It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used as colonies. Inhabitants of the space colonies would live on the inside surface of the cylinder. Inertial effects would resemble gravity's influence and keep them 'plastered to the surface.' Suppose that you are an inhabitant of a space colony which is 1070 miles in length and 4.86 miles in diameter. How many revolutions per hour must the cylinder have in order for the occupants to experience a centripetal acceleration equal to the acceleration of gravity

The stiffness constant of the spring is 68,290.3 N/m

Apply the principle of conservation of energy;

U = K.E

¹/₂kx² = ¹/₂mv²

kx² = mv²

k = mv²/x²

where;

k = (1300 x 16.67²)/(2.3²)

k = 68,290.3 N/m

Thus, the stiffness constant of the spring is 68,290.3 N/m.

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The  energy she has at each position shown are:

Velocity is known to be a term that connote the direction of any kind of a moving body or an object.

Note that the Speed is known to be a a scalar quantity and as such,  Velocity is said to be a vector quantity.

Note also that from the question given,  Jessica's height of the swing 3 meters above the ground, therefore:

Jessica Position at maximum height :

PE = mgh

PE = 44kg x  9.8m/s² x 3

PE = 1,294 J

Jessica  Position at minimum height:

PE = 0 J

Jessica  Position at maximum velocity:

KE = 1/2 x mv²

KE = 1/2 x 44kg x (5m/s²)²

KE = 550 J

Jessica  Position at minimum velocity:

KE = 0 J

Therefore, The  energy she has at each position shown are:

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The mass of the airplane of area of two wings 395m², and the average speed of lower and upper surface of the wings are, 259 and 288m/s is 386.8×10^3 kg.

To find the answer, we need to know about the Bernoulli's principle.

                   P+ [tex]\frac{1}{2}[/tex] ρv²+ρgh=a constant.

Instead of ρ we take d as density.

                    [tex]A= 395 m^2\\v_l=295 m/s\\v_u=288m/s\\density=1.21kg/m^3[/tex]

                     [tex]P_1+\frac{1}{2}v_1^2d+dgh_1=P_2+ \frac{1}{2}v_2^2d+dgh_2\\here\\h_1=h_2\\thus\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\P_1-P_2=\frac{1}{2}*1.21(288^2-259^2)=9597.12 atm\\[/tex]

                       [tex]\frac{F}{A}=9597.12 atm\\\\ F=9597.12*395m^2=37.9*10^5 N\\\\mg=37.9*10^5 N\\\\m=\frac{37.9*10^5 N}{9.8}\\\\ m=386.8*10^3kg[/tex]

Thus, we can conclude that, the mass of the airplane is 386.8×10^3 kg.

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The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.

We need to be aware of the Bernoulli principle in order to determine the solution.

                             P+ [tex]\frac{1}{2}[/tex]dv²+dgh = constant.

We substitute d for to represent density.

                           [tex]v_1=259m/s\\v_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]

                            [tex]P_1+\frac{1}{2}dv_1^2+dgh= P_2+\frac{1}{2}dv_2^2+dgh\\\\P_1+\frac{1}{2}dv_1^2=P_2+\frac{1}{2}dv_2^2\\\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\\\P_1-P_2=9597 atm[/tex]

                        [tex]F/A= 9597 atm\\mg=9597*395 =38*10^5N\\m=\frac{38*10^5}{9.8} = 387*10^3kg[/tex]

Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.

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Answer:

1.85c

Explanation:

a photon moves at c, the electron is moving at 0.85c, and since they are moving in opposing directions, the relative speed would be 1.85c

The relative velocity of the electron and photon is equal to the speed of light in a vacuum, which is 3 × 10⁸ m/s The negative sign indicates that they are moving in opposite directions.

The formula for adding velocities in special relativity is given by:

V(relative) = (V₁ - V₂) ÷ (1 - (V₁ × V₂ ÷ c²))

where:

V₁ = velocity of the electron

V₂ = velocity of the photon

c = speed of light in a vacuum

The relative velocity is:

V(relative) = (0.85c - c) ÷ (1 - (0.85c × c ÷ c²))

V(relative) = (0.85c - c) ÷ (1 - 0.85)

V(relative) = (-0.15c) ÷ (0.15)

V(relative) = -c

Therefore, The relative velocity of the electron and photon is equal to the speed of light in a vacuum, which is 3 × 10⁸ m/s The negative sign indicates that they are moving in opposite directions.

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Answer:

12 grams of the isotope carbon-12.

Explanation:

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K is cation by losing of electron whereas Br is anion due to accepting of electrons.

Yes, the positive ion appears on K and become cation whereas the negative ion bears on Br which make it anion because of losing and gaining of electron by these atoms. This transferring of electrons leads to formation of ionic bonds between them.

So we can conclude that K is cation by losing of electron whereas Br is anion due to accepting of electrons.

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(a) The speed of a satellite on a low lying circular orbit around this planet is  7,338.93 m/s.

(b) The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

(c) The radius of the synchronous orbit of a satellite is 69,801 km .

v = √GM/r

where;

v = √[(6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 7,338.93 m/s

v = √2GM/r

v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]

v = 10,378.82 m/s

v = 2πr/T

r = vT/2π

r = (7,338.93 x 16.6 x 3600 s) / (2π)

r = 69,801 km

Thus, the speed of a satellite on a low lying circular orbit around this planet is  7,338.93 m/s.

The minimum speed required for a satellite in order to break free permanently from the planet is 10,378.82 m/s.

The radius of the synchronous orbit of a satellite is 69,801 km .

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Answer:

Maximum height = 2040 m

Explanation:

We can solve the problem using kinematics.

Consider the vertical motion of the object and use the equation:

[tex]\boxed{v^2 = u^2 + 2as}[/tex]

where:

• v = final velocity      (0 m/s, because when the object is at max. height, it has no vertical velocity)

• u = initial velocity    (400sin30° m/s ⇒ vertical component of 400 m/s at 30° to horizontal)

• a = acceleration      (-9.81 m/s²; considering upward acceleration to be negative)

• s = displacement    (? m; this represents the max. height of the object),

Substitute the values into the equation and solve for s :

[tex]0^2 = (400 sin (30 \textdegree))^2 + 2(-9.81)(s)[/tex]

⇒ [tex]2(9.81)(s) = (400 sin (30 \textdegree))^2[/tex]

⇒  [tex]s = 2040 \space\ m[/tex]     (3 s.f.)

The mechanical energy lost is 353.28 Joules . The distance he slid off is 0.16m.

we know:-

Mass = 69 kg

Speed = 3.2 m/s

Coefficient of friction ( ratio of friction force and normal force ) = 0.70

Acceleration due to gravity, g = 9.8 m/s^2

(a) To determine the amount of mechanical energy that is lost because of friction acting on the runner, we would calculate the change in kinetic energy:

[tex]KE = \frac{1}{2} m (v-u)^{2}[/tex]

      [tex]= \frac{1}{2} 69 ( 3.2-0 )^{2}[/tex]

      [tex]= 353.28[/tex] Joules

Mechanical energy = 353.28 Joules

(b) To determine how far (distance) the runner slide:

acceleration = ug

                     [tex]= 3.2[/tex] × [tex]9.8[/tex]

                     [tex]= 31.36[/tex] [tex]m/s^{2}[/tex]

distance ,

[tex]V^{2} = U^{2} + 2aS[/tex]

[tex]( 3.2)^{2} = 0 + 2 ( 31.36) S[/tex]

[tex]10.24 = 62.72 S[/tex]

[tex]S = {\frac{10.24}{62.72} }[/tex]

Distance, S = 0.16 m  

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The shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.

We know that proton is positively charge particle while on the other hand, electron is negatively charge particle. Due to opposite charges, these particles attract each other.

So we can conclude that the shared property of electron and proton that belongs in the region marked "B” is the presence of charge on both of them.

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Answer:

electrically charged

Explanation:

got it right

The total electric potential energy is [tex]\frac{kq_{1} q_{3} }{r_{13} } + \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1} q_{2} }{r_{12} }[/tex].

Electric Potential Energy of a System of Charges :

The system's electric potential energy is equal to the amount of work necessary to create a system of charges by guiding them toward their designated locations from infinity against the electrostatic force without accelerating them. The symbol for it is U.U=W=qV. Electrostatic fields are conservative, therefore the work is independent of the path.

Assume three charges q₁ , q₂ and q₃ bring from infinity to point P.

To bring  q₁ no work is done,

[tex]V_{p} = \frac{kq_{1} }{r_{1} }[/tex]

where, V = electric potential energy.

            q = point charge.

            r = distance between any point around the charge to the point charge.

           k = Coulomb constant; k = 9.0 × 109 N.

Now bring q₂,

[tex]V_{2} = \frac{kq_{2} }{r_{2} }[/tex]

Work done by q₁ ;

[tex]W_{1} = V_{p} q_{2} = \frac{kq_{1}q_{2} }{r_{12} }[/tex]

Now bring  q₃,

[tex]V_{3} = \frac{kq_{3} }{r_{3} }[/tex]

Work done on q₃ by q₁ and q₂

[tex]W= q_{3} [ V_{1} + V_{2} ][/tex]

    [tex]=\frac{kq_{1} q_{3} }{r_{13} }[/tex][tex]+ \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1}q_{3} }{r_{12} }[/tex]

This work done is stored in the form of potential energy.

∴U=W= potential energy of three systems.

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The total electric potential energy is [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]

Electric Potential Energy of a System of Charges

The labor required to establish a system of charges by guiding them toward their intended positions from infinity against the electrostatic force without accelerating them is equal to the electric potential energy of the system. Its identifier is U.U=W=qV. Due to the conservatism of electrostatic fields, the work is independent of the path.

Consider having three charges. Q1, Q2, and Q3 bring point P from infinity.

No work has been done to bring q1,

[tex]V_{1} = \frac{kq1}{r1}[/tex]

where, V = electric potential energy.

           q = point charge.

           r = distance between any point around the charge to the point charge.

          k = Coulomb constant; k = 9.0 × 109 N.

Now bring q₂,

[tex]V_{2} = \frac{kq2}{r2}[/tex]

Work done by q₁ ;

W1 = [tex]V_{p} q2[/tex] = [tex]\frac{kq1q2}{r12}[/tex]

Now bring  q₃,

[tex]V_{3} = \frac{kq3}{r3}[/tex]

Work done on q₃ by q₁ and q₂

W= q3{[tex]V_{1} + V_{2}[/tex]}

W = [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]

This work done is stored in the form of potential energy.

∴U=W= potential energy of three systems.

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A. During long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.

Inertia is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

When an athlete runs before taking the jump, he is trying to increase his inertia, that is his reluctance to stop, thereby increasing his forward motion or jump.

Thus, during long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.

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Answer:

It doubles as well.

Explanation:

Assuming that the area on which the force acts remains constant, and knowing that by definition:

P = F/A

We see that if we have a double force:

p = 2 F/A

Then the pressure is doubled.

Answer:

pressure doubles if area is constant

Explanation:

we know,

p=F/A

when force is double by keeping area constant

F=2F

Then,

Change in pressure (p') will be

P'=2F/A

or,p'=2×F/A

or,p'=2p(since, P=F/A)

therefore, when force is double by keeping area constant the pressure will increase by 2 times

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Answer:

A and C

Explanation:

drag (the area of lower air pressure behind the car when moving) and mostly air resistance (the work to push the air in front of us away to move through - the faster we go, the stronger the air resists to move aside).

The angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.

The principle of conservation of angular momentum states that the sum of initial angular momentum is equal to final angular momentum.

Li = Lf

Li = Ii ωi  + Ic ωc

Li = Iiωi  +  MR²(V/R)

Li =  Iiωi + MRV ----- (1)

Lf = If ωf

ωf = Lf/If

If = Im  + MR²

ωf = Lf / ( Im  + MR²)

Recall, Lf = Li

ωf = (Iiωi + MRV) / ( Im  + MR²)

where;

ωf = (Iiωi + MRV) / ( Im  + MR²)

ωf = (0 + 40 x 1.5 x 8) / (600 + 40(1.5)²)

ωf = (480) / (690)

ωf = 0.696 rad/s

Thus, the angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.

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hydrogen

Atomic fusion occurs when 2 or more atoms combine to create a new element.

Helium

Before figuring out what elements create helium, we need to understand its properties. Helium is the second element on the periodic table. It has 2 protons and usually 2 electrons in the base state.

There are technically ions but base helium will have 2 electrons, especially because it is a noble gas.

Fusion

In most cases, including this one, when 2 atoms fuse together you can add their protons together to find the atomic number of the new element. Since helium only has 2 protons, the elements fused to form helium must both have 1 proton.

The element with 1 proton is hydrogen. So, 2 hydrogen atoms fuse to form helium.

Real World Applications

Fusion can be seen across the universe. One of the most common examples of hydrogen fusing to become helium is in stars. Due to the heat within stars, 2 hydrogen atoms can fuse together to create helium. This process also creates energy in the form of heat and light that the star is made of.

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

In Newtonian mechanics, linear momentum, or simply momentum, is the product of the mass and velocity of an object.

It is a vector quantity, possessing a magnitude and a direction.

The mathematical expression for momentum is:

p = m . v

where,

An object has a velocity (v) of 9 m/s and its linear momentum (p) is 72 kg.m/s. We will use the definition of linear momentum to calculate the mass of the object.

p = m . v

m = p / v

m = (72 kg.m/s) / (9 m/s) = 8 kg

An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.

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26.10 N is the vertical component of the force.

Rx  represents the Horizontal component of force

Ry represents The Vertical component of force

According to the given diagram

Rx - Tcosθ = 0

Rx = Tcosθ

And,

Ry + Tsinθ = mg

Ry = mg - Tsinθ

The horizontal component of force =The Vertical component of force  

Rx = Ry

Tcosθ = mg - Tsinθ

T(cosθ + sinθ) = 29 × 9.8 = 284.2 N

T√2 cosθ = 284.2 N

T × √2 ×0.544 = 284.2 N

T × 0.769 = 284.2 N

T = 370 N (app)

So,

Ry = 284.2 - 370 (sin 57°)

    = 284.2 - 310.3 = -26.10 N

Hence, 26.10 N is the vertical component of the force exerted.

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Answer:

Tension in the vertical rope: approximately [tex]613 \; {\rm N}[/tex].

Tension in the horizontal rope: approximately [tex]3.74 \times 10^{3}\; {\rm N}[/tex].

Assumption: [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].

Explanation:

Since the system is not moving, the tension in the vertical rope would be equal to the weight of the crate:

[tex]\begin{aligned}\text{weight of crate} &= m\, g \\ &= 62.5\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 613.25 \; {\rm N}\end{aligned}[/tex].

Note that [tex]\theta = 57.7^{\circ}[/tex] is the angle between the beam (the lever) and the vertical rope. The torque that this vertical rope exert on the beam would be:

[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &=(7.65\; {\rm m}) \, (613.25\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.965 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].

This torque is in the clockwise direction.

The weight of the beam ([tex]m = 116\; {\rm kg}[/tex]) would be:

[tex]\begin{aligned}\text{weight of beam} &= m\, g \\ &= 116 \; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \\ &= 1.138 \times 10^{3}\; {\rm N}\end{aligned}[/tex].

Note that [tex]\theta = 57.7^{\circ}[/tex] is also the angle between the beam and the direction of the (downward) gravitational pull on this. Since this beam is uniform, it would appear as if the weight of this beam is applied at the center of this beam (with a distance of [tex](7.65\; {\rm m}) / 2[/tex] from the pivot.) Thus, the torque gravitational pull exerts on this beam would be:

[tex]\begin{aligned} \tau &= r\, F\, \sin(\theta) \\ &= \genfrac{(}{)}{}{}{7.65\; {\rm m}}{2} \, (1.138 \times 10^{3}\; {\rm N})\, (\sin(57.7^{\circ})) \\ &\approx 3.679 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].

This torque is also in the clockwise direction.

The tension in the horizontal rope would need to supply a torque in the counterclockwise direction. The magnitude of that torque would be approximately:

[tex]\begin{aligned} & 3.965\times 10^{3}\; {\rm N \cdot m} + 3.679\times 10^{3}\; {\rm N \cdot m} \\ \approx \; & 7.645 \times 10^{3}\; {\rm N \cdot m} \end{aligned}[/tex].

Note the angle between the direction of this tension and the beam is [tex](90^{\circ} - \theta) = 32.3^{\circ}[/tex]. This force is applied  [tex](7.65\; {\rm m}) / 2[/tex] from the pivot. Hence, achieving that torque of [tex]7.645 \times 10^{3}\; {\rm N \cdot m}[/tex] would require:

[tex]\begin{aligned} F &= \frac{\tau}{r\, \sin(90^{\circ} - \theta)} \\ &\approx \frac{7.645\times 10^{3}\; {\rm N \cdot m}}{((7.65\; {\rm m}) / 2) \times \sin(32.3^{\circ})} \\ &\approx 3.74 \times 10^{3}\; {\rm N} \end{aligned}[/tex].

The ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.

An emf is induced in a coil placed in a magnetic field when a current carrying conductor moves in the field.

emf = NdФ/dt

where;

Thus, the ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.

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The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

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(a) The normal force exerted by the road on the car is 8,506.2 N.

(b) The normal force exerted by the car on the driver is 394.9 N.

(c) The speed of the car at which the normal force on driver is zero is 29.2 m/s.

The normal force exerted by the road on the car is calculated as follows;

Normal force = weight of the car - centripetal force of car

Weight of the car = (1400 x 9.8) = 13,720 N

Centripetal force of the car = (1400 x 18²)/87 = 5,213.8 N

Normal force = 13,720 N -  5,213.8 N

Normal force = 8,506.2 N

Normal force = weight of driver - centripetal force of driver

Weight of driver = (65 x 9.8) = 637 N

Centripetal force of driver = (65² x 18²)/(87) = 242.07 N

Normal force =  637 N - 242.07 N = 394.9 N

N = mg - mv²/r

0 = mg - mv²/r

mv²/r = mg

v²/r = g

v² = rg

v = √rg

v = √(87 x 9.8)

v = 29.2 m/s

Thus, the normal force exerted by the road on the car is 8,506.2 N.

The normal force exerted by the car on the driver is 394.9 N.

The speed of the car at which the normal force on driver is zero is 29.2 m/s.

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D. The canister with the fastest moving particle is Z due to concentration of particle.

The speed of the particles depend on the mean distance of the particles.

The canister with the largest concentration per particle will contain particles with the greatest speed.

If the particle concentration is increasing from W to Z, then Z will have fastest moving particle.

Thus, the canister with the fastest moving particle is Z due to concentration of particle.

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Answer:

d. Z

Explanation:

If all the canisters had the same amount of particles, which canister would have the fastest moving particles?

Z

The initial speed of the sled at the given height is 4.85 m/s.

Apply the principle of conservation of energy;

K.E = P.E

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

v = √(2 x 9.8 x 1.2)

v = 4.85 m/s

Thus, the initial speed of the sled at the given height is 4.85 m/s.

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(a) The time taken by the ball to reach its greatest height is 2 seconds.

(b) The time taken by the ball to travel from A to B, C is 4.08 m/s.

(c) The speed with which the ball hits the hoarding is 69.3 m/s.

H = u²sin²θ/2g

H = (40² x (sin 30)²) / (2 x 9.8)

H = 20.4 m

h = ut - ¹/₂gt²

20.4 = (40 x sin 30)t - (0.5)(9.8)t²

20.4 = 20t - 4.9t²

4.9t² - 20t + 20.4 = 0

solve the quadratic equation using formula method;

t = 2 seconds

This is the time of motion of the ball;

T = 2usinθ/g

T = (2 x 40 sin 30)/9.8

T = 4.08 s

v = u + gt

vy = 40 sin30  + (9.8 x 4.08)

vy = 59.98 ms

vx = 40 x cos30

vx = 34.64 m/s

vf = √(vy² + vx²)

vf = √(59.98² + 34.64²)

vf = 69.3 m/s

Thus, the time taken by the ball to reach its greatest height is 2 seconds.

The time taken by the ball to travel from A to B, C is 4.08 m/s.

The speed with which the ball hits the hoarding is 69.3 m/s.

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The horizontal distance traveled by the venom before it hits the ground is 0.6 m

A stone or ball or anything projected is known as a projectile.

Given that  a spitting cobra rears up to a height of 0.420 m above the ground and launches venom at 2.50 m/s, directed 35.0° above the horizon.

Assuming air resistance is neglected, the parameters to be considered are;

The ball we reach the ground at the same time when it is dropped vertically or horizontally.

In a vertical direction,

h = 1/2g[tex]t^{2}[/tex]

Substitute all the parameters into the formula

0.42 = 1/2 x 9.8 x [tex]t^{2}[/tex]

0.42 = 4.9[tex]t^{2}[/tex]

[tex]t^{2}[/tex] = 0.42/4.9

[tex]t^{2}[/tex] = 0.0857

t = [tex]\sqrt{0.0857}[/tex]

t = 0.29 s

The horizontal distance traveled by the venom before it hits the ground will be

Distance = ucosФ x t

Distance = 2.5cos35 x 0.29

Distance = 0.599 m

Distance = 0.6 m

Therefore,  the horizontal distance (in m) traveled by the venom before it hits the ground is 0.6 m

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Planet Y has rotated by 135.5° through during this time.

To find the answer, we need to know about the relation between angle and radius of orbit.

            angle= (orbital velocity × time)/radius

   = (4/3)^(3/2) × 88°

   = 135.5°

Thus, we can conclude that Planet Y has rotated by 135.5° through during this time.

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Answer: 1. The movie one is virtual and the bathroom mirror is real

2. The image is distorted in a way

3. Behind the spoon

Explanation:

The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.

In virtual desktops, the desktop environment is segregated from the physical device being used to access it. They are preloaded images of operating systems and apps. Over a network, users can remotely view their virtual desktops.

A virtual desktop can be accessed from any endpoint device, including a laptop, smartphone, or tablet. The user interacts with the client software that was installed on the endpoint device by the virtual desktop provider.

A virtual desktop mimics the appearance and feel of a real workstation. Because robust resources, like storage and back-end databases, are easily accessible, the user experience is frequently even better than that of a physical workstation.

Therefore, The movie one is virtual and the bathroom mirror is real and The image is distorted in a way and Behind the spoon.

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The tension in the string holding the tassel and the vertical will the tension in the string

Generally, the equation for Tension is  mathematically given as

[tex]TCos\theta = mg[/tex]

Therefore

[tex]TCos6.58^{o} = 19.8*10^{-3}*9.8[/tex]

T = 0.1953 N

b).

Where

[tex]T* sin \theta = ma[/tex]

[tex]0.1953*Sin6.58 \textdegree = 19.8*10^{-3}*a[/tex]

a = 1.13 m/s^2

In conclusion

T* sinФ = ma

2msinФ = ma

2sinФ = a

[tex]sin\theta = \frac{a}{2}[/tex]

[tex]\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}[/tex]

Ф = 34.4 °

In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string

T = 0.1953 N

Ф = 34.4 °

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The force exerted per square centimeter is 126 N/cm².

Pressure is the force acting per unit area.

Based on the data given:

volume increase, ΔV/V0 = 6 * 10⁻³

Bulk Modulus, B = 2.1 * 10⁹ N/m²

Bulk modulus B of a material is ratio of change in pressure and change in volume as given below:

Solving for ΔP;

ΔP = B * [(ΔV/V)]

ΔP = (2.1 * 10⁹ N/m²) * (6 * 10⁻³)

ΔP = 1.26 * 10⁶ N/m²

Converting to per square centimeter

ΔP = (1.26 * 10⁶ N/m²)/10⁴

ΔP = 126 N/cm²

In conclusion, the force exerted per square centimeter is a measure of the pressure.

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