Consider a golf club hitting a golf ball that results in the following graph of force versus time on a 45 gram golf ball. If the final velocity of the ball has a magnitude of 41 m/s, determine the value of Fmax

Answer:

4.4 m/s

Explanation:

momentum is always conserved so we can use conversation of momentum to solve the question, also momentum is a vector quantity ( it has magnitude and direction) which is the product of the bodies mass and velocity.

conservation law of momentum relates by the formula below:

momentum before collision = momentum after collision

M1U1 + M2U2 = M1V1 + M2V2

in the case of this two, the formula becomes

M1U1 + M2U2 = V (M1 + M2) since she jumped into his arm

there masses are M1 = 75.6 kg M2 = 59 kg and their velocities are  U1 = 3.7 m/s and U2 = 5.4 m/s, their common velocity after collision = V since their motion is backward the formula becomes

-M1U1 - M2U2 = V(M1 + M2)

substitute the values into the equations

(-75.6 × 3.7 ) + (- 59 × 5.4) = V ( 75.6 + 59)

- 598.32 = 134.6 V

divide both side by 134.6

V = - 598.32 / 134.6 = -4.445 m/s  = -4.4 m/s to nearest tenth the negative means in the same backward direction

Answer:F= (GM1M2)/r^2

F= ((6.67•10^-11)(5.98•10^24)(70))/ (6.40•10^6)^2 = 681.6557617

Explanation:physics

The force of gravitational attraction between the Earth and a student can be calculated using Newton's law of universal gravitation. By substituting the known values into the equation, we can find the answer.

The subject of this question is the force of gravity between two bodies. The force of gravity can be calculated using the formula from Newton's law of universal gravitation: F = G * (m1 * m2) / r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two bodies, and r is the distance between the centers of the two bodies.

Using the values given in the question, we get: F = (6.67 x 10^-11 Nm^2/kg^2) * ((5.98 x 10^24 kg * 70 kg) / (6.40 x 10^6 m)^2).

Carrying out this calculation gives you the force of gravitational attraction between the student and the Earth. This Physics concept helps us understand how forces work in the universe and is further used in studies related to astronomy, engineering, and many other fields.

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1. Answer: You notice that your grandmother's silver is very dark in places and needs polish: Chemical Change.

Your cup of hot chocolate gives off steam: Chemical Change

When you mix baking soda and vinegar, the two substances fizz and produce bubbles: Chemical change

An ice cube melts: Physical change

You burn an oak log in a fireplace: Chemical change

You activate a heat pack to warm your hands: Chemical change.

2. Answer: 0.96kg

Sorry

didn't have enough time to finish the whole thing.

In the given physics problem, the ball comes to a stop at the highest point. This is because it has transformed all of its kinetic energy into potential energy, and has no more energy to continue moving upwards, as there is no energy transfer with either the track or the air.

In this physics problem, the ball on the track is an example of a system that conserves energy. Since there is no energy transfer with either the track or the surrounding air, the total mechanical energy of the ball, which comprises potential and kinetic energy, remains constant throughout its motion. Potential energy is energy due to height, and kinetic energy is energy due to motion.

To place the dot at the highest point the ball will reach, you identify where the ball has maximum potential energy and minimum kinetic energy because at the highest point, the ball momentarily stops and hence, kinetic energy is zero.

Therefore, the ball will stop at the highest point because it has converted all its kinetic energy (energy of motion) into potential energy (energy stored due to its position) and has no more energy left to continue moving upwards. At this point, it starts converting its potential energy back into kinetic energy as it moves downwards, hence it starts moving back down the track.

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The ball will reach the highest position at the same height as position B because of the conservation of mechanical energy. At this point, all kinetic energy converts back to potential energy, causing the ball to momentarily stop before reversing direction. This principle assumes no energy losses in the system.

To determine the highest position the ball will reach on the track before stopping and reversing direction, we need to consider the conservation of energy principle. In this scenario, the total mechanical energy (sum of potential and kinetic energy) remains constant because no energy is transferred between the ball and the track or the air.

When the ball starts from rest at position B, it possesses only potential energy and no kinetic energy. As the ball moves downward, it loses potential energy but gains kinetic energy, and as it moves upward, it loses kinetic energy and gains potential energy. The ball will reach the same height on the opposite side of the track as it started, assuming no energy losses. Therefore, the highest position the ball will reach will be the same height as position B.

The ball stops at this highest position because all of its kinetic energy is converted back to potential energy at that point, making its speed momentarily zero before it reverses direction. This illustrates the conservation of mechanical energy where the initial and final heights are the same when no external forces are acting on the system.

Answer:

Option D

0.83 m/s2

Explanation:

Time is assumed as 10 seconds

First, convert the speeds from km/h to m/s

20 km/h\times \frac {1000 m}{3600 s}=5.555555556 m/s \approx 5.56 m/s

50 km/h\times \frac {1000 m}{3600 s}=13.88888889  m/s \approx 13.89 m/s

Acceleration, [tex]a=\frac {v-u}{t}[/tex] where u and v are the initial and final velocities respectively, t is the time taken to accelerate.

Substituting 13.89 m/s for v, 5.56 m/s for u and 10 s for t then

[tex]a=\frac {13.89-5.56}{10}=0.8333333 m/s^{2}\approx 0.83 m/s^{2}[/tex]

Answer:

Speed at the bottom of the roller coaster = 49.79 m/s

Explanation:

A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically drops that distance before the track curves forward.

We have to find the speed at the bottom.

Here the gravitational energy fully converts to kinetic energy, so we equate it.

Gravitational energy = [tex]m\times g\times h[/tex]

Kinetic energy = [tex]0.5 \times m\times v^{2}[/tex]

[tex]m\times g\times h[/tex] = [tex]0.5 \times m\times v^{2}[/tex]

[tex]9.8\times 126.5 = 0.5\times v^{2}[/tex]

[tex]v^{2}[/tex] = 2479.4

Velocity, v = 49.79 m/s

Explanation:

a. 5 m/s

b. 0 m/s

c. If we neglect air resistance, the horizontal velocity remains constant.  At the highest point, the vertical velocity is 0.

At maximum height, the horizontal and vertical component of the velocity will be 5 m/s and 0 m/s respectively.

Projectile motion is a form of motion experienced by an object or particle that is projected near Earth's surface and moves along a curved path under the action of gravity only.

Given is the initial velocity of a projectile has a horizontal component equal to 5 m/s and a vertical component equal to 6 m/s

a -  At the maximum height, the horizontal component will be 5m/s.

b -  At the maximum height, the vertical component will be 0 m/s.

c -  The horizontal velocity remains remains constant through out the motion as there is no acceleration in the horizontal direction. The vertical component at the maximum height is 0 m/s because the projectile will no longer move in the upward direction. At maximum height the projectile will only move horizontally.

Therefore, at maximum height, the horizontal and vertical component of the velocity will be 5 m/s and 0 m/s respectively.

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Final answer:

To build a 1.1 Hz oscillator with a 1.0 F capacitor, you would need an inductor of approximately 3.16 × 10⁻⁴ H.

Explanation:

To calculate the length of the inductor needed to build a 1.1 Hz oscillator with a 1.0 F capacitor, we can use the formula:

f = 1 / (2π√(LC))

Where f is the frequency, L is the inductance, and C is the capacitance.

Here, f = 1.1 Hz and C = 1.0 F. Rearranging the formula, we get:

L = (1 / (4π²f²C))

Plugging in the values, we find:

L = (1 / (4π²(1.1 Hz)²(1.0 F)))

L ≈ 3.16 × 10⁻⁴ H

Therefore, the length of the inductor needed would be approximately 3.16 × 10⁻⁴ H.

The length of the inductor must be approximately 1.4 meters.

To build a 1.1 Hz oscillator with a 1.0 F capacitor, we need to determine the inductance required for the LC circuit to oscillate at the desired frequency. The formula for the resonant frequency of an LC circuit is given by:

[tex]\[ f = \frac{1}{2\pi\sqrt{LC}} \][/tex]

We can rearrange this formula to solve for [tex]\( L \)[/tex]:

[tex]\[ L = \frac{1}{(2\pi f)^2 C} \][/tex]

Given that [tex]\( f = 1.1 \)[/tex] Hz and [tex]\( C = 1.0 \)[/tex] F, we can calculate [tex]\( L \)[/tex]:

[tex]\[ L = \frac{1}{(2\pi \times 1.1 \text{ Hz})^2 \times 1.0 \text{ F}} \] \[ L \approx \frac{1}{(2\pi)^2 \times 1.1^2 \times 1.0} \] \[ L \approx \frac{1}{4\pi^2 \times 1.21 \times 1.0} \] \[ L \approx \frac{1}{15.228451} \] \[ L \approx 0.06569 \text{ H} \][/tex]

Now that we have the required inductance, we need to calculate how much wire is needed to create an inductor with this inductance. The inductance of a solenoid (which is the shape we are wrapping the wire into) is given by:

[tex]\[ L = \frac{\mu_0 N^2 A}{l} \][/tex]

[tex]\[ l = \frac{\mu_0 N^2 A}{L} \][/tex]

The cross-sectional area [tex]\( A \)[/tex] of the coil can be calculated using the diameter of the plastic cylinder:

[tex]\[ A = \pi \left(\frac{d}{2}\right)^2 \] \[ A = \pi \left(\frac{4.0 \text{ cm}}{2}\right)^2 \] \[ A = \pi \left(2.0 \text{ cm}\right)^2 \] \[ A = \pi \times 4.0 \text{ cm}^2 \] \[ A = 12.566371 \text{ cm}^2 \] \[ A = 0.012566371 \text{ m}^2 \][/tex]

Since the wire is wrapped in 2 layers, we need to consider the number of turns [tex]\( N \)[/tex] per layer and multiply by 2. The number of turns per layer can be estimated by dividing the circumference of the cylinder by the diameter of the wire:

[tex]\[ N_{\text{per layer}} = \frac{\pi d_{\text{cylinder}}}{d_{\text{wire}}} \] \[ N_{\text{per layer}} = \frac{\pi \times 4.0 \text{ cm}}{0.25 \text{ mm}} \] \[ N_{\text{per layer}} = \frac{\pi \times 4.0 \text{ cm}}{0.00025 \text{ m}} \] \[ N_{\text{per layer}} = \frac{\pi \times 4000 \text{ mm}}{0.25 \text{ mm}} \] \[ N_{\text{per layer}} = \pi \times 16000 \] \[ N_{\text{per layer}} \approx 50265.4824 \][/tex]

Now, we can calculate the total number of turns [tex]\( N \)[/tex]:

[tex]\[ N = 2 \times N_{\text{per layer}} \] \[ N \approx 2 \times 50265.4824 \] \[ N \approx 100530.965 \][/tex]

Finally, we can calculate the length [tex]\( l \)[/tex] of the inductor:

[tex]\[ l = \frac{\mu_0 N^2 A}{L} \] \[ l = \frac{4\pi \times 10^{-7} \text{ H/m} \times (100530.965)^2 \times 0.012566371 \text{ m}^2}{0.06569 \text{ H}} \] \[ l \approx \frac{4\pi \times 10^{-7} \times 1010619310.025 \times 0.012566371}{0.06569} \] \[ l \approx \frac{5079198.577}{0.06569} \] \[ l \approx 1415457.869 \text{ m} \][/tex]

Therefore, the inductor must be approximately 1.4 meters long to create a 1.1 Hz oscillator with a 1.0 F capacitor using the given wire and cylinder.

Answer:

They act on different objects, so they would not appear together.

Explanation:

The student has made an incorrect conclusion that the speed of sound in water must be slower than in air because the sound is muffled.

In fact, sound waves travel faster through water than through air because water is denser (the particles are closer together). The muffled sound is due to the fact that only around 0.01% of sound is transmitted from the air into water, as water particles are harder to displace, causing most of the sound to be reflected back off the surface.

Moreover, sound under water is often heard through vibrations in the mastoid bone, behind the ear, rather than through the ear canal as on land. The use of earplugs reduces the intensity of sound and water carries low frequencies better than air, which is why aquatic animals like whales and dolphins utilize low-frequency sounds for communication.

In summary, the muffled sound is not because the speed of sound is slower in water but due to the efficient reflection at the water surface and the reduced transmission of sound from air to water.

Answer:

8.25 m/s

Explanation:

From the fundamental equation of motion, velocity is rate of change of displacement per unit time

[tex]Velocity=\frac {Displacement}{time}[/tex]

Given information

Displacement=2862 m

Time=347 s

Substituting the given information we obtain

[tex]Velocity=\frac {2862 m}{347 s}=8.247838617\approx 8.25 s[/tex]

Answer:

61 units

Explanation:

Since OA=60 units and OB= 11 units, the resultant is the hypotenuse of the two vectors

Therefore, [tex]resultant=\sqrt {(OA)^{2}+(OB)^{2}}[/tex]

Substituting the given values then

[tex]Resultant=\sqrt{60^{2}+11^{2}}=61[/tex]

Therefore, resultant between A and B is 61 units

To calculate the Sun's mass, we use the formula for gravitational force, substitute the given and known values, and then rearrange the equation to solve for the Sun's mass.

The question involves the calculation of the Sun's mass using the known gravitational force between Earth and the Sun. Given that, the formula used to calculate the gravitational force between two objects is: F = G * m1 * m2 / r^2, where.

Substitution and rearranging the formula results in: m2 = F * r^2 / (G * m1) which substituting the given and known values, delivers the estimated mass of the Sun.

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Answer:

6.  Acceleration = 4.74 m/s^2

7.  Centripetal force = 40.5 N

Explanation:

Problem 6.

Recall that the centripetal acceleration is defined as: [tex]a_c=\frac{v^2}{r}[/tex], where V is the object's tangential velocity, and r the radius of the circular motion. Therefore, in or case, the centripetal acceleration would be:

[tex]a_c=\frac{v^2}{r}\\a_c=\frac{3.77^2}{3}\,\frac{m}{s^2} \\a_c=4.7376 \frac{m}{s^2}[/tex]

which we can round to 4.74 m/s^2 (option b in your list)

Problem 7.

Now we need to find not just the centripetal acceleration using the same formula as above, but then the centripetal force.

[tex]a_c=\frac{v^2}{r}\\a_c=\frac{4.5^2}{2.5}\,\frac{m}{s^2} \\a_c=8.1 \frac{m}{s^2}[/tex]

Now we calculate the centripetal force by multiplying this acceleration times the mass of the object following the definition of force as mass times acceleration:

Centripetal force = 5.0 kg * 8.1 m/s^2 = 40.5 N

The answers comes in Newtons (N)

Answer:

m, g

Explanation:

G=m.g

m mass

g gravity

Answer:

530 g

Explanation:

Ideal gas law:

PV = nRT

(4.0×10⁵ Pa) (0.20 m³) = n (8.31 Pa m³ /mol /K) (20+273.15 K)

n = 32.8 mol

Find the mass using the molar mass:

32.8 mol × (16 g / mol) = 525 g

Rounded to two significant figures, the mass is 530 g.

Note: I used the molar mass provided, but oxygen is usually in a diatomic state (O₂), which would make the molar mass 32 g/mol.

43.48 mph

We are given;

First instance;

Second instance;

We are required to determine the average speed for the entire journey.

Time = Distance ÷ speed

First instance

Time = 50 miles ÷ 40 mph

        = 1.25 hours

Second instance;

Time = 20 miles ÷ 55 mph

        = 0.36 hours

Therefore;

Total distance = 70 miles

Total time = 1.25 hrs + 0.36 hr

               = 1.61 hrs

Thus;

Average speed = 70 miles ÷ 1.61 hrs

                          = 43.478 miles per hour

                          = 43.48 mph

Scientists measure the time between the arrival of an earthquake's __P____ and ___S____ waves to help determine the distance between the recording seismograph and the earthquake epicenter.

Explanation:

P- (compressional) and S- (shear) waves produced in earthquakes travel at different speeds. P waves are faster than S waves and hence will be detected first by a seismograph after an earthquake. The further away a seismograph is from the epicenter of an earthquake,  the longer the time difference between the two (2) waves will be.

Using several, at least 3, seismographs located at different geoghraphical locations and detecting earthquakes, geologists can extrapolate the epicenter of an earthquake using the time differences in arrivals of the two waves in each of the seismographs, using the mathematics of triangulation.

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The time between the arrival of P waves and S waves, recorded by seismographs, is used to determine the distance to an earthquake's epicenter, with the precision being affected by the accuracy of wave speed and arrival time measurements.

Scientists gauge the distance to an earthquake's epicenter by measuring the time between the arrival of an earthquake's P waves and S waves. Seismographs can record these waves with a precision of 0.100 seconds. Given that P waves travel at 7.20 km/s and S waves travel at 4.00 km/s, the time difference recorded on seismograms allows scientists to calculate the distance to an epicenter with considerable accuracy. However, the precision of this distance measurement can be limited by the exactitude of the wave speed measurements and the timing of their arrivals.

Furthermore, the tracking of seismic waves can also be significant for monitoring underground nuclear tests. If there is uncertainty in the speeds of S and P waves, or in the measurement of their arrival times, this poses a limitation in accurately determining the source of seismic energy, which could hinder detection capabilities for such underground activities.

Ball Y's motion after the elastic collision with ball X should be such that it ensures the law of conservation of momentum. The direction and speed would ensure the vector sum of the momenta of the two balls equals the initial momentum of ball X.

The question is regarding an elastic collision between two identical balls on a frictionless surface, ball X moving initially with a velocity of 10 m/s, and ball Y at rest. After the collision, the velocity of ball X changes to 6 m/s at an angle of 53 degrees to its original direction. In an elastic collision, total kinetic energy and momentum are conserved.

Using the conservation of momentum: Initial momentum = Final momentum. The initial momentum was all with ball X (10*m), where m is the mass of the ball. The final momentum is a combination of the momentum of ball X and ball Y. Ball X would have a momentum of 6*m (mass*new speed), in a direction 53 degrees from its initial direction. Hence, the remaining momentum must be carried by ball Y to satisfy the law of conservation of momentum.

The exact motion of ball Y after the collision would then be determined by the pathway that would ensure the vector sum of the momenta of the two balls equals the initial momentum of ball X. This can be determined by drawing a vector diagram or by using some trigonometry.

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According to the Law of Conservation of Mass, the mass of the rust should be greater than the original mass of the iron nail, because the mass of the oxygen and water that form the rust are added to the mass of the iron nail.

The principle at play here is the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. In this scenario, an iron nail is reacting with oxygen and water from the environment to form rust, or iron(III) oxide.

If the nail were left to rust completely, the mass of the rust would be greater than the original mass of the nail. This is because the mass of the oxygen and water that combines with the iron is added to the mass of the nail to form the rust. The additional mass comes from the oxygen atoms in the air and in the water that combine with the iron atoms in the nail.

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The second ionization energy is the energy required to remove the second electron after a valence one has been removed.

For an element, the first ionization energy is defined as the amount of energy required to remove one electron from the outermost valence shell of a neutral atom. Removing one electron increases the number of protons, making it a 1+ ion.  

The nucleus (protons) has more bonding to the electrons with negative charge and thus more energy is required if another electron needs to be removed. This higher energy required to remove second electron from a 1+ ion (after the first one has been removed) is termed as the second ionization energy. Second ionization energy leads to formation of a 2+ ion. Similarly, third ionization energy is higher than second ionization energy.

The distances are:

- E and F: 4 units.

- C and D: 5 units.

- G and H: 3 units.

- A and B: 6 units.

Why?

We can find the distance between each pair of points using the following formula:

[tex]d(P_1,P_2)=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}[/tex]

So, calculating we have:

- E(-2, -1) and F(-2, -5):

[tex]d(E,F)=\sqrt{(-2-(-2))^{2}+(-5-(-1))^{2}}\\\\d(E,F)=\sqrt{0+(-5+1)^{2}}=\sqrt{0+(-4)^{2}}=\sqrt{16}=4units[/tex]

- C(-4, 1) and D(1, 1):

[tex]d(C,D)=\sqrt{(1-(-4))^{2}+(1-(1))^{2}}\\\\d(C,D)=\sqrt{(5)^{2}+0^{2}}=\sqrt{25+0}=\sqrt{25}=5units[/tex]

- G(3, -5) and H(6, -5):

[tex]d(G,H)=\sqrt{(6-(3))^{2}+(-5-(-5))^{2}}\\\\d(G,H)=\sqrt{(3)^{2}+0^{2}}=\sqrt{9+0}=\sqrt{9}=3units[/tex]

- A(5, 4) and B( 5, -2):

[tex]d(A,B)=\sqrt{(5-(5))^{2}+(-2-(4))^{2}}\\\\d(A,B)=\sqrt{0+(-6)^{2}}=\sqrt{0+36}=\sqrt{36}=6units[/tex]

Have a nice day!

Answer:

G(3, -5) and H(6, -5) ------>3 units

arrowRight

E(-2, -1) and F(-2, -5)------->4 units

arrowRight

A (5, 4) and B( 5, -2)------->6 units

arrowRight

C(-4, 1) and D(1, 1)---------->5 units

arrowRight

Explanation:

The length of the brass rod must be 1.00 m

Explanation:

The change in length of a metal rod due to the increase in temperature is given by

[tex]\Delta L = L_0 \alpha \Delta T[/tex]

where

[tex]L_0[/tex] is the initial length of the rod

[tex]\alpha[/tex] is the linear expansivity of the metal

[tex]\Delta T[/tex] is the increase in temperature

For the iron rod, we can write the following

[tex]\Delta L = L_i \alpha_i \Delta T[/tex]

Where

[tex]L_i = 1.58 m[/tex] is the initial length of the iron rod

[tex]\alpha_i = 1.2\cdot 10^{-5} K^{-1}[/tex] is the iron linear expansivity

For the brass rod, we can write

[tex]\Delta L = L_b \alpha_b \Delta T[/tex]

Where

[tex]L_b = [/tex] is the initial length of the brass rod

[tex]\alpha_b = 1.9\cdot 10^{-5} K^{-1}[/tex] is the brass linear expansivity

We want the two changes in length to be the same for the same change in temperature [tex]\Delta T[/tex], so we can write

[tex]L_i \alpha_i \Delta T=L_b \alpha_b \Delta T[/tex]

And solving for [tex]L_b[/tex], we find the length of the brass rod:

[tex]L_b = \frac{L_i \alpha_i}{\alpha_b}=\frac{(1.58)(1.2\cdot 10^{-5})}{1.9\cdot 10^{-5}}=1.00 m[/tex]

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It is really important to cite units while engaging with scientific problems as they help in evaluating the right solution without any mistake.

Explanation:

While solving analytical questions to evaluate the right answers, the units play a crucial role in the equations. With the help of units in the formula, we can easily judge whether we are placing the values in the same parameters and hence, lessen the probabilities of wrong answers.

For example, while adding two measurements i.e. 4 m and 36 cm; if we don't consider the units and move on with the addition, the answer will be 40. Now, the first thing is that we are adding two measurements of different parameters. Besides this, the answers will be wrong i.e. 4.36 m is the correct answer instead of 40 and that too without mentioning the unit.

The chances of selecting the wrong answer are more when we need to choose options out of multiple choices because here we often get confused. That's why we should always make sure that we approach the scientific questions along with the units.

Answer:

W located in the middle of the ruler, R located at 10cm from the pivot point.

See the attached image.

Explanation:

To solve this problem we have to perform a free body diagram symbolizing each of the forces that act on the bar in balance.

In the attached image we can see the free body diagram.

Answer:

The light ray goes inside with angle of refraction 30° and gets reflected at an angle 30° and then comes out refracted at an 45° with normal of third face

Explanation:

At the first face of prism where the light is first incident,

we can use snell's law to determine the angle of refraction

given refractive index of prism , μ'= √2.

and angle if incidence , i = 45°

now from snell's law,   sin(i) = μ sin(r),    r= angle of refraction

                                                                 μ= relative refractive index

⇒ sin45° = √2 sinr

⇒ sinr = [tex]\frac{1}{2}[/tex] ⇒ r = 30°

Now, let top corner of prism's triangular face be A

         point where light is incident be B

         point where light gets reflected on silvered surface be C

so that ABC forms a triangle

now as r = 30°, ∠ABC = 90°-30° = 60°

As sum of angles in a triangle is 180°,

∠ACB = 60° (since ∠BAC = 60°)

⇒ angle of incidence on silvered face = 90°-60°= 30°

⇒ angle of reflection = 30°

Using same theory as above,

the angle of incidence on third face = 30°

Again using snell's law, sin(i) = μ sin(r)

⇒ sin(30) = [tex]\frac{1}{\sqrt{2} }[/tex] sin(r)

here relative refractive index ,μ = [tex]\frac{1}{\sqrt{2} }[/tex], since light travels from prism to air

⇒ r= 45°

⇒The light ray goes inside with angle of refraction 30° and gets reflected at an angle 30° and then comes out refracted at an 45° with normal of third face.

Answer:

amount of space occupied

Within the provided options, 'the amount of space a substance's matter occupies' best describes volume. Volume is an extensive property and the SI unit is a cubic meter (m³). It is different from mass and density, where mass refers to the amount of matter a substance contains, and density is the ratio of mass to volume.

The best description of volume among the given choices would be 'the amount of space a substance's matter occupies.' Volume, in the physical or scientific context, is an extensive property which means that its value depends on the amount of matter being considered. It is directly proportional to the amount of substance, hence, the more substance there is, the greater the volume. The standard unit of volume in the International System of Units (SI) is the cubic meter (m³), which is defined as the space occupied by a cube with sides of one meter in length.

It's important to differentiate volume from other properties like mass and density. Mass refers to the amount of matter a substance contains and density is an intensive property defined as the ratio of a substance's mass to its volume.

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Answer:

The weight of the astronaut on the surface of moon due to the force of gravitation becomes one-sixth of the size of the gravitational acceleration on Earth.

Explanation:

Let the mass of the astronaut is, m = M

The acceleration due to gravity on Earth is, a = g

The acceleration due to gravity on the moon is a' = g/6

The weight of the astronaut on Earth, W = Mg

The weight of the astronaut on the moon is, w = Mg/6

                                                                           W = w /6

Hence, the weight of the astronaut on the surface of moon due to the force of gravitation becomes one-sixth of the size of the gravitational acceleration on Earth.

Answer:

The astronaut’s weight decreases because the moon’s gravitational acceleration is less than Earth’s

Explanation:

I took it on Edge.

Answer:

3

3.75

Explanation:

First ball

Givens

a = 5m/s^2

m = ??

F = 15N

Formula

F = m*a

Solution

15 = m * 5

15/5 = m

m = 3 kg

Second ball

Givens

a = 4m/s^2

m = ??

F = 15N

Formula

F = m*a

Solution

15 = m * 4

15/4 = m

m = 3.75 kg