someone solvw this for me please

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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Given the fact that energy conversion is not entirely efficient, it is impossible to produce a perpetual motion machine.

The perpetual motion machine in one that is able to work continuously without stopping. This would mean that the efficiency of this machine must that the machine is 100% efficient which violates the second law of thermodynamics.

Thus, given the fact that energy conversion is not entirely efficient and energy looses cause machines not function effectively, it is impossible to produce a perpetual motion machine.

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Answer:

And so calculating we get the maximum speed at each proton will reach To be 1.36 Times 10 to the four m the second

The middle ear provides impedance matching between air and fluid by the reduction in area between the tympanic membrane and the stapes footplate.

Middle ear is part of ear that separates the outer ear from the inner ear.

The middle ear consists of the following bones:

The middle ear provides impedance matching between air and fluid by

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(a) The maximum potential difference across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(c) The rms current through the resistor is 0.16 A.

(d)  The average power dissipated by the resistor is 38.4 W.

Vrms = 0.7071V₀

where;

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

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Sedimentation ponds are useful for reducing the effects of soil erosion because : ( D ) They provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers

Sedimentation ponds are ponds that hold sediments and suspended solids discharged by runoff waters thereby preventing the soil sediments from reaching water bodies such as ponds, lakes and rivers.

The sediment ponds hold the water while allowing the debris and solids contained in the water to settle down.

Hence we can conclude that sedimentation ponds provide a place to collect soil sediments so that it doesn't reach ponds, lakes and rivers

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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 manager is both most and least likely to lead seminars and provide everyone skills.

When a store has a “seasonal aisle,” it suggests it has extra money to spend on items that are in season. Festivals, special days, seasons, and other things are examples of seasonal components.

An operational strategy focused on creating profit. Seasonal goods are required by a supply chain. A manager wants everyone to prioritize setting up the seasonal aisle displays because of this.

Hence, the significance of the seasonal aisle is aforementioned.

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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

The acceleration of the body is obtained as 0.33 m/s^2.

Now we know that the object is acted upon by a system of forces. The net force is the effective force that acts on the body.

Hence;

ma = F -Ff

The applied force is F while the frictional force is Ff. The net force is represented as ma

Thus;

12 * a = 6 - 2

a = 4/12

a = 0.33 m/s^2

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(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

Torque equation for frictional force;

[tex]f = (\frac{r}{R} T) - (\frac{I}{R^2} )a[/tex]

solve (1) and (2)

[tex]a = \frac{TR(R - r)}{I + MR^2}[/tex]

since the yoyo is pulled in vertical direction, T = mg [tex]a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2[/tex]

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

W = mg

W = 0.15 x 9.8 = 1.47 N

T = mg = 1.47 N

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

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Velocity and displacement are the two quantities that increases as the other decreases during simple harmonic motion.

If a body's velocity at any given moment is inversely proportional to its displacement from the mean position, its motion is said to be simple harmonic.

Angular harmonic frequency and time t are functions that can be used to explain simple harmonic motion:

Displacement from equilibrium has amplitude A:

x = Asin(ωt)

and velocity has amplitude ωA

v = dx/dt = ωAcos(ωt)

At ωt = 0, x = 0 while v = Aω (max).

At ωt = π/2, x = A (max) while v = 0.

As a result, v reaches it's maximum /2 (90°) before x does. Therefore, displacement follows velocity at a 90° angle.

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The two forces are 12N and 10N (option B)

The net force is the force that has the same effect in magnitude and direction as the two forces acting together.

This problem can be solved using a right angled triangle;

Let the forces be P and Q

sin 40 = P/16

P= 16 sin 40

P = 10 N

Cos 40 = Q/16

Q = 16 cos 40

Q = 12 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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Answer:

acceleration = 2.25 m/s²

Explanation:

[tex]a = \frac{v - u}{t}[/tex]

where:

a = acceleration (?)

v = final velocity (5.5 m/s)

u = initial velocity (0 m/s, because the ball starts from rest)

t = time taken (2s).

Using the formula:

acceleration = [tex]\frac{5.5 - 0 }{2}[/tex]

                     = 2.75 m/s²

The power required from a 75-kg patient when the treadmill is sloping at required speed is 131.58 W.

The parallel force on the patient is calculated as follows;

F = mg sinθ

F = (75 x 9.8) x sin(12)

F = 152.82 N

P = FV

where;

F = 152.82 x 0.861

F = 131.58 W

Thus, the power required from a 75-kg patient when the treadmill is sloping at required speed is 131.58 W.

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(a) The object moves with uniform velocity from A to B.

(b) The object moves with constant velocity from B to C.

(c) The object moves with increasing velocity from C to D.

V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s

V(B to C) = (6 - 6)/(11 - 4) = 0 m/s

V(C to D) = (7 - 6)/(12 - 11) = 1 m/s

final velocity = 1 + 1.5 m/s = 2.5 m/s

Thus, we can conclude the following;

The object moves with uniform velocity from A to B.

The object moves with constant velocity from B to C.

The object moves with increasing velocity from C to D.

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Answer:The object moves with

positive accelaration

from A to B. It

speeds up

from B to C. It moves with

constant velocity

from C to D.

Explanation:I took the test

Considering that the moment you provide the action force, the car will certainly apply the response force. You will consequently come into an opposing force.

According to Newton's third law of motion, every action has an equal and opposite reaction. As well as the action and reaction are always acted in pairs.

You manually move a big automobile. You are then pushed back by the automobile with an opposing but equivalent force.

Because the automobile will undoubtedly apply the response force when you apply the action force. As a result, you will encounter an opposing force.

Hence, when you push a heavy car by hand. The car, in turn, pushes back due to the action-reaction analogy.

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The mass of block 1 is 1.30176 kg.

A tension force in physics is a force developed in a rope, string, or cable when stretched under an applied force.

Tension is acted along the length of the rope/cable in a direction that is opposite to the force applied on it.

According to the question,

Mass of block 2= 11.3 kg

Kinetic friction μ=0.200

The component of the weight in the direction of the incline

W = m₂gsinθ

T = [tex]F_{f} - W[/tex]........(1)

When the force of friction

[tex]F_{f}[/tex] = μ[tex]_k[/tex]N

[tex]F_{f}[/tex] = μ[tex]_k[/tex](m₂gcosθ)

Here , Put the value in equation 1

T = m₂gcos 31.5° - μ[tex]_k[/tex]m₂gcos 31.5°

We know that ,

T - m₁g = 0

T = m₁g

So,

m₁g = m₂g (Sin 31.5° - 0.200 x cos 31.5°)

m₁ = 11.3 x (0.0840 - 0.2 x 0.996)

m₁ = 11.3 x (0.1152)

m₁ = 1.30176 kg

Hence, the mass of block 1 is 1.30176 kg.

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A. The force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.

Centrifugal force is an inertial force that appears to act on all objects when viewed in a rotating frame of reference.

This force is directed away from the center around which the body is moving.

This is force that acts on a body moving in a circular path and is directed towards the center around which the body is moving.

While centripetal force is directed towards to the center, the centrifugal force is directed away.

Thus, the force that causes the water on the lettuce to come off the lettuce and go to the walls of the bowl is centrifugal force.

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Centripetal force causes the water on the lettuce to come off the lettuce and go to the walls of the bowl.

Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation. This force can move the body in the circular direction.

So we can conclude that Centripetal force causes the water on the lettuce to come off the lettuce and go to the walls of the bowl.

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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 reverberation time with 800 audiences is 0.875 seconds.

R₁V₁ = R₂V₂

where;

R₂ = R₁V₁/V₂

R₂ = (1.4 x 500) / 800

R₂ = 0.875 seconds

Thus, the reverberation time with 800 audiences is 0.875 seconds.

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(a) The magnitude of the static friction force is 145 N

(b) The minimum possible value of static friction is 0.44.

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).

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

Given data;

The horizontal force exerted, F =145 N

Mass of crate,m =33.2 kg.

The velocity of the crate,v =0 m/sedc

The magnitude of the static friction force, [tex]\rm f_s[/tex] =? N

The minimum possible value of the coefficient of static friction between the crate and the floor is,

Normal reaction,N = mg

a)

The forces acting on the crate are balanced since it is at rest; as a result, the horizontal force is equal to the frictional force, f:

F = [tex]\rm f_s[/tex] = 145 N

(b)

The following equations provide the greatest allowable force of friction between the floor and the crate:

[tex]\rm f_{min}= \mu_s N \\\\ \rm f_{min}= \mu_s mg \\\\[/tex]

The force applied to the box must be less than or equal to the maximum force of friction in order for it to stay at rest.

[tex]\rm f \leq f_{min} \\\\ f \leq \mu_s mg \\\\ 145 \leq \mu_s (33.2 \times 9.81 ) \\\\ \mu_s \geq 0.44[/tex]

Hence, the magnitude of the static friction force and minimum possible value of the coefficient of static friction between the crate and the floor will be 145 N and 0.44.

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The banking angle of curved road to allow to move without slipping is 11.53 ⁰.

The banking angle of curved road is calculated as follows;

V(max) = √(rg tanθ)

where;

V² = rg tanθ

tanθ = V²/rg

tanθ = (61²)/(1860 x 9.8)

tanθ = 0.204

θ = arc tan(0.204)

θ = 11.53 ⁰

Thus, the banking angle of curved road to allow to move without slipping is 11.53 ⁰.

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Answer: Answer.

1)Concave lens is transparent whereas the concave mirror is opaque. 2)concave mirrors cause reflection whereas concave lense cause refraction of light.

3)concave mirror forms both real, inverted images of various sizes and virtual, erect, and enlarged images whereas a concave lens form only virtual images.

Explanation:

Answer:

frequency and amplitude increases

The altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

The given parameters are:

The formula to calculate the pressure altitude is:

Altitude = (29.92 – Altimeter reading) * 1,000 + Ground level

Substitute the known values

1450 = (29.92 - Altimeter reading) * 1000 + 700

Evaluate the like terms

750 = (29.92 - Altimeter reading) * 1000

Divide both sides by 1000

0.75 = 29.92 - Altimeter reading

Evaluate the like terms

Altimeter reading = 29.17

Hence, the altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

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Since the tension in the string over the top of the circle is negligible, the tension in the string at the bottom of the circle will be zero.

Circular motion is the same as rotational motion. That is, motion moving in a circle.

Given that a small ball is tied to a string and set rotating with negligible friction in a vertical circle. If the ball moves over the top of the circle at its slowest possible speed (so that the tension in the string is negligible)

Since the tension in the string over the top of the circle is negligible, the tension in the string at the bottom of the circle, assuming there is no additional energy added to the ball during rotation will be zero.

The tension is negligible at the top in order to allow the ball rotate. And since the tension is negligible, there no tension force pulling inward

Therefore, the numerical value of the answer is Zero.

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A) When the cord is stretched 16.7 m, the tension in the cord is 18537 N.

b) The work that must be done against the elastic force of the bungee cord to stretch it 16.7 m is 15.48 kJ.

The tension or force in an elastic string is given by the formula:

a) The force or tension in the cord = 111 N/m × 16.7

Tension = 18537 N

b) Work done in a string = ke²/2

where k = 111 N/m, e = 16.7

Work done = 111 × 16.7²/2 = 15,478.395 J

Work done = 15.48 kJ

In conclusion, the force constant of the cord is required to calculate the tension and work done against the elastic force of the cord.

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