Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.

Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)

331.35 N is incorrect.

Answer:

Approximately [tex]31\; {\rm m\cdot s^{-1}}[/tex] assuming that there is no friction on the rollercoaster, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].

Explanation:

Make use of the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where:

If there is no friction on the rollercoaster, the acceleration of the rollercoaster would be equal in magnitude to the gravitational field strength, [tex]g[/tex]:

[tex]a = g = 9.81\; {\rm m \cdot s^{-2}}[/tex].

The initial velocity of this rollercoaster is [tex]0\; {\rm m\cdot s^{-1}}[/tex] (that is, [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) since the rollercoaster was initially stationary. The displacement of this rollercoaster would be [tex]x = 49\; {\rm m}[/tex] (same as the height of the hill.)

Rearrange the equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find an expression for the final velocity [tex]v[/tex] of this rollercoaster:

[tex]\begin{aligned}v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 9.81\; {\rm m\cdot s^{-2} \times 49\; {\rm m} + (0\; \rm m\cdot s^{-1}})^{2}}\\ &\approx 31\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Answer: I don't understand if it is a question

Explanation:

More than 700 people in the United States have been infected with monkeypox virus American vector: the epidemic is spreading like a snowball.

The  position of the car at the given time is 18 m.

The position of the car at the given time is calculated as follows;

x = vt

where;

x = 3 m/s x 6.0 s

x = 18 m

Thus, the  position of the car at the given time is 18 m.

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The temperature to which it must be heated in order to fit the shaft is 73.33 ⁰C.

The temperature to which it must be heated in order to fit the shaft is calculated as follows;

[tex]\frac{\Delta L}{L} = \alpha \Delta T\\\\\Delta T = \frac{\Delta L}{\alpha L}[/tex]

where;

ΔT = (0.04)/(50 x 15 x 10⁻⁶)

ΔT = 53.3 ⁰C

T₂ - T₁ = ΔT

T₂ = ΔT + T₁

where;

T₂ = 53.3 + 20

T₂ = 73.33 ⁰C

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To create a tragic past for your character, you would have to write a plot that shows a sad ending to a good story.

This refers to a type of story that has negative emotions such as sadness, death, etc.

Hence, we can see that for example, you could write about two lovers and how despite all the struggles they faced, eventually died in mysterious circumstances.

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The length of the y-component of the vector plotted below is 1 unit.

We know this because this vector can be descomposed in two vectors (blue) that form an angle of 90° beetween them. The measures of this vectors must be able to form a right-angle triangle with the descomposed vector (see image).

The sound waves with the frequency 20 Hz, 200 Hz, 2,000 Hz, and 20,000 Hz have a velocity of 344 m/s.

A sound wave is a sample resulting from the motion of strength traveling thru a medium (inclusive of air, water, or every other liquid or stable matter) because it propagates from the supply of the sound. Sound waves are created and convey strain waves, for example, a cellphone.

The frequency can be given as:

Frequency = Velocity/ Wavelength

Velocity = Frequency * Wavelength

The velocity of the following sound waves is given as:

Wavelength = 17.2 m

Velocity = 20 * 17.2 m/s

Velocity = 344 m/s

Wavelength = 1.72 m

Velocity = 200 * 1.72 m

Velocity = 344 m/s

Wavelength = 17.2 cm = 0.172 m

Velocity = 2,000 * 0.172 m/s

Velocity = 344 m/s

Wavelength = 1.72 cm = 0.0172 m

Velocity = 20,000 * 0.0172 m/s

Velocity = 344 m/s

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(1) The intensity of a sound wave is linearly proportional to the change in the pressure squared.

(2) You are able to hear sounds from farther away on a cool night because sound waves are refracted toward warmer air near the ground.

(3) When the trough of one wave collides at the same time & place of the crest of an identical wave, destructive interference occurs.

(4) The time taken for the heater to provide the energy is 300 secs.

(5) The total amount of heat provided to the ice sample is 7,980 cal.

(6) The amount of heat required to increase the temperature of water is 33,472 J.

The intensity of a sound wave is linearly proportional to the change in the pressure squared and inversely proportional to the density and the speed.

You are able to hear sounds from farther away on a cool night because sound waves are refracted toward warmer air near the ground.

When the trough of one wave collides at the same time & place of the crest of an identical wave, destructive interference occurs.

E = Power x Time

t = E/P

t = 60,000/200

t = 300 secs

Q = 79.8 cal/g x 100 g

Q = 7,980 cal

Q = 0.1 kg x 4184 J/kgC x (100 - 20)

When water vapor condense into liquid water, temperature = 100⁰C

Q = 33,472 J

Thus, the intensity of a sound wave is linearly proportional to the change in the pressure squared.

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(a) The angle the plane makes with respect to the horizontal is 27⁰.

(b) The magnitude of the normal force acting on the block is 135.3 N.

(c) The component of the force of gravity along the plane, fgx is 68.96 N.

(d) The expression for the net force on the block is mgsinθ - μmgcosθ = 0.

sin θ = h/L

where;

sin θ = 3.05/6.8

sin θ = 0.4485

θ = arc tan(0.4485)

θ = 26.7⁰

θ ≈ 27⁰

Fₙ = mgcosθ

Fₙ = (15.5)(9.8)cos(27)

Fₙ = 135.3 N

Fx = mgsinθ

Fx = (15.5)(9.8) sin(27)

Fx = 68.96 N

∑F = 0

Fx - Ff = 0

mgsinθ - μmgcosθ = 0

Thus, we can conclude the following;

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Answer:

• black hole     -   one of the most dense objects in the universe

• red giant       -    a very large star

• white dwarf   -   an old, very dense hot star that is cooling

• nebula           -   a mass of gas and dust

The direction of torque τ this method is mathematically given as

D=X  

Option A is correct.

Generally, the equation for torque is  mathematically given as

τ = r X F

Hence to decipher the torque direction with respect to the center of mass of the body due to force F acting on the body at a location indicated by the vector r

In conclusion, the direction of this method is

D=X   Option A.

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The tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 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 = (21 x 9.8)/(2 x sin66)

T = 112.64 N

T + F = W

F = W - T

F = (9.8 x 21) - 112.64

F = 93.16 N

Thus, the tension in the cable is 112.64 N and the vertical component of the force exerted by the hi.nge on the beam is 93.16 N.

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The speed will be 0.2 m/s and the velocity will be 0 m/s.

Speed = Total Distance / Total time

We have given total distance as ( 50 + 95 ) metres and total time as 12 minutes or we can say 720 seconds.

Speed = 145/ 720 m/s

Speed = 0.2 m/s

Velocity = Total Displacement / Total time

As the initial and final is the home, hence the net displacement is 0 in that case.

In this case also the total time we have given is 12 minutes or we can say 720 seconds.

Velocity = 0 / 720 m/s

Velocity = 0 m/s

So to conclude with we can say that the speed is 0.2 m/s and the velocity is 0 m/s.

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(a)The initial velocity of the shell will be 594.27 m/sec

(b)The maximum height it reaches will be 9000 m.

c)101.249 m meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface of the ship.

d)The error could be significant compared to the size of a target. Option C is correct.

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

m is the mass of the block = Kg.

u is the initial velocity of fall = m/sec

h is the distance of fall =  m

g is the acceleration of free fall = m/sec²

v is the hitting velocity of =?

a)

The range of the projectile is;

[tex]\rm R = \frac{u^2 sin 2 \theta }{g} \\\\ 36 \times 10^ 3 = \frac{U^2 sin 45^0}{9.81} \\\\ U= 594.27 \ m/sec[/tex]

b)

The maximum height of the projectile is;

[tex]\rm H = \frac{u^2 sin 2 \theta }{2g} \\\\ H = \frac{(594.27)^2\times (sin 45)^2}{2 \times 9.81 } \\\\ H = 9000 \ m[/tex]

c)

The distance between its surface and the ship, measured in a horizontal arc parallel to the surface, will be 36.0 kilometers. The distance from the lower surface is found as;

[tex]\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m[/tex]

d)

An error is a mistaken or erroneous action. In some contexts, an error is interchangeable with a mistake.

The difference between the calculated value and the original value is known as the error. The inaccuracy may be large in comparison to the target's size. Option C is correct.

Hence the initial velocity of the shell, maximum height, and the distance from the lower surface will be 594.27 m/sec,9000 m, and 101.249 m and option c for question d are correct respectively.

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Answer:

1.37 seconds before.... splat !

Explanation:

y = y0 + v0 t + 1/2 a t^2      y0 = 30       y = 0

                    v0 = 0 ( from rest)

                            a = - 32.2 ft/s^2

0 = 30 - 1/2 * 32.2 * t^2

t=1.366 seconds                    Note that   mass   is irrelevant.

Answer:

P

Explanation:

,............................................ I have. a doubt too on D

Since the diameter of the star suddenly shrinks to 0.350 times its present size, the new period of the star is 4.2 days

To find the new period, we need to know Kepler's third law of planetary motion

Kepler's third law of planetary motion states that the square of the period of orbit of a planet, T equals the cube of its radius, R.

So, T² ∝ R³

and T₂²/T₁² = R₂³/R₁³ where

Given that the diameter suddenly shrinks to 0.350 times its present size and its radius is proportional to its diameter. So, its radius shrinks 0.350 times its initial size. Also, since there is a uniform mass distribution before and after, the mass remains the same.

So, R₂ = 0.350R₁

Making T₂ subject of the formula, we have

T₂ = [√(R₂/R₁)³]T₁

Substituting the values of the variables into the equation, we have

T₂ = [√(R₂/R₁)³]T₁

T₂ = [√(0.350R₁/R₁)³]20.30 days

T₂ = [√(0.350)³]20.30 days

T₂ = [√0.042875]20.30 days

T₂ = [0.2071]20.30 days

T₂ = 4.2 days

So, the new period of the star is 4.2 days

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Material P does not always lead to the glowing of the bulb hence the material P is an insulator.

The term insulator refers to an object that does not permit current to pass through it. Insulators include materials such as wood, plastic etc.

Looking at the two circuits, we can see that the material P does not always lead to the glowing of the bulb hence the material P is an insulator.

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A characteristic of the part of the atom marked "B” is that it has a large mass. Option B. This is further explained below.

Generally,  Atom, the lowest possible unit into which matter may be split without releasing electrically charged particles.

In conclusion, the protons and neutrons are represented by the red and white spheres, respectively. Since it indicates both protons and neutrons, the line preceding the bracket around the balls (designated A) stands in for the atom's nucleus.The mass of the "B" section of the atom is very high.

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Answer:

it's b

Explanation:

i did the ed quiz

+ X axis.

The equation for torque is:
[tex]\tau = r \times F[/tex]

Torque is the CROSS-PRODUCT of 'r' and 'F'.

We can use the right-hand rule to determine the direction of torque.

Looking at the diagram, with an open hand, point your fingers towards the top of the picture (Along the +Z axis in the direction of the 'r' vector.)

Then, curl your fingers in the direction of the force vector, (along the -Y axis).

Your thumb will stick out in the direction of the torque, or the +X axis.

You can also just eliminate any options that involve axes that have already been used for 'r' and 'F'. In this case, there are components in the Z and Y axis, so this leaves the 'X' axis to be considered.

The butcher should balance both the meat and 0.5 kg mass on one arm of the beam balance and 2 kg mass on the other arm of the beam balance.

A beam balance can be defined as an apparatus which comprises a lever with two (2) equal arms and a bowl (pan) that is suspended from each arm, so as to measure the mass of an object.

In this scenario, the butcher should balance both the meat and 0.5 kg mass on one arm of the beam balance and 2 kg mass on the other arm of the beam balance.

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Answer:

the watt is the unit of power or radiant flax

The speed after it has traveled 0.400 m up the wall will be 1.77 m/sec.

From the work-energy theorem, it is stated that the net work done is equal to the change in the kinetic energy.

Given data;

Mass of book,m =5.00

Force,F = 96.0 N

Θ is the angle above the horizontal =  60°

The coefficient of kinetic friction between the book and the wall is μ =0.300.

R is the normal reaction

W₁ = Fd cosΘ

W₁=(96.0 N) (0.4 m) in 60°

W₁= 33.25 J

Work done by gravity is,

W₂ = mgdcos180°

W₂ = (5.00 kg) (9.81 m/s²)(0.4 m) cos180°

W₂ = -19.6J

The Normal force on the book by the wall is,

N = FcosΘ

N = (96 N) cos 60°

N=48 J

Work done by friction is;

W₃=μkRdcos180°

W₃=0.3)(48 J)(0.4 m) cos180°

W₃ =-5.76 J

The net work done is;

W =W₁+W₂+W₃

W=33.25-19.6-5.76

W=7.89 N

From the work-energy theorem, we have

W = ΔK

[tex]\rm W = \frac{1}{2} mv^2 - 0[/tex]

The speed is;

[tex]\rm v = \sqrt \frac{2 W}{m} \\\ v = \sqrt{\frac{2 \times 7.89} {5.00}}\\\\ v = 1.77 \ m/sec[/tex]

Hence, the speed after it has traveled 0.400 m up the wall will be 1.77 m/sec.

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(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

Apply Newton's second law of motion;

∑F = ma

where;

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

           →                 →              Ф                         ←

         1st child      friction       wagon                2nd child

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

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The energy lost to the air due to friction is 37.7 J.

The energy lost to the air due to friction is determined from the change in kinetic energy of the ball.

ΔK.E = K.Ef - K.Ei

ΔK.E = ¹/₂mv² - ¹/₂mu²

ΔK.E =  ¹/₂m(v² - u²)

ΔK.E = ¹/₂(0.45)(19.89² - 15.1²)

ΔK.E = 37.7 J

Thus, the energy lost to the air due to friction is 37.7 J.

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The speed of cars at A and B will be 25 m/sec.

Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.

Given data;

Mass of roller coaster,m =750 kg

Height at location 1,h₁ = 40 m

Height at location 2,h₂ = 302 m

The velocity of car A,[tex]\rm v_A[/tex]

The velocity of the car B, [tex]\rm v_B[/tex]

The velocity of the car A and B is the same is found as;

[tex]\rm v_A = v_B = v = \sqrt{v_0^2 + 2g(h_1-h_2} )\\\\ v = \sqrt{15^2 + 2 \times 10 (40-30)} \\\\ v = \sqrt{225+400} \\\\ v = 25 \ m/sec[/tex]

Hence the speed of cars at A and B will be 25 m/sec

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The new period will be 2.486 days.

The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.

Given data;

Mass of a star,m= 1.210×10³¹ kg

The time period for one rotation of the star, T = 20.30 days

D' = 0.350 D

R' = 0.350 R

From the law of conservation of angular momentum;

[tex]\rm I \omega = I' \omega' \\\\ \frac{2}{5} MR^2 \times \frac{2 \pi }{T}=\frac{2}{5} MR'^2 \\\\ R^2 \times \frac{1}{T}= R'^2 \times \frac{1}{T} \\\\ T' = \frac{R'^2T}{R^2} \\\\ T' = \frac{(0.350 R)^2 \times 26.1 }{R^2} \\\\T' = 0.1225 \times 20.30 \\\\ T'= 2.486 \ days[/tex]

Hence, the new period will be 2.486 days.

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