A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg.
If the tension in rope A is 7.42x10^2 N and the tension rope B carries is 7.52x10^2 N, what is the tension in rope C?

The height of the building in meters, given the data is 108 m

The height of the building can be obtained as illustrated below:

h = ½gt²

h = ½ × 9.8 × 4.7²

h = 108 m

Thus, the height of the building is 108 m

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The braking distance is the distance traveled by a car experiencing a braking force until it comes to rest.

Our initial energy is solely kinetic:
[tex]E_i = \frac{1}{2}mv^2[/tex]

And, since the car goes to rest, it is no longer in motion. It will have no kinetic energy.
[tex]E_f = 0[/tex]

Therefore, there was work done by the braking force.

[tex]W_B = E_f - E_i = -\frac{1}{2}mv^2[/tex]

Recall the definition of work:
[tex]W = F\cdot \Delta x[/tex]

Or in this case, since the displacement and breaking force are antiparallel:
[tex]W = -F_B\Delta x[/tex]

This is equivalent to the dissipation of kinetic energy:
[tex]W = -F_B\Delta x = -\frac{1}{2}mv^2[/tex]

Now, to visualize this, let's rearrange the equation to solve for displacement.

[tex]\Delta x =\frac{mv^2}{2F_B}[/tex]

There is a direct, SQUARE relationship between necessary braking distance speed.

If the speed was reduced by 10.3 percent, its new speed is only 89.7% percent of the original, so:
[tex]\Delta x' =\frac{m(0.897v)^2}{2F_B}[/tex]

[tex]\Delta x' = 0.8046\Delta x[/tex]

The reduction by a percentage is:
[tex]1 - 0.8046 = 0.1954 \\\\\boxed{= 19.54\%}[/tex]

The tension in rope C is determined as 122.23 N.

The tension in rope C is calculated as follows;

B² = A² + C²

C² = B² - A²

where;

C² = (752²) - (742²)

C² = 14,940

C = √14,940

C = 122.23 N

Thus, the tension in rope C is determined as 122.23 N.

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The work done in dragging an object along a straight path is  173.2 J.

Work done is equal to product of force applied and distance moved.

Given You are trying to drag an object 5.0 m along a straight path. You are pulling with a force of 40 N along a rope that is inclined 30° to the horizontal.

Work = Force x Distance x cos(angle)

W= 40 x 5 x cos 30°

W = 173.2 Joules

Thus, the work done is  173.2 Joules

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The magnitude of the tension in the upper rope is 24.93 N.

The tension in the lower rope is calculated as follows;

T1 + T2 = W

where;

12.8 + T2 = (3.85)(9.8)

12.8  + T2 = 37.73

T2 = 37.73 - 12.8

T2 = 24.93 N

Thus, the  magnitude of the tension in the upper rope is 24.93 N.

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Answer:

5

Explanation:

Use the Pythagorean theorem. 3^2 + 4^2 = 25, which is the square of 5.

Brainliest, please :)

The direction of the vector would eventually depend on the horizontal component and vertical component which are 4 and 3 respectively. So, it would be calculated as 5 according to the Pythagoras theorem.

In physics, the direction of the vector may be defined as the orientation of the vector, that is, the angle it makes with the X-axis. A vector is drawn by a line with an arrow on the top and a fixed point at the other end.

The direction in which the arrowhead of the vector is directed gives the direction of the vector. The direction of the vector measures the angle it makes with a horizontal line.

According to the question,

The horizontal component of the vector = 4

The vertical component of the vector = 3

The direction of the vector =  square of the horizontal component + square of the vertical component.

                                             = [tex]4^2+3^2[/tex] = 16 + 9 = 25 i.e. = [tex]5^2.[/tex]

Therefore, the direction of the vector would eventually depend on the horizontal component and vertical component which are 4 and 3 respectively. So, it would be calculated as 5 according to the Pythagoras theorem.

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Answer:

122.17 m/s

Explanation:

x cos 40 = horizontal velocity

1500 m / x cos 40   = time in the air = 1958.11 / x

x sin 40 = vertical velocity

 find when shell vertical velocity = 0 (this is max height....1/2 way through its flight)  , the time when it hits the ground will be twice this...  

   0 =  x sin 40 - 9.8 t

  t =  x sin40 / 9.8        time in the air is twice this = .13118 x

Equate the two times from above to solve for x

1958.11/ x  =  .13118 x

x = 122.17 m/s

Answer:

All people with the observant trait are often a steadying force that always wants to get things done. Their energy is very elegant in the sense of working on real things in real time.

Explanation:

Hope I helped

Rightward and upward

Whenever any object moves in one direction only, the motion is known as one-dimensional motion.

Two-dimensional motion is the movement of the object in two directions simultaneously.
Example: A ball thrown at an angle is a two-dimensional motion.

In two-dimensional motion, there are two axes used, generally the x-axis and the y-axis.

Generally, the x-axis is positive towards the rightward direction and negative towards the leftward direction.

Similarly, the y- axis is positive towards the upward direction and negative toward the downward direction.

So, the rightward and upward directions are chosen as positive.

Therefore:

When working with two-dimensional motion, choose the direction of  , the rightward and upward to be a positive direction.

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Answer:

H = L sin θ / 2    height of center of mass

E = m g H = m g L sin θ / 2

I = m L^2 / 3        moment of inertia about end of rod

1/2 I ω^2 = m g L sin θ / 2      KE of rod when it hits surface

1/2 m L^2 / 3  ω^2 = m g L sin θ / 2

ω^2 = 3 g sin θ / L

ω = (3 g sin θ / L)^1/2      angular speed of rod striking surface

P = I α       if P = torque    

α = m g L/2 / (m L^2 / 3) = 3 g / (2 L)       angular acceleration of rod

Answer:

A is the correct answer

Explanation:

resistance force is the force that opposing to effort

The tension in the string marked A is determined as 331.35 N.

tanθ = Δy/Δx

tanθ = (6 - 1)/(7 - 1) = 0.8333

θ = arc tan(0.8333) = 39.81⁰

with respect to horizontal = 90 - 39.81 = 50.19⁰

tanθ = Δy/Δx

tanθ = (9 - 6)/(7 - 5) = 1.5

θ = arc tan(1.5) = 56.31 ⁰

tanθ = Δy/Δx

tanθ = (6 - 4)/(14 - 7) = 0.2857

θ = arc tan(0.2857) = 15.95 ⁰

Bcosθ + Aosθ = Ccosθ

Bcos(56.31) + A[cos(50.19)] = 56.3cos(15.95)

0.55B + 0.64A = 54.13 ----- (1)

Bsinθ + Asinθ = Csinθ

Bsin(56.31) + A[sin(50.19)] = 56.3sin(15.95)

0.832B + 0.77A = 15.47---- (2)

Solve (1) and (2)

0.2A = 66.27

A = 66.27/0.2

A = 331.35 N

Thus, the tension in the string marked A is determined as 331.35 N.

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The work done on  both Elastic X and Elastic Y is 25.2 J.

The work done on the Elastic X is calculated as follows;

W = ¹/₂fx

where;

W = ¹/₂(42)(1.2)

W = 25.2 J

The work done on the Elastic Y is calculated as follows;

W = ¹/₂fx

where;

W = ¹/₂(24)(2.1)

W = 25.2 J

Thus, the work done on  both Elastic X and Elastic Y is 25.2 J.

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The moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

The moment of inertia is the amount of rotation obtained  by an object when it is in state of motion or rest.

Three small objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 1.09 kg, mB = 2.83 kg, mC = 3.39 kg.

The coordinates  of Ball A :(2,1) Ball B :(8,2) and Ball C: (5,8) and axis is at (4,3)

Here, for the moment of inertia of each ball

For ball A

IA = 1.09 x ((4 - 2)^2 + (3-1)^2)

IA = 8.72 Kg.m^2

for the ball B

IB = 2.83 * ((4 - 8)^2 + (3 - 2)^2)

IB = 48.11 Kg.m^2

for the ball C

IC = 3.39 * ((4 - 5)^2 + (3 - 8)^2)

IC = 88.14 Kg.m^2

Total moment of inertia = 8.72 + 48.11 +  88.14

Total moment of inertia = 144.97 Kg.m^2

Thus, the moment of inertia of this set of objects with respect to the axis perpendicular to the the x-y plane passing through location x = 4.00 m and y = 3.00 m is 144.97 Kg.m^2.

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The distance traveled by car A when they pass each other is 3,428.56 ft.

(Va)t - (Vb)t = d

where;

80t - (-60)t = 6000

80t + 60t = 6000

140t = 6000

t = 6000/140

t = 42.857 s

s = vt

s = 80 x 42.857

s = 3,428.56 ft

Thus, the distance traveled by car A when they pass each other is 3,428.56 ft.

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Beats can be defined as the periodic fluctuations in the frequency of sound waves. That is option D

Sound waves are those waves that are produced by the vibration of an object whose energy is usually propagated through a medium.

When two sound waves of different frequencies meets, a periodic variation that occurs is called beats.

Therefore, Beats can be defined as the periodic fluctuations in the frequency of sound waves.

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The frequency the student detect from waves omitted from the fork and waves coming from the wall is 348.1 Hz.

The observed frequency is determined by applying Doppler effect;

f' = f₀(v + v₀)/(v)

where;

f' = 342(6 + 336)/(336)

f' = 348.1 Hz

Thus, the frequency the student detect from waves omitted from the fork and waves coming from the wall is 348.1 Hz.

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The total kinetic energy  of the sphere when it reaches the bottom of the incline is  44.57 J.

The total kinetic energy of the sphere = translational + rotational kinetic energy.

K.E = ¹/₂mv²

K.E(bottom) = P.E(top) = mgh

K.E = (2.35)(9.8)(1.29)

K.E = 29.71 J

K.E(rot) = ¹/₂Iω²

K.E(rot) = ¹/₂I(v²/R²)

where;

I = ¹/₂MR²

K.E(rot) = ¹/₂( ¹/₂MR²)(v²/R²)

K.E(rot) = ¹/₄Mv²

K.E(rot) = ¹/₄M(2gh)

K.E(rot) = ¹/₂Mgh

K.E(rot) = ¹/₂(29.71) = 14.855 J

Total kinetic energy = 29.71 + 14.855 = 44.57 J

Thus, the total kinetic energy  of the sphere when it reaches the bottom of the incline is  44.57 J.

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The average acceleration of snowboarder's is 0.19 m/s² and distance travel by snowboarder is 6.70 metres.

We have given that snowboarder start moving from rest means

Initial speed (u) = 0 m/s

Final speed (v) = 1.6 m/s

time taken (t) = 8.4 s

Applying first equation of motion

v = u + at

1.6 = 0 + a×8.4

1.6 = 8.4a

a = 1.6/8.4 m/s²

a = 0.19 m/s²

So average acceleration is 0.19 m/s².

Now Applying second equation of motion

s = ut + 1/2at²

s = (0) × (8.4) + (1/2)(0.19)(8.4)² metre

s = (1/2)(0.19)(8.4)²  metre

s = 6.70 metre

So that distance covered by snowboarder is 6.70 metre.

So we find out the average acceleration by using first equation of motion and the average acceleration came out to be 0.19 m/s² and to find out the distance covered by snowboarder we use second equation of motion and the distance came out to be 6.70 metre.

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The percentage decrease in the braking distance of the car is 19.5%.

v² = u² + 2as

where;

s = (v² - u²)/(2a)

v = 100%u - 10.3%u = 89.7% u = 0.897u

s = [(0.897u)² - u²]/(2a)

s = -0.195u²/2a

percent decrease = 0.195 x 100% = 19.5%

Thus, the percentage decrease in the braking distance of the car is 19.5%.

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The equivalent capacitance between A and B points is 2.5F.

The two parallel plates placed at a distance apart used to store charge when electric supply is on.

The capacitance of a capacitor is given by

C = ε₀ A/d

From the given circuit C1, C2 and C3, C4 are in parallel  C1=4F, C2=4F, C3=2F, C4=4F, C5= 9.2 F

C1, C2 = 4 +4 =8F

C3, C4 = 2 +4 =6F

Now , all capacitors are in series.

Total equivalent capacitance is
1 / Ceq = 1/ 8 +1/6 +1/ 9.2

Ceq = 2.5 F

Thus,  the equivalent capacitance between A and B points is 2.5F.

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Electric current is the vector quantity it has both magintude and direction.

The pace at which electrons travel through a conductor is known as electric current. The ampere is the SI unit for electric current. Electrons are tiny particles that reside within a substance's molecular structure.

Because it has both a magnitude and a direction, the current is a vector. However, a vector always abides by the law of vector addition.

Current is a scalar since it follows algebraic addition and doesn't obey it. Additionally, the current is represented as the dot product of the area vector and the current density vector.

Hence, electric current is the vector quantity it has both magnitude and direction.

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The ratio of the kinetic energy of the block/bullet system immediately after the collision to the initial kinetic energy of the bullet is 0.78 %.

Apply the principle of conservation of energy to determine the final velocity of the block/bullet system.

K.E = P.E

¹/₂mv² = mgh

¹/₂v² = gh

v² = 2gh

v = √2gh

where;

v = √(2 x 9.8 x 1.1)

v = 4.64 m/s

Apply the principle of conservation of linear momentum.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

(0.0075)u₁ + (0.95)(0) = 4.64(0.0075 + 0.95)

0.0075u₁ = 4.4428

u₁ = 4.4428/0.0075

u₁ = 592.37 m/s

K.Ei = ¹/₂m₁u₁²

K.Ei = ¹/₂(0.0075)(592.37)²

K.Ei = 1,315.88 J

K.Ef = ¹/₂(m₁ + m₂)v²

K.Ef = ¹/₂(0.0075 + 0.95)(4.64)²

K.Ef = 10.31 J

= K.Ef/K.Ei x 100%

= (10.31 / 1,315.88) x 100%

= 0.78 %

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The boom makes an angle of θ = 60.2° with the vertical wall and the tension in the horizontal rope is mathematically given as

[tex]T_1=730.85 \mathrm{~N}[/tex]

[tex]T_1'=\frac{1980.51 \mathrm{~N}}{}[/tex]

Generally, the equation for is  mathematically given as

1) Tension in vertical rope,

[tex]T_{1}=74.5 \times 9.81\\T_1=730.85 \mathrm{~N}[/tex]

2) Tension in horizontal rope,

[tex]\sum{Mg} =0\\Mg\frac{l}{2} \sin \theta+T_{1} \frac{l}{2} \sin \theta &=T_{2} l \cos \theta \\[/tex]

[tex]T_{2} &=\frac{M g(l / 2) \sin \theta+T_{1}(l / 2) \sin \theta}{l \cos \theta} \\\\&=\frac{M g \tan \theta}{2}+\frac{T_{1} \tan \theta}{2} \\\\T_{2} &=\frac{142 \times 9.81 \tan 61.8}{2}+\frac{730.85 \times \tan 61.8}{2} \\\\=& 1299+681.51[/tex]

[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]

In conclusion, the tension in the horizontal rope is

[tex]T_1=\frac{1980.51 \mathrm{~N}}{}[/tex]

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Answer:

Current = 5.45amps

Power = 654 watts

Explanation:

E =120V

I =?

R = 22.02 Ohms

I= E/R

I= 120/22.02

I = 5.45AmPs

P = ?                    P= E x R

E = 120V              P= 120x5.45

I = 5.45 AmPs      P=  654W

Answer:

The long pointer measures altitude in intervals of 10,000 feet (2 = 20,000 feet). The short, wide pointer measures altitude in intervals of 1,000 feet (2 = 2,000 feet). The medium, thin pointer measures altitude in intervals of 100 feet (2 = 200 feet).

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation, which are represented in the given function as 6 units and 98 rad/s respectively.

y = A sinωt

where;

A = 6 units

ω = 98 rad/s

A simple harmonic motion is defined by the amplitude and angular frequency of the oscillation. Thus, the wave is executing simple harmonic motion.

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The gravitational attraction between electron and proton is 10−40 whereas electrostatic force of attraction between a proton and an electron is 10-8.

The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10−40 while on the other hand, the electrostatic force of attraction between a proton and an electron in a hydrogen atom is 10- 8 which is 9 times.

The electric charge of the electron and proton are the same i.e. -1.60x10-19C whereas their gravitational force is different due to difference in mass.

So we can conclude that gravitational attraction between electron and proton is 10−40 whereas electrostatic force of attraction between a proton and an electron is 10-8.

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The mouse will be 4.61 meters distant horizontally. and the owl won't fall into the nest.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Diameter of nest,d = 30 cm

The velocity of flying, V =  3.50 m/s toward the east

The angle of flying below the horizontal, Θ = 32°

Position of owl,12.0 m above the middle of the 30 cm-diameter nests, at 4.00 m west.

Height of owl above the ground,h = 12 m

Distance from the west direction, a = 4.00 m

The vertical component of velocity,vₐ

The horizontal component of velocity,vₓ

After covering a vertical distance of 12 m, the vertical component of the mouse's velocity, v, is given by

vₐ-u²=2gh

u = asinΘ

Substitute the above value;

vₐ²- (a sin Θ )² = 2gh

Substitute the given value;

vₐ² - (4 sin 32)² = 2 ×9.81 ×12

vₐ² - (4 × 0.53)² = 235.44

vₐ² - 4.494 = 235.44

vₐ² = 235.44 + 4.494

vₐ² = 239.934

vₐ = 15.49 m/s

The time required to descend the aforementioned 12 meters;

[tex]\rm t = \frac{v_{rel}}{g}[/tex]

t = (vₐ - u )/g

t =( vₐ - a sin Θ )/g

Substitute the given value;

t =(15.49 - 4 ×sin 32°) / 9.81

t= (15.49 - 4 * 0.53) / 9.81

t =(15.49 - 2.12) / 9.81

t = 13.37 / 9.81

t= 1.36 s

Horizontal distance traveled for the given period;

x=uₓ × t

x =a cosΘ × t

Substitute the given value;

x = 4×(cos 32°)×1.36

x= 4×0.848×1.36

x= 4.61 m

Hence, the mouse will be 4.61 meters distant in the horizontal direction. and the owl won't fall into the nest,

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Some household appliances and their fuse ratings are given below:

A fuse is a device which is designed to protect an appliance by cutting off the electric circuit when current greater than its rating flows through it.

Some household appliances and their fuse ratings are given below:

In conclusion, a fuse is designed to protect an electrical appliance from excess current.

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