An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is spun at 646.1 rad/s646.1 rad/s while it is being read, and then is allowed to come to rest over 0.234 seconds0.234 seconds , what is the magnitude of the average angular acceleration of the disk?

A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.

A book slides across a level, carpeted floor with an initial speed (u) of 3.25 m/s. It comes to rest (final speed = v = 0m/s) after 3.25 m (s).

Since this is a uniformly decelerated rectilinear motion, we can calculate the acceleration (a) using the following kinematic expression.

[tex]v^{2} = u^{2} + 2as\\\\(0m/s)^{2} = (3.25m/s)^{2} + 2a(3.25m)\\\\a = -1.63 m/s^{2}[/tex]

The negative sign only explains that the friction opposes the movement. We can calculate the force exerted by the friction (F) using Newton's second law of motion.

[tex]F = m \times a[/tex]    [1]

where,

We can also calculate the force of friction using the following expression.

[tex]F = k \times N = k \times m \times g[/tex]    [2]

where,

Given [1] = [2], we get,

[tex]m \times a = k \times m \times g\\\\k = \frac{1.63 m/s^{2} }{9.81 m/s^{2}} = 0.166[/tex]

A book that slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m, has a coefficient of kinetic friction of 0.166.

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This question involves the concepts of the equations of motion, Newton's Second Law, and Frictional Force.

The coefficient of kinetic friction is found to be "0.15".

First, we will use the third equation of motion to find out the acceleration of the book:

[tex]2as = v_f^2-v_i^2\\[/tex]

where,

a = acceleration = ?

s = distance covered = 3.25 m

vf = final speed  = 0 m/s

vi = initial speed = 3.25 m/s

Therefore,

[tex]2a(3.25\ m) = (0\ m/s)^2-(3.25\ m/s)^2\\\\a=\frac{-10.56\ m^2/s^2}{7\ m/s}[/tex]

a = - 1.51 m/s² (negative sign indicates deceleration)

Now the force can be calculated using Newton's Second Law of motion:

F = ma

This force is also equal to the frictional force:

F = μmg

comparing both forces, we get:

ma = μmg

a = μg

[tex]\mu = \frac{a}{g} = \frac{1.51\ m/s^2}{9.81\ m/s^2}[/tex]

μ = 0.15

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The attached picture shows the equations of motion.

Answer:

The baseball does not reach the wall, because the ball falls at 125.33 meters and the wall is at 150 meters

Explanation:

V= 37.5 m/s

α= 30º

g= 9.8 m/s²

Vx= V * cos(30º) = 32.47 m/s

Vy= V * sin(30º) = 18.75 m/s

flytime of the baseball:

t= 2 * Vy/g

t= 3.86 sec

distance of baseball fall:

d= Vx * t

d= 125.33 m

Answer: 18.67 m wall

Explanation:this is projectile.

For Max height,

H = v²sin²©/2g

H= (37.5)²sin²(30)/(2*9.81)

H = 17.919 m

But he is standing 0.751 m above ground, so total height = 17.919+0.751= 18.67m

Answer: false

Explanation: the air will increase if it is compressed by an adiabatic compressor

Answer:

a) Magnitude of the average velocity from t = 0 to t = 2 s = 2.24 m/s

b) Magnitude of the average velocity from t = 0 to t = 5 s = 3.16 m/s

Explanation:

a) Velocity is rate of change of position.

A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0

A particle's position coordinates (x, y) are (5.0 m, 5.0 m) at t = 2.0 s

Displacement = (5-1)i + (5-3)j = 4i + 2j

Change in time = 2s

Velocity

      [tex]v=\frac{4i+2j}{2}=2i+j[/tex]

Magnitude of velocity

      [tex]v=\sqrt{2^2+1^2}=2.24m/s[/tex]

b) Velocity is rate of change of position.

A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0

A particle's position coordinates (x, y) are (14.0 m, 12.0 m) at t = 5.0 s

Displacement = (14-1)i + (12-3)j = 13i + 9j

Change in time = 5s

Velocity

      [tex]v=\frac{13i+9j}{5}=2.6i+1.8j[/tex]

Magnitude of velocity

      [tex]v=\sqrt{2.6^2+1.8^2}=3.16m/s[/tex]

Answer:

Part a)

[tex]q_2 = -6.8 \mu C[/tex]

Part b)

[tex]F = 0.700 N[/tex]

direction = downwards

Explanation:

As we know that the negative charge will experience the force due to some other charge below it

the force is given as

[tex]F = 0.700 N[/tex]

now we know that

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

now plug in all data

[tex]0.700 = \frac{(9 \times 10^9)(0.550\mu C)q_2}{0.220^2}[/tex]

[tex]0.700 = 1.022\times 10^5 q_2[/tex]

[tex]q_2 = -6.8 \mu C[/tex]

since this is a repulsion force so it must be a negative charge

Part b)

As per Newton's III law it will exert equal and opposite force on it

So here the force on the charge below it will be same in magnitude but opposite in direction

so here we have

[tex]F = 0.700 N[/tex]

direction = downwards

It takes 225 Earth days for Venus to orbit the Sun. Interestingly, Venus spins on its axis so slowly that its day (243 Earth days) is longer than its year. The Sun takes 117 Earth days to return to the same place in Venus' sky.

The time it takes for Venus to make one orbit around the sun, also known as its orbital period, is actually 225 Earth days. This is quite different compared to Earth which takes 365.25 days to orbit the Sun. Additionally, Venus spins on its axis very slowly, with its rotational period being 243 Earth days. As a result, a day on Venus - considering its rotation - is longer than its year! Also, this leads to an unusual phenomenon where the sun takes 117 Earth days to return to the same place in the Venusian sky. This suggests that Venus' rotation and orbit display unique characteristics compared to other planets in our solar system, likely due to factors from its formation.

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Venus completes one orbit around the Sun in approximately 225 Earth days. This is shorter than Earth's orbital period due to Venus's closer distance to the Sun.

Orbit of Venus Around the Sun-

The distance between the Earth and the Sun is approximately 1.50 x 1011 meters, while Venus is closer, at 1.08 x 1011 meters. This difference in distance influences the orbital period of each planet. Venus takes about 225 Earth days to complete one full orbit around the Sun.

Venus has a nearly circular orbit and, being closer to the Sun than Earth, receives almost twice as much light and heat. The elliptical orbits mean that different planets have different orbital periods, with those closer to the Sun having shorter years.

Answer:

Tangential speed, v = 314.15 m/s

Explanation:

It is given that,

Mass of the object, m = 15 kg

It is moving in a circle of radius, r = 10 m

Time period, t = 0.2 seconds

We need to find the tangential velocity of the object. It is given by :

[tex]v=\dfrac{2\pi r}{t}[/tex]

Where

v = tangential speed

[tex]v=\dfrac{2\pi \times 10\ m}{0.2\ s}[/tex]

v = 314.15 m/s

So, the tangential speed of the object is 314.15 m/s. Hence, this is the required solution.

Answer:

37.725 A

Explanation:

B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T

r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m

i = magnitude of current flowing through the wire = ?

Magnetic field by a long wire is given as

[tex]B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}[/tex]

Inserting the values

[tex]0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}[/tex]

i = 37.725 A

Answer:

Part a)

[tex]k = 1632 J[/tex]

Part b)

[tex]KE = 47 J[/tex]

Part c)

[tex]m = 7.9 kg[/tex]

Part d)

[tex]v = 2.57 m/s[/tex]

Part e)

[tex]KE = 26.1 J[/tex]

Part f)

[tex]PE = 20.9 J[/tex]

Explanation:

Total Mechanical energy is given as

[tex]E = 47.0 J[/tex]

Its maximum displacement from mean position is given as

[tex]A = 0.240 m[/tex]

Part a)

Now from the formula of energy we know that

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

[tex]47 = \frac{1}{2}k(0.240)^2[/tex]

[tex]k = 1632 J[/tex]

Part b)

At the mean position of SHM whole mechanical energy will convert into kinetic energy

so it is given as

[tex]KE = 47 J[/tex]

Part c)

As per the formula of kinetic energy we know that

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]47 = \frac{1}{2}m(3.45^2)[/tex]

[tex]m = 7.9 kg[/tex]

Part d)

As we know by the equation of the speed of SHM is given as

[tex]v = \sqrt{\frac{k}{m}(A^2 - x^2)}[/tex]

[tex]v = \sqrt{\frac{1632}{7.9}(0.24^2 - 0.16^2)}[/tex]

[tex]v = 2.57 m/s[/tex]

Part e)

As we know by the formula of kinetic energy

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]KE = \frac{1}{2}(7.9)(2.57^2)[/tex]

[tex]KE = 26.1 J[/tex]

Part f)

As per energy conservation we know

KE + PE = Total energy

[tex]26.1 + PE = 47[/tex]

[tex]PE = 20.9 J[/tex]

(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]

(b) The KE at equilibrium point =47j

(c) The mass =7.9kg

(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]

(e)The KE of block 0.160m will be 26.1

(f) The PE will be 20.9j

(a) for finding spring constant

KE=47 j

A=0.240m

By using the formula

[tex]E=\dfrac{1}{2} kx^2[/tex]

[tex]47=\dfrac{1}{2} k(0.240)^2[/tex]

[tex]k=1632\frac{N}{m^2}[/tex]

(b) At the mean position the whole mechanical energy will be equal to KE so

KE=47j

(c) The mass of the system

[tex]KE =\dfrac{1}{2} mv^2[/tex]

[tex]47=\dfrac{1}{2} m(3.45^2)[/tex]

[tex]m=7.9kg[/tex]

(d)Now the speed of the block

[tex]v=\sqrt{\dfrac{k}{m} (A^2-x^2)}[/tex]

[tex]v=\sqrt{\dfrac{1632}{7.9} (0.24^2-0.16^2)}[/tex]

[tex]v=2.57\frac{m}{s}[/tex]

(e) The KE of the block

[tex]KE=\dfrac{1}{2} mv^2=\dfrac{1}{2} 7.9(2.57)^2=26.1J[/tex]

(f) The PE of the system

[tex]Total Energy = KE+PE[/tex]

[tex]PE= 47-26.1 =20.9J[/tex]

Thus

(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]

(b) The KE at equilibrium point =47j

(c) The mass =7.9kg

(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]

(e)The KE of block 0.160m will be 26.1

(f) The PE will be 20.9j

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Answer:

The new force F' will be same of the original force F.

Explanation:

Given that,

Charges = 0.3

We need to calculate the force between the charges

Suppose that the distance between the charges is r.

The force between the charges

[tex]F =\dfrac{kq_{1}q_{2}}{r^2}[/tex]

Put the value into the formula

[tex]F=\dfrac{k\times0.3\times0.3}{r^2}[/tex]

[tex]F=\dfrac{k\times(0.09)}{r^2}[/tex]

If the charges are doubled, and the distance between them increased by 100%.

So, The charges are 0.6 and the distance is 2r.

Then,

The force between the charges

[tex]F'=\dfrac{k\times0.6\times0.6}{(2r)^2}[/tex]

[tex]F'=\dfrac{k\times0.09}{r^2}[/tex]

[tex]F'=F[/tex]

Hence, The new force F' will be same of the original force F.

Answer:

12 m

Explanation:

f = 0.2 Hz, v = 2.4 m/s

v = f x λ

Where, λ is the wavelength

λ = v / f = 2.4 / 0.2 = 12 m

The wavelength is defined as the distance between two consecutive crests or troughs.

So, the distance between two consecutive crests is 12 m.

Answer:

a)

600 N

b)

1.2 x 10⁵ Pa

Explanation:

(a)

d₁ = diameter of small piston = 8 cm = 0.08 m

d₂ = diameter of large piston = 40 cm = 0.40 m

F₂ = force applied to large piston = 15000 N

F₁ = force applied to small piston = ?

Using pascal's law

[tex]\frac{F_{1}}{(0.25)\pi d_{1}^{2}} = \frac{F_{2}}{(0.25)\pi d_{2}^{2}}[/tex]

Inserting the values

[tex]\frac{F_{1}}{(0.08)^{2}} = \frac{15000}{(0.40)^{2}}[/tex]

F₁ = 600 N

b)

Pressure applied is given as

[tex]P = \frac{F_{1}}{(0.25)\pi d_{1}^{2}}[/tex]

[tex]P = \frac{(600)}{(0.25)(3.14) (0.08)^{2}}[/tex]

P = 1.2 x 10⁵ Pa

To determine the force needed on the small piston of a hydraulic lift, the ratios of the pistons' areas are used in combination with the load on the large piston, applying Pascal's Principle. The pressure in the hydraulic fluid is then found by dividing the force by the area of the piston where the force is applied.

Given the diameters of the pistons, we can find the areas by using the formula for the area of a circle, A = πr², where r is the radius of the piston.

For the provided diameters of 8.0 cm and 40 cm, the areas of the pistons are:

With a load of 15,000 N on the large piston:

Once we perform these calculations, we can determine the required force to apply to the small piston and the pressure applied to the fluid within the lift.

Answer:

The acceleration of the block is 6.35 m/s².

Explanation:

It is given that,

A force is applied to a block to move it up a 30° incline. The applied force is, F = 65 N

Mass of the block, m = 5 kg

We need to find the acceleration of the block. From the attached figure, it is clear that.

[tex]F_x=ma_x[/tex]

[tex]F\ cos\theta-mg\ sin\theta=ma_x[/tex]

[tex]a_x=\dfrac{F\ cos\theta-mg\ sin\theta}{m}[/tex]

[tex]a_x=\dfrac{65\ N\ cos(30)-5\ kg\times 9.8\ m/s^2\ sin(30)}{5\ kg}[/tex]

[tex]a_x=6.35\ m/s^2[/tex]

So, the acceleration of the block is 6.35 m/s². Hence, this is the required solution.

Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².The magnitude of the acceleration of the block will be 6.35 m/sec².

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and accelertaion in the different components and balancing the equation gets.Components in the x-direction.

The given data in the problem,

F is the applied force =65 N

Θ is the angle of inclined plane=30°

m is the mass of the block= 5 kg

We need to find the acceleration of the block in the x-direction

[tex]\rm F_x=ma_x\\\\\rm F_x=Fcos\theta-mgsin\theta\\\\\rm Fcos\theta-mgsin\theta=ma_x\\\\\rm a_x=\frac{Fcos\theta-mgsin\theta}{m}\\\\ \rm a_x=\frac{65\timescos30^0-mgsin30^0}{5} \\\\\rm a_x=6.35 m/sec^2[/tex]

Hence the magnitude of the acceleration of the block will be 6.35 m/sec².

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Answer:

d = 0.50 m

Explanation:

Let say the speed at the top and bottom of the window is

[tex]v_1 \: and \: v_2[/tex] respectively

now we have

[tex]d = \frac{v_1 + v_2}{2}t[/tex]

[tex]1.90 = \frac{v_1 + v_2}{2} (0.380)[/tex]

[tex]v_1 + v_2 = 10 [/tex]

also we know that

[tex]v_2 - v_1 = 9.8(0.380)[/tex]

[tex]v_2 - v_1 = 3.72[/tex]

now we have from above equations

[tex]v_2 = 6.86 m/s[/tex]

[tex]v_1 = 3.14 m/s[/tex]

now the distance from which it fall down is given as

[tex]v_f^2 - v_i^2 = 2ad[/tex]

[tex]3.14^2 - 0^2 = 2(9.8)d[/tex]

[tex]d = 0.50 m[/tex]

Answer:

Part a)

Final charge on C : q = 1.875

Part b)

Ratio for A = 6 : 1.25

Ratio for B = -1 : 1.875

Ratio for C = 0

Explanation:

When two identical metal sphere are connected together then the charge on them will get equally divided on both after connecting them by conducting wire

So here we have

[tex]q_A = + 6q[/tex]

[tex]q_B = -q[/tex]

[tex]q_c = 0[/tex]

Step 1: We connected A and B and then separate them

so we have

[tex]q_A' = q_B' = 2.5q[/tex]

Step 2: We connected A and C and then separate them

so we have

[tex]q_A'' = q_c' = 1.25q[/tex]

Step 3: We connected B and C and then separate them

so we have

[tex]q_c'' = q_b'' = 1.875q[/tex]

The final charge on sphere C is 2.5q. Before touching, the total charge is +5q, which remains the same after all interactions, demonstrating conservation of charge.

When identical conductive spheres come into contact, the charges redistribute evenly across both spheres. If Sphere A is initially charged with +6q and Sphere B has a -q charge, touching them together allows their total charge to be shared, resulting in each sphere having (6q - q)/2 = 2.5q. After separation, both spheres A and B would have a charge of 2.5q. By touching Sphere C, which is uncharged, to A and then B in sequence, C gains a fraction of the charge from each, ending up with (2.5q)/2 from A and (2.5q)/2 from B, which totals 2.5q, since touching B does not change the charge obtained from A.

Before contact, the total charge is +5q (+6q from A and -q from B). After all the interactions, the total charge remains the same, +5q, but redistributed: A and B with 2.5q each and C with 2.5q.

In general, the total charge before and after remains constant, demonstrating conservation of charge.

Answer:

[tex]v_f = 27.9 m/s[/tex]

Explanation:

Component of the weight of the truck along the inclined plane is given as

[tex]F_1 = W sin\theta[/tex]

[tex]F_1 = 15000 sin15[/tex]

[tex]F_1 = 3882.3 N[/tex]

also the engine is providing the constant force to it as

[tex]F_2 = 8000 N[/tex]

now the net force along the the plane is given as

[tex]F_{net} = 8000 + 3882.3[/tex]

[tex]F = 11882.3 N[/tex]

mass of the truck is given as

[tex]m = \frac{w}{g} = 1529 kg[/tex]

now the acceleration is given as

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

[tex]a = 7.77 m/s^2[/tex]

now the speed of the truck after travelling distance of d = 50 m is given as

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

[tex]v_f^2 = 0 + 2(7.77)(50)[/tex]

[tex]v_f = 27.9 m/s[/tex]

Answer:

09m

Explanation:

yes the current flow in clockwise direction

Explanation:

Given that,

Diameter of the metal ring, d = 4.2 cm

Radius, r = 2.1 cm

Initial magnetic field, B = 1.12 T

The magnetic field is decreasing at the rate of, [tex]\dfrac{dB}{dt}=0.24\ T/s[/tex]

Due to change in magnetic field, an emf is induced in it. And hence, electric field is induced. It is given by :

[tex]\int\limits {E.dl}=\dfrac{d}{dt}(\int\limits{B.dA)}[/tex]

[tex]E.(2\pi r)=\dfrac{dB}{dt}(\pi r^2)[/tex]

[tex]E=\dfrac{dB}{dt}\times \dfrac{r}{2}[/tex]

[tex]E=0.24\times \dfrac{2.1\times 10^{-2}}{2}[/tex]

[tex]E=0.00252\ N/C[/tex]

So, the magnitude of the electric field induced in the ring has a magnitude of 0.00252 N/C.

The direction of electric field will be counter clock wise direction as viewed by someone on the south pole of the magnet.

Answer:

The final temperature is 79.16°C.

Explanation:

Given that,

Heat [tex]Q=1.50\times10^{5}\ J[/tex]

Temperature = 20.0°C

Entropy = 465 J/k

We need to calculate the average temperature

Using relation between entropy and heat

[tex]\Delta S=\dfrac{\Delta Q}{T}[/tex]

[tex]T=\dfrac{\Delta Q}{\Delta S}[/tex]

Where, T = average temperature

[tex]\Delta Q[/tex]= transfer heat

[tex]\Delta S[/tex]= entropy

Put the value into the formula

[tex]T=\dfrac{1.50\times10^{5}}{465}[/tex]

[tex]T=322.58\ K[/tex]

We need to calculate the final temperature

Using formula of average temperature

[tex]T = \dfrac{T_{i}+T_{f}}{2}[/tex]

[tex]T_{f}=2T-T_{i}[/tex]....(I)

Put the value in the equation (I)

[tex]T_{f}=2\times322.58-293[/tex]

[tex]T_{f}=352.16\ K[/tex]

We convert the temperature K to degrees

[tex]T_{f}=352.16-273[/tex]

[tex]T_{f}=79.16^{\circ}\ C[/tex]

Hence, The final temperature is 79.16°C.

Answer:

option C

multiplication factor n = 1 when volume change to half and pressure become double

Explanation:

we know by Ideal Gas law:

P1V1 = nRT1

P2V2 = nRT2

according to the question

pressure is doubled and volume is reduced to half  

so we have

new pressure = 2*P1

new volume = V1/2

hence,

(2P1) * (V1/2) = nRT2

P1V1 = nRT2

we have now

nRT1 = nRT2

we get

T1 = T2

thus no change in temperature

we know that internal energy is given as

internal energy = nCvT,

since temperature is directly proportional to internal energy and since temperature remains constant therefore internal energy remains constant

So there is no change in internal energy

thus, multiplication factor n = 1 when volume change to half and pressure change to double

The light leaves the glass at about 42° relative to its normal

[tex]\texttt{ }[/tex]

Let's recall Snell's Law about Refraction as follows:

[tex]\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}[/tex]

where:

n₁ = incident index

θ₁ = incident angle

n₂ = refracted index

θ₂ = refracted angle

[tex]\texttt{ }[/tex]

Given:

incident angle = θ₁ = 30°

index of refraction of air = n₃ = 1.0

index of refraction of glass = n₂ = 1.5

index of refraction of water = n₁ = 1.33

Asked:

refracted angle = θ₃ = ?

Solution:

Surface between Glass - Water:

[tex]n_1 \sin \theta_1 = n_2 \sin \theta_2[/tex]

[tex]\sin \theta_2 = \frac{n_1}{n_2} \sin \theta_1[/tex] → Equation A

[tex]\texttt{ }[/tex]

Surface between Air - Glass:

[tex]n_2 \sin \theta_2 = n_3 \sin \theta_3[/tex]

[tex]\sin \theta_3 = \frac{n_2}{n_3} \sin \theta_2[/tex]

[tex]\sin \theta_3 = \frac{n_2}{n_3} (\frac{n_1}{n_2} \sin \theta_1)[/tex] ← Equation A

[tex]\sin \theta_3 = \frac{n_1}{n_3} \sin \theta_1[/tex]

[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \sin 30^o[/tex]

[tex]\sin \theta_3 = \frac{1.33}{1.0} \times \frac{1}{2}[/tex]

[tex]\sin \theta_3 \approx 0.665[/tex]

[tex]\boxed{\theta_3 \approx 42^o}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Light

The angle at which the light leaves the glass is approximately 19.5°.

When a light ray crosses from one medium to another, such as from water to glass, it undergoes refraction, which causes the ray to bend. The angle between the incident ray and the normal to the surface of the glass is called the angle of incidence. To find the angle of incidence, we can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the indices of refraction of the two media.

In this case, the index of refraction of water (n1) is 1.33 and the index of refraction of glass (n2) is 1.5.

The angle of incidence (θ1) is given as 30°. We can use the formula: n1*sin(θ1) = n2*sin(θ2) to solve for θ2, the angle at which the light leaves the glass.

Plugging in the values, we get: 1.33*sin(30°) = 1.5*sin(θ2).

Solving for θ2 gives us θ2 = sin^(-1)((1.33*sin(30°))/1.5).

Evaluating this expression, we find that θ2 is approximately 19.5°.

Answer:

Explanation:

Comet is made by dust particles, icy particles, gases etc.

A comet has a fixed time to complete a revolution around the sun.

As a comet comes nearer to the sun, due to the heat of the sun the vapour and the icy particles becomes gases and due to the radial pressure of energy od sun, we observe a tail of comet which has vapours mainly. SI the comet is visible easily.

Explanation:

It is given that,

Mass of SUV, m = 900 kg

It is moving with a constant speed of 1.44 m/s due south. We need to find the total force on the vehicle. The second law of motion gives the magnitude of force acting on the object. It is given by :

F = m a

Since, [tex]a=\dfrac{dv}{dt}[/tex] i.e. the rate of change of velocity

As SUV is travelling at a constant speed. This gives acceleration of it as zero.

So, F = 0

So, the total force acting on SUV is 0 N. Hence, this is the required solution.

Answer:

The total force acting on the vehicle is 0 N.

Explanation:

Given that,

Mass of SUV = 900 kg

Velocity = 1.44 m/s

SUV is travelling at a constant speed of 1.44 m/s due north.

We need to calculate the force on the vehicle.

Using newton's second law

[tex]F = ma[/tex]....(I)

We know that,

The acceleration is the first derivative of the velocity of the particle.

[tex]a = \dfrac{dv}{dt}[/tex]

Now,put the value of a in equation (I)

[tex]F=m\dfrac{dv}{dt}[/tex]

SUV is traveling at a constant speed it means acceleration is zero.

So, The force will be zero.

Hence, The total force acting on the vehicle is 0 N.

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = [tex]\frac{1}{3}\pi r^2 h[/tex]

thus,

volume of the cone = [tex]\frac{1}{3}\pi 1.2^2\times 4[/tex] = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the [tex]\frac{1}{4}[/tex]times the total height

thus,

center of mass lies at,  h' = [tex]\frac{1}{4}\times4=1m[/tex]

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

The work against gravity required to build a cone of height 4 m and base of radius 1.2 m out of a material of density 600 kg/m3 is 35,467.278 J.

Given to us

Height of the cone = 4m

The radius of the cone = 1.2 m

Density of the material = 600 kg/m³

We know that the work gone against the gravity is given as,

[tex]W = mgh'[/tex]

where W is the work, m is the mass, g is the acceleration due to gravity, and h' is the center of mass.

Also, the mass of an object is the product of its volume and density, therefore,

[tex]m = v \times \rho[/tex]

We know the value of the volume of a cone,

[tex]m = \dfrac{1}{3}\pi r^2 h \times \rho[/tex]

The center of mass of a cone lies at the center of its base at 1/4 of the total height from the base,

[tex]h' = \dfrac{h}{4}[/tex]

Substitute all the values we will get,

[tex]W = mgh'\\\\W = (v \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi r^2h \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi (1.2)^2 \times4 \times 600) \times 9.81 \times \dfrac{4}{4}\\\\W = 35467.278\rm\ J[/tex]

Hence, the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3 is 35,467.278 J.

Learn more about Work:

https://brainly.com/question/81684

Answer:

[tex]v = 4.51 \times 10^3 m/s[/tex]

Explanation:

electric field = 2170 N/C

now the speed of the charge particle is given as

[tex]v = 5.45 \times 10^3 m/s[/tex]

here we know that charge particle moves without any deviation

so we will have

[tex]qvB = qE[/tex]

now magnetic field in this region is given as

[tex]B = \frac{E}{v}[/tex]

[tex]B = \frac{2170}{5.45 \times 10^3}[/tex]

[tex]B = 0.398 T[/tex]

Now another charge particle enters the region with different speed and experience the force upwards

[tex]F = qE - qvB[/tex]

[tex]1.54 \times 10^{-9} = (4.10\times 10^{-12})[2170 - v(0.398)][/tex]

[tex]375.6 = 2170 - v(0.398)[/tex]

[tex]v = 4.51 \times 10^3 m/s[/tex]

The answer is: [tex]9.43 \times 10^5 \, \text{m/s}[/tex].

To determine the speed of the different particle that enters the velocity selector, we need to consider the forces acting on it due to the electric and magnetic fields. The net force acting on a charged particle in an electric field [tex]\( E \)[/tex] and a magnetic field [tex]\( B \)[/tex] is given by the Lorentz force equation:

[tex]\[ \vec{F} = q\vec{E} + q(\vec{v} \times \vec{B}) \][/tex]

where [tex]\( q \)[/tex] is the charge of the particle, [tex]\( \vec{v} \)[/tex] is the velocity of the particle, [tex]\( \vec{E} \)[/tex] is the electric field, and [tex]\( \vec{B} \)[/tex] is the magnetic field.

For the particles that pass through the velocity selector without being deflected, the forces due to the electric and magnetic fields must cancel each other out. This means that:

[tex]\[ qE = qvB \][/tex]

where [tex]\( v \)[/tex] is the speed of the particles that pass through undeflected.

Given that the electric field [tex]\( E \)[/tex] is 2170 N/C directed vertically upward, and the particles are traveling east, the magnetic field [tex]\( B \)[/tex] must be directed south to produce a force that cancels the electric force. The speed [tex]\( v \)[/tex] of the undeflected particles is given as [tex]\( 5.45 \times 10^3 \)[/tex] m/s.

Now, for the different particle with a charge of [tex]\( +4.10 \times 10^{-12} \)[/tex] C, the net force [tex]\( F \)[/tex] acting on it is [tex]\( 1.54 \times 10^{-9} \)[/tex] N, pointing directly upward. This means that the magnetic force is not sufficient to cancel the electric force, and the net force is due to the electric force only:

[tex]\[ F = qE \][/tex]

We can use this equation to find the speed [tex]\( v' \)[/tex] of the different particle. Since the net force is equal to the electric force, the magnetic force must be zero. This implies that the velocity of the particle is such that the magnetic force component is equal and opposite to the electric force component, but since the net force is not zero, the particle is not moving at the correct speed to pass through undeflected.

Let's solve for the speed [tex]\( v' \)[/tex] of the different particle:

[tex]\[ F = qE \][/tex]

[tex]\[ 1.54 \times 10^{-9} \, \text{N} = (4.10 \times 10^{-12} \, \text{C})(2170 \, \text{N/C}) \][/tex]

[tex]\[ v' = \frac{F}{qB} \][/tex]

We know [tex]\( F \), \( q \), and \( E \)[/tex], but we need to find [tex]\( B \)[/tex] from the information given for the undeflected particles:

[tex]\[ qvB = qE \][/tex]

[tex]\[ vB = E \][/tex]

[tex]\[ B = \frac{E}{v} \][/tex]

[tex]\[ B = \frac{2170 \, \text{N/C}}{5.45 \times 10^3 \, \text{m/s}} \][/tex]

[tex]\[ B = 3.98 \times 10^{-4} \, \text{T} \][/tex]

Now we can find [tex]\( v' \)[/tex]:

[tex]\[ v' = \frac{1.54 \times 10^{-9} \, \text{N}}{(4.10 \times 10^{-12} \, \text{C})(3.98 \times 10^{-4} \, \text{T})} \][/tex]

[tex]\[ v' = \frac{1.54 \times 10^{-9}}{1.6322 \times 10^{-15}} \][/tex]

[tex]\[ v' = 9.43 \times 10^5 \, \text{m/s} \][/tex]

Therefore, the speed of the different particle is [tex]\( 9.43 \times 10^5 \) m/s[/tex].

Answer:

v = 10.1 m/s

Explanation:

As we know that by the law of conservation of volume the rate of volume flowing through the pipe will remain conserved

so here we have flow rate given as

[tex]Q = Area\times velocity[/tex]

now we have

[tex]A_1 v_1 = A_2 v_2[/tex]

now we have

[tex]A_1 = 3.70 \times 10^{-2} m^2[/tex]

[tex]v_1 = 0.260 m/s[/tex]

[tex]A_2 = 9.50 \times 10^{-4} m^2[/tex]

now from above equation we have

[tex]v_2 = \frac{A_1}{A_2} v_1[/tex]

[tex]v_2 = \frac{3.70\times 10^{-2}}{9.50\times 10^{-4}}(0.260)[/tex]

[tex]v_2 = 10.1 m/s[/tex]

Answer:

a). The potential is highest at the center of the sphere

Explanation:

We k ow the potential of a non conducting charged sphre of radius R at a point r < R is given by

[tex]E=\left [ \frac{K.Q}{2R} \right ]\left [ 3-(\frac{r}{R})^{2} \right ][/tex]

Therefore at the center of the sphere where r = 0

[tex]E=\left [ \frac{K.Q}{2R} \right ]\left [ 3-0 \right ][/tex]

[tex]E=\left [ \frac{3K.Q}{2R} \right ][/tex]

Now at the surface of the sphere where r = R

[tex]E=\left [ \frac{K.Q}{2R} \right ]\left ( 3-1 \right )[/tex]

[tex]E=\left [ \frac{2K.Q}{2R} \right ][/tex]

[tex]E=\left [ \frac{K.Q}{R} \right ][/tex]

Now outside the sphere where r > R, the potential is

[tex]E=\left [ \frac{K.Q}{r} \right ][/tex]

This gives the same result as the previous one.

As [tex]r\rightarrow \infty , E\rightarrow 0[/tex]

Thus, the potential of the sphere is highest at the center.

Answer:

13.79 revolutions

Explanation:

Velocity of ball = v = 34.5 m/s

Spin rate of ball = 26 revolutions/s

Distance between home plate and pitching mound = s = 18.3 m

Time taken by the ball to reach the home plate = t

[tex]t=\frac{18.3}{34.5}\ s[/tex]

Number of revolutions the ball completes is

[tex]26\times \frac{18.3}{34.5}=13.79[/tex]

∴Number of revolutions the ball completes before reaching home plate is 13.79

Answer:

The number of revolutions is 13.78

Explanation:

Given data:

v = speed = 34.5 m/s

d = distance = 18.3 m

spin rate = 26 rev/s

The time taken is equal to:

[tex]t=\frac{d}{v} =\frac{18.3}{34.5} =0.53s[/tex]

The number of revolutions is equal:

N = 0.53 * 26 = 13.78 rev

Answer:

a) 800 keV

b) 25 km

Explanation:

[tex]Strength\ of\ Electric\ field=2\times 10^6\ V/m\\a)\ Potential\ Difference=Strength\ of\ Electric\ field\times Distance\\\Rightarrow Potential\ Difference=Kinetic\ Energy\ =2\times 10^6\times 0.4\\\therefore Energy=0.8\times 10^6\ eV=800\ keV\\[/tex]

[tex]b)\ Potential\ difference=50\ GeV=50\times 10^9\ eV\\Distance=\frac{Potential\ difference}{electric\ field}\\\Rightarrow Distance=\frac{50\times 10^9}{2\times 10^6}\\\Rightarrow Distance=25\times 10^3\ m\\\therefore Distance=25\ km[/tex]

Answer:

a) 800 keV

b)  24.996 km.

Explanation:

(a) we have

[tex]\large \Delta K.E=q\Delta V[/tex]  .............(1)

where,

[tex]\large \Delta K.E[/tex] = Change in kinetic energy

[tex]q[/tex] = charge of an electron

[tex]\Delta V[/tex] = Potential difference

also

[tex]\large E=\frac{V}{d}[/tex]       .......(2)

E = electric field

d = distance traveled

Now from (1) and (2) we have,

[tex]\large \Delta K. E=qV=qEd[/tex]

substituting the values in the above equation, we get

[tex]\large \Delta K. E=(1.6\times 10^{-19}C)(2\times 10^6V/m)(0.400m)(\frac{1eV}{1.6\times 10^{-19}J})(\frac{1keV}{1000eV})[/tex]

[tex]\large \Delta K. E=800keV[/tex]

Thus, the energy gained by the electron is 800 keV if it is accelerated over a distance of 0.400 m.

(b) Using the equation (1), we have

[tex]\large d=\frac{\Delta K.E}{qE}[/tex]

[tex]\large d=\frac{(50\times 10^9eV)}{(1.6\times 10^{-19C})(2\times 10^6V/m)}(\frac{1.6\times 10^{-19}J}{1eV})[/tex]

or

[tex]\large d=2.4996\times 10^4m[/tex]

or

[tex]\large d=24.996\times 10^3m=24.996km[/tex]

Thus, to gain 50.0 GeV of energy the electron must be accelerated over a distance of 24.996 km.

Answer:

42.3 MV

Explanation:

d = diameter of the metal sphere = 2.15 m

r = radius of the metal sphere

diameter of the metal sphere is given as

d = 2r

2.15 = 2 r

r = 1.075 m

Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C

Potential near the surface is given as

[tex]V = \frac{kQ}{r}[/tex]

[tex]V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}[/tex]

V = 4.23 x 10⁷ volts

V = 42.3 MV

Answer:

55.79 minutes

Explanation:

Volume of water = 5200 L = 5.2 m^3

Diameter = 3 cm

radius, r = 1.5 cm = 0.015 m

v = 2.2 m/s

The volume of water coming out in 1 second = velocity x area

                                                       = 2.2 x 3.14 x 0.015 x 0.015

                                                       = 1.55 x 10^-3 m^3

1.55 x 10^-3 m^3 water flows = 1 second

5.2 m^3 water flows = 5.2 / (1.55 x 10^-3) = 3345.57 second

                                                                    = 55.79 minutes