Piston 1 in the figure has a diameter of 1.87 cm.
Piston 2 has a diameter of 9.46 cm. In the absence of friction, determine the force F, necessary to support an object with a mass of 991 kg placed on piston 2. (Neglect the height difference between the bottom of the two pistons, and assume that the pistons are massless).

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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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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Tension in the rope C is 1.24× 10² N.

To find the answer, we need to know about the horizontal component of tension in the rope B.

Horizontal component= tension in rope B × cos80.5°

= 7.52×10² N × cos80.5°

= 1.24×10² N

Thus, we can conclude that the tension in the rope C is 1.24× 10² N.

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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 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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1.06 is the maximum refractive index that the liquid may have for the light to be totally reflected.

Only when a light source passes from a denser to a rarer medium can it completely reflect.

When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.

sin [tex]i_{c}[/tex] = μ[tex]_{r}[/tex] / μ[tex]_{d[/tex]

From the diagram

Angle of incidence = 60°

sin60°  ≥ sin[tex]i_{c}[/tex] = μ[tex]_{r}[/tex]/μ[tex]_{d}[/tex]

μ[tex]_{r}[/tex] ≤ μ[tex]_{d}[/tex] sin60°

μ[tex]_{r}[/tex] ≤ √1.5 × √3/2

   = 1.06

Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06

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Answer:

more durability and easier to pound a hammer with the heavy weight.

Explanation:

One should use an iron hammer instead of the wooden hammer because the force developed with the help of the iron hammer is far greater than the wooden hammer

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.

The mathematical expression for Newton's second law is as follows

F = m*a

where F represents the force applied

m is the mass of the object

a is the acceleration of the object

As we know that force is a product of mass and acceleration, let us assume that both iron and wooden hammers have the same dimension , considering the acceleration applied is the same while hammering ,

As for the same dimension the mass of iron hammer would be greater as result force applied through the iron hammer would be grater

Since the iron hammer's force is so much stronger than the wooden hammer's force, one should use an iron hammer instead of a wooden hammer.

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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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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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(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

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No. of turns in the solenoid is an option (b) 144.

The self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current.

Self-inductance of solenoid = 39.1 μH

                                           = 39.1 × [tex]10^-^6[/tex] H

Length of the solenoid = 4.0 cm

Cross-sectional area = 0.60 cm²

Expression for the self-inductance of a coil ;

L = µ₀N²A / [tex]l[/tex]

where,

L = Self- Inductance

N = No. of turns.

A = Cross-sectional area

[tex]l=[/tex] Length of the solenoid

L =( 4π × [tex]10^-^7[/tex] × N² × 0.60 ) / 4.0

39.1× 4.0 / 4π × [tex]10^-^7[/tex] × 0.60 = N²

N² = 2.07 × [tex]10^6[/tex]

N = 144

Therefore, the no. of turns of the solenoid is 144.

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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.

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:

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).

A round cactus with many spines is the adaptation of a cactus that protects it from predators.

A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.

The one adaptation that protects the cactus from predators is spines.

A cactus does not have any parts that resemble leaves if you could look at one closely.

Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.

Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.

Other than protection, Spines perform many functions like

1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.

2)  The spines also impede airflow and prevent evaporation by trapping air.

3) Collecting dew from the early-morning fog is another crucial job that the spines do.

The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.

Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.

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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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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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The minimum speed required for a satellite in order to break free permanently from the planet and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.

To find the answer, we need to know about the escape velocity and time period of revolving satellite.

          = 7.3 Km/s

        = 3.1 × 10⁴ km

Thus, we can conclude that the escape velocity and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.

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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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These given conditions satisfy the second law of thermodynamics.

As the process is isobaric

So there will be a straight line of P= 200kPa in P-v and P-T planes

P1 = P2 = 100kPa

For perfect ideal gas, v-T plane:

[tex]v = (\frac{R}{P}) T[/tex]

[tex]v_{1} = (\frac{R}{P_{1} }) T_{1}[/tex] = 287 × 500/200000 = 0.717 m³/kg

[tex]v_{2} = (\frac{R}{P_{2} }) T_{2}[/tex] = 287 × 600/200000 = 0.861m³/kg

As it is the calorically perfect gas

de = [tex]c_{v}[/tex]dT

Integration on both sides

e2 - e1 = [tex]c_{v}[/tex](T2 - T1)

           = ( 716.5J/kg/K) (600-500)

           = 71650 J/kg

also,

Tds = de + Pdv

Tds = [tex]c_{v}[/tex]dT +Pdv

For ideal gas

V = RT/P        

dv = Rdt/P - RTdp/P²

Tds = [tex]c_{v}[/tex]dT + Rdt - RTdp/P

ds = ([tex]c_{v}[/tex] + R)dT/T - RdP/P

ds = ([tex]c_{v} + c_{p} -c_{v}[/tex])dT/T - RdP/P

ds = [tex]c_{p}[/tex]dT/T - RdP/P

Integration on both sides

s2 - s1 = [tex]c_{p}[/tex]ln (T2/T1) - R ln (P2/P1)

Since P is constant

s₂ - s₁ = [tex]c_{p}[/tex] ln (T2/T1)

           = 1003.5 ln (600/500)

           = 1003.5 × 0.182

           = 182.95 J/kg/K

w = Pdv

[tex]w_{12}[/tex] = P(v₂ - v₁)

     = 2,00,000 ( 0.861 - 0.717)

     = 28,800 J/kg

de = δq -δw

δq = de + δw

q₁₂ = (e₂ - e₁) +  w₁₂

    =  71,650 + 28,800 = 1,00,450 J/kg

Now in this process, the gas is heated from 500 K to 600 K. We would expect at a minimum that the surroundings were at 600 K.

Let’s check for second law satisfaction.

s₂ - s₁ ≥ q₁₂ / Tₓ

182.95 ≥ 1,00,450 / 600 K

182.95 J/kg/K ≥ 167.41 J/kg/K

Hence this condition satisfies the second law of thermodynamics

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The minimum  necessary height of the IV bag above the position of the needle is 0.37 m.

we know that,

P = ρgh

where,

P = 35.7mmHg

  = 4759.609 Pa

g = 9.8[tex]\frac{m}{s^{2} }[/tex]

ρ = 1.308 kg / m^3

now, substituting all the values, we get,

4759.609 = 1.308 × 9.8 × h

h = 0.37 m

The minimum  necessary height of the IV bag above the position of the needle = 0.37 m.

what is an IV bag ?

A reagent, also called as an analytical reagent, is a substance or compound that is added to a system in chemistry to bring about a chemical reaction or examine to see if one happens. Even though the terms "reagent" and "reactant" are frequently used synonymously, "reactant" refers to a substance that is consumed during a chemical reaction.

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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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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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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 direction of magnetic field is south-east and the magnitude is

[tex]23.66[/tex] × [tex]10^{3} N/C[/tex].

Here, the magnitude of CD and BC will be cancelled, as they both are in the opposite direction and equal to each other.
the magnitude, towards the diagonal AC will result in CP, which is the direction of the electric field.

magnitude of electric field can be defined as :- The force per charge on the test charge is a straight forward way to define the size of the electric field.

To find the magnitude of the electric field use the formula

[tex]E = kq/ r^{2}[/tex]

inserting the values,

[tex]E = 9. 10^{9}[/tex] × [tex]4.05[/tex] × [tex]10^{-6} / 1.1 \sqrt{2}[/tex]

[tex]E= 36.45[/tex] × [tex]10^{3}[/tex] / [tex]1.54[/tex]

[tex]E = 23.66[/tex] × [tex]10^{3}[/tex] N/C

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The number of half-lives that would go by in 44 h is 7 half-lives

This is the time taken for half a substance to decay

The number of half-lives that will elaspe after 44 h can be obtained as illustrated below:

n = t / t½

n = 44 / 6.01

n = 7

Thus, 7 half-lives will elaspe after 44 h

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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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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 magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

                           [tex]N_x=86.62N[/tex]

                           [tex]N_y=F_V=mg-Tsin59\\[/tex]

                           [tex]F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N[/tex]

                    [tex]N_y= (40*9.8)-(169.8*sin59)=246.4N[/tex]

                 [tex]N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N[/tex]

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

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The hi.nge will be subjected to a force of 261.12N from the beam.

We must understand the tension in order to choose the solution.

                       [tex]N_y=F_v=mg-Tsin59[/tex]

                         [tex]F_H=N_x=Tcos59\\T=\frac{F_H}{cos59} =169.84N[/tex]

                  [tex]N_y=(40*9.8)-(169.84*sin59)=246.4N[/tex]

                        [tex]N=\sqrt{N_x^2+N_y^2} =261.12N[/tex]

Thus, we can infer that the force the beam applies to the height is 261.12N in size.

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Answer:

12 grams of the isotope carbon-12.

Explanation:

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