 and q2 are the charges, [tex]\kappa_{0}[/tex] is the proportionality constant and r is the distance between the two charges.

Hence, the electron in the outermost orbital of Ne is submitted to a greater electric force according with equation 1, the required energy to remove it (ionization energy⁵) will be greater than in the case of Na (for that case will be the first ionization energy).

¹Atomic number: The number of protons or electrons in an atom.

²Electricaly neutral: All the charges are balanced (same number of positive charges and negative charges).

³Atomic radius: Distance between the center of the nucleus and an electron placed in the outermost orbital for a specific atom.

⁴Electronic configuration: Show how the electrons of an atom will be arranged in different orbitals according with the fact that each orbital has a specific number of electrons that can be held.

⁵Ionization energy: Energy required to remove an electron from an atom.

Key values:

First ionization energy of Na: 495 kJ/mol

First ionization energy of Ne: 2080 kJ/mol

Atomic radius of Na: 2.27 Å

Atomic radius of Ne: 1.54 Å

Atomic number of Na: 11

Atomic number of Ne: 10

Answer:

(B) Na has a lower first ionization energy than Ne.

Explanation:

As we know that ionization potential is the energy required to remove one electron from outermost shell. So if we found the ionization potential of Na and then ionization potential of Ne we will find the difference.

Experimentally it is found that Na has smaller value of ionization potential compare to Ne. So we can say the effective charge of Nucleus of Na must be less than the effective charge of nucleus of Ne.

Because if effective charge of nucleus is more than it will have strong attraction on outer electron and it must have more ionization potential.

So here correct answer will be

(B) Na has a lower first ionization energy than Ne.

Answer:

a = 252 m / s²

Explanation:

The initial speed Vο = 0

The speed of baseball with witch it is thrown  Vf = 42 m/s

Displacement covered = 3.5 m

Hence acceleration will be constant at this moment, so as per the equation of constant acceleration : ( Vf² - Vο² ) / 2d = a

                                                            a = 252 m / s²

Light striking a mirror at a 45° angle will be reflected at a 45° angle.

This is because the law of reflection states that the angle of incidence is equal to the angle of reflection, where both angles are measured with respect to the normal (the imaginary line perpendicular to the surface at the point of incidence).

Therefore, if light hits a mirror with a 45-degree angle of incidence, the angle of reflection will also be 45 degrees, assuming the medium on both sides of the mirror is the same, typically air in common scenarios.

Answer:

Yes, it is right-angled

Explanation:

Two vectors are orthogonal if the scalar product between them is zero. Then, we will match each pair of vertices with a vector, wich is formed with the following formula:

Given two points A and B, the vector going from A to B is

[tex]AB=B-A=(B_{x}-A_{x},B_{y}-A_{y},B_{z} -A_{z})[/tex]

So, we calculate each component separately.

[tex]PQ=Q-P=(2-1,1-(-2),-3-(-1))=(1,3,-2)[/tex]

[tex]QR=R-Q=(6-2,-1-1,-4-(-3))=(4,-2,-1)[/tex]

[tex]RP=P-R=(1-6,-2-(-1),-1-(-4))=(-5,-1,3)[/tex]

Finally, using the scalar product formula

[tex]A*B=A_{x}* B_{x}+ A_{y}* B_{y}+ A_{z}* B_{z}[/tex]

we see if the products is zero

[tex]PQ*QR=1*4+3(-2)+(-2)*(-1)=0[/tex]

In this case we don't even have to calculate the other products as we've found that PQ and QR form a right angle.

Final answer:

By calculating the dot product of vectors representing the sides of the triangle, it is found that the dot product of PQ and QR is zero, indicating that these vectors are orthogonal. Therefore, the triangle PQR is right-angled.

Explanation:

To determine whether the triangle with vertices P(1, -2, -1), Q(2, 1, -3), and R(6, -1, -4) is right-angled, we can use the concept of the dot product between two vectors. A triangle is right-angled if one of the pairs of vectors corresponding to its sides is orthogonal, meaning their dot product is zero.

First, we find the vectors representing the sides of the triangle:

⇨ PQ = Q - P = (2, 1, -3) - (1, -2, -1) = (1, 3, -2)

⇨ PR = R - P = (6, -1, -4) - (1, -2, -1) = (5, 1, -3)

⇨ QR = R - Q = (6, -1, -4) - (2, 1, -3) = (4, -2, -1)

Next, calculate the dot products of each pair of vectors:

PQ ⋅ PR = (1, 3, -2) ⋅ (5, 1, -3) = 1*5 + 3*1 - 2*(-3) = 5 + 3 + 6 = 14

PQ ⋅ QR = (1, 3, -2) ⋅ (4, -2, -1) = 1*4 + 3*(-2) - 2*(-1) = 4 - 6 + 2 = 0

PR ⋅ QR = (5, 1, -3) ⋅ (4, -2, -1) = 5*4 + 1*(-2) - 3*(-1) = 20 - 2 + 3 = 21

As we see, the dot product of PQ and QR is zero, which means these two vectors are orthogonal, and hence the triangle PQR is right-angled.

Answer:

(a) the approximate mass of a column of air is: 1225 (Kg), (b) the weight of this amount of air is: 12017.25 (N) and (c) the pressure at the bottom is: 120172.5 (KPa)

Explanation:

We need to remember that the equation that related mass and volume is the density as:[tex]p=\frac{m}{V}[/tex], where p is the density, m is the mass and V is the volume and assuming that air density is 1.225 (Kg/m3) and knowing that exosphere is the upper atmosphere and its height is 10000 (Km). We now can get the volume as:[tex]Volume=A*h_{upper atmosphere} =0.0001*10000000=1000(metre^{3})[/tex], so that the approximate mass of a column of air is: [tex]m=p*V=1.225*1000=1225 (Kg)[/tex]. Then the weight of this amount of air is:[tex]W=m*g[/tex], where W is the weight, m is the mass and g the gravity acceleration, so:[tex]W=1225*9.81=12017.25 (N)[/tex]. Finally to find the pressure we know that gauge pressure equation [tex]P=p*g*h[/tex] where P is the pressure, p is the density of the column and h is the height of the column, so [tex]P=1.225*9.81*10000000=120172.5 (KPa)[/tex].

The approximate mass of the air column can be calculated using the density of air and the volume of the column. The weight of the air column can be calculated by multiplying the mass by the acceleration due to gravity. The pressure at the bottom can be calculated by dividing the weight by the area.

The approximate mass of a column of air with an area of 1cm² that extends from sea level to the upper atmosphere can be calculated using the density of air. The density of air near the Earth's surface is about 1.29 kg/m³. Using this density, we can calculate the mass by multiplying the density with the volume of the column of air. The volume can be calculated by multiplying the area of 1 cm² with the height of the column, which is the distance from sea level to the upper atmosphere.

To calculate the weight of this amount of air, we need to multiply the mass of the air column by the acceleration due to gravity. The acceleration due to gravity is approximately 9.81 m/s². Multiplying the mass by the acceleration due to gravity will give us the weight of the air column in newtons.

The pressure at the bottom, or at sea level, can be calculated using the weight of the air column and the area of 1 cm². The pressure is equal to the weight divided by the area.

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Answer:

D=7.385km

Explanation:

First we calculate how much it moves vertically (north and south), we will call this distance Y,.

taking into account that the car moves to the southwest and that it has a slope of 45 degrees, we can say that the component that moved south can be subtracted from the distance that moved north.

Y=6.71-1.33sen45=5.77km

then the distance that moves in the east we will call it X

we can add the component of the distance that moves to the southwest with the distance that moves to the east

X=3.67+1.33cos45=4.61Km

then we use the distance equation

[tex]D=\sqrt{x^{2} +y^{2} }[/tex]

[tex]D=\sqrt{4.61^{2}+5.77^{2} }[/tex]

D=7.385km

A 0.40-kg mass attached to the end of string swings in a vertical circle having a radius of 1.8 m,  the magnitude of the tension in the string at this instant is

T= 8.1 N Option C. This is further explained below.

Generally, the tension equation is

[tex]T - mg cos40 = F_c[/tex]

Where

[tex]F_c=mv^2/L[/tex]

[tex]F_c=(0.40*v^2)/L\\\\F_c=(0.40*5^2)/L\\\\F_c=(0.40*5^2)/1.8\\\\F_c= 5.55........equ1\\\\[/tex]

Therefore

[tex]T - mg cos40 = F_c[/tex]

[tex]T-(0.4*9.81) cos40 = F_c\\\\T-(0.4*9.81) (0.7660444431) = F_c[/tex]

When (0.4*9.81) * (0.7660444431) we have

3.005958395

Hence

T =3.005958395  +(0.40*v^2)/L

From equ 1

T = 3.005958395  +5.55

T = 8.1 N

In conclusion, the magnitude of the tension

T = 8.1 N

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To find the tension in the string of a swinging mass, we consider both the gravitational and centripetal forces. At a 40-degree angle below the horizontal and a speed of 5.0 m/s, the calculated tension is approximately 8.56 N, closest to option c (8.1 N).

The student's question involves calculating the tension in the string of a mass swinging in a vertical circle. To find the tension at a specific moment, we must consider the forces acting on the mass: the gravitational force (weight) and the centripetal force required to keep the mass moving in a circular path.

At the moment the string makes a 40-degree angle below the horizontal, the tension in the string (T) is the sum of the radial (centripetal) component of the tension needed to keep the mass moving in a circle (Fc) and the component of the gravitational force (weight) acting along the string.

The gravitational force is Fg = m*g = 0.40 kg * 9.8 m/s² = 3.92 N. The centripetal force is Fc = (m*v²)/r. Plugging in the values, we get Fc = (0.40 kg * (5.0 m/s)²) / 1.8 m = 5.56 N. These forces combine to give the total tension, but we need to account for the angle when calculating the component of gravitational force along the string's direction.

So we have T = Fg*cos(40°) + Fc. Using cosine of 40 degrees which is approximately 0.766, Fg*cos(40°) = 3.92 N * 0.766 ≈ 3.00 N. Now we add the centripetal force to get T = Fc + Fg*cos(40°) = 5.56 N + 3.00 N = 8.56 N.

Therefore, the exact tension would be 8.56 N, which is closest to the provided answer of 8.1 N (Option c), assuming some rounding differences.

Answer:

Instantaneous velocity is [tex]v=63[/tex] at t=1.

Explanation:

The height in feet after t second is given by y(t)=95t−16t2.

Average velocity is defined by:

[tex]v_{ave}=\frac{x_{f} - x_{i} }{t_{f} - t_{i}  }[/tex].

i)

[tex]t_{i}=1\\x_{i}=y(1)=79\\t_{f}=1+0.1\\x_{f}=y(1+0.1)=85.14\\[/tex]

⇒ [tex]v_{ave}=61.4[/tex].

ii)

[tex]t_{i}=1\\x_{i}=y(1)=79\\t_{f}=1+0.01\\x_{f}=y(1+0.01)=79.6284\\[/tex]

⇒ [tex]v_{ave}=62.84[/tex].

iii)

[tex]t_{i}=1\\x_{i}=y(1)=79\\t_{f}=1+0.001\\x_{f}=y(1+0.001)=79.062984\\[/tex]

⇒ [tex]v_{ave}=62.98[/tex].

Instantaneos velocity is defined by: [tex]v=\lim_{\triangle t \to 0} \frac{\triangle x}{\triangle t}[/tex]

So we have seen "manually" that when [tex]\triangle t \rightarrow 0[/tex], [tex]v \rightarrow 63[/tex].

So instantaneous velocity must be [tex]v=63[/tex] at t=1.

To find the average velocity for different time intervals, we need to calculate the displacement of the ball and divide it by the time interval.

To find the average velocity for the time period beginning when t=1 and lasting (i) 0.1 seconds, (ii) 0.01 seconds, and (iii) 0.001 seconds, we need to calculate the displacement of the ball during each time interval and then divide it by the corresponding time interval.

(i) For the time interval of 0.1 seconds:

The displacement is given by:

d = y2 - y1

where y2 = 95(1.1) - 16(1.1)^2 and y1 = 95(1) - 16(1)^2

Substituting the values:

d = 104.5 - 79 = 25.5 ft

The average velocity is given by:

v = d / t

:p>where d is the displacement and t is the time interval.

Substituting the values:

v = 25.5 / 0.1 = 255 ft/s

(ii) For the time interval of 0.01 seconds:

The displacement is given by:

d = y2 - y1

where y2 = 95(1.01) - 16(1.01)^2 and y1 = 95(1) - 16(1)^2

Substituting the values:

d = 100.95 - 78.84 = 22.11 ft

The average velocity is given by:

v = d / t

where d is the displacement and t is the time interval.

Substituting the values:

v = 22.11 / 0.01 = 2211 ft/s

(iii) For the time interval of 0.001 seconds:

The displacement is given by:

d = y2 - y1

where y2 = 95(1.001) - 16(1.001)^2 and y1 = 95(1) - 16(1)^2

Substituting the values:

d = 100.595 - 78.596 = 21.999 ft

The average velocity is given by:

v = d / t

where d is the displacement and t is the time interval.

Substituting the values:

v = 21.999 / 0.001 = 21999 ft/s

Based on the above results, the instantaneous velocity of the ball when t=1 will be the same as the average velocity for a very small time interval, which approaches 22000 ft/s.

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

[tex]v = \frac{2 \pi r}{T}[/tex]  

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

[tex]T^{2} = r^{3}[/tex]

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

[tex]T^{2} = r^{3}[/tex]

[tex]T = \sqrt{(133370 Km)^{3}}[/tex]

[tex]T = \sqrt{(2.372x10^{15} Km)}[/tex]

[tex]T = 4.870x10^{7} Km[/tex]

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( [tex]1.50x10^{8} Km[/tex] )

1 AU is defined as the distance between the earth and the sun.

[tex]\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU[/tex]

[tex]T = 0.324 AU[/tex]

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

[tex]T = \frac{0.324 AU}{1 AU} . 1 year[/tex]

[tex]T = 0.324 year[/tex]

That can be expressed in units of days

[tex]T = \frac{0.324 year}{1 year} . 365.25 days[/tex]  

[tex]T = 118.60 days[/tex]

Circular velocity for the particle in the Encke Division:

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

[tex]v = \frac{2 \pi (133370 Km)}{(118.60 days)}[/tex]

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

[tex]118.60 days .\frac{86400 s}{1 day}[/tex] ⇒ 10247040 s

[tex]133370 Km .\frac{1000 m}{1 Km}[/tex] ⇒ 133370000 m

[tex]v = \frac{2 \pi (133370000 m)}{(10247040 s)}[/tex]

[tex]v = 81.778 m/s[/tex]

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

[tex]T^{2} = r^{3}[/tex]

[tex]T = \sqrt{(69000 Km)^{3}}[/tex]

[tex]T = \sqrt{(3.285x10^{14} Km)}[/tex]

[tex]T = 1.812x10^{7} Km[/tex]

[tex]\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU[/tex]

[tex]T = 0.120 AU[/tex]

[tex]T = \frac{0.120 AU}{1 AU} . 1 year[/tex]

[tex]T = 0.120 year[/tex]

[tex]T = \frac{0.120 year}{1 year} . 365.25 days[/tex]  

[tex]T = 43.83 days[/tex]

Circular velocity for the particle in D Ring:

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

[tex]v = \frac{2 \pi (69000 Km)}{(43.83 days)}[/tex]

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

[tex]43.83 days . \frac{86400 s}{1 day}[/tex] ⇒ 3786912 s

[tex]69000 Km . \frac{1000 m}{ 1 Km}[/tex] ⇒ 69000000 m

[tex]v = \frac{2 \pi (69000000 m)}{(3786912 s)}[/tex]

[tex]v = 114.483 m/s[/tex]

[tex]\frac{114.483 m/s}{81.778 m/s} = 1.399[/tex]            

The particle in the D ring is 1399 times faster than the particle in the Encke Division.  

Answer:

it's A

Explanation:

the masses of each car must be known

According to momentum conservation law, the total initial momentum of the colliding objects is equal to the total final momentum. To find the momentum, we have to know their masses. Hence, option A is correct.

Momentum of an object is its ability to bring the applied force to make a maximum displacement. It is the product of mass and velocity of the object. Momentum is a vector quantity.

In all collisions momentum is conserved. Hence, total initial momentum is equal to the total final momentum of the colliding objects. Let m1 and m2 be the masses of the objects and u1 and u2 be their initial velocities then v1 and v2 are taken as their final velocities.

Then,

initial momentum = m1 u1 + m2 u2

final momentum =m1 u1 + m2 u2

conservation of momentum is written as:

m1 u1 + m2 u2 = m1 u1 + m2 u2.

Hence, by knowing the initial motion we can predict their final motion and momentum but we must know the masses of colliding objects. Hence, option A is correct.

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Answer:

Option C

Explanation:

When this type of situation occurs where the price falls and the sale of a product increases then a shift is observed in the supply curve towards right as a result of this shift there is a need to shift down the demand curve

Thus When supply curve is shifted to the right the demand curve is moved down accordingly.