Find the direction of the sum of
these two vectors:
B 3.14 m 2.71 m 30.0°
-60.0°A direction (deg)

The tension in the cable is 4248 N.

The force transmitted through a rope, string, or wire when two opposing forces pull on it is known as tension. The tension force pulls energy equally on the bodies at the ends and is applied along the entire length of the wire.

Given parameters:

Weighs of the elevator, M = 300.0 kg.

weighs of Joey, m = 60.0 kg.

Upward acceleration of the elevator, a = 2.0 m/s².

And, acceleration due to gravity, g = 9.8 m/s².

So, net tension in the cable, T = (M + m)g + (M + m)a

= ( 300.0 + 60.0 ) × 9.8 N + (300.0 +60.0 )× 2.0 N

= 4248 N.

Hence, the tension in the cable that pulls the elevator upwards is 4248 N.

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The neighborhood can help children build new relationships by holding a winter festival with activities for children and families.

In today's world, where everyone is quite busy in their own world, children have academic pressure and many more.

They tend to detach themselves from people, and hence many relationships ruin.

Building relationship helps the children grow and makes them more sociable. It boots their intelligence and makes them aware of the importance of having a relationship and building bonds with people.

The neighborhood holds a winter festival with activities for children and families is a great way to build new relationships.

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Answer: (C)

Explanation:  (B) is the weight of the box acting downwards

(A) is the frictional force acting on the box because of moving forward against the plane

(D) is the force that moves the box forward

( C) is perpendicular to the box which means it is at 90 degrees with the box. Normal usually means at 90 degrees

Tension is defined as the pulling force transmitted axially using a string, a rope, chain, or similar object, or by each end of a rod, truss member, or similar 3-D object; tension might also be expressed as the action-reaction pair of forces acting at each end of said elements.

Maximum tension:

T - mg = mv²/L

By energy conservation:

mgh = ¹/₂ mv²

v² = 2gh

Now the tension force in the vine at this position is given as:

T = mg +  mv²/L

Now substitute the values in the above equation:

T = 688 N + m(2gh)/L

T = 688 + 2×3.2(688)/18

T = 932.6 N

As the force is less than the limit of 950 N so the vine didn't break.

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Two projectiles of mass and are fired at the same speed but in opposite directions from two launch sites separated by a distance, will land at a place x = D/2 [ 1+{ (m₁ - m₂) / m₁ + m₂)}

This is calculated using the conservation of linear momentum in the horizontal direction as ,

(vm₁vₓ₁ - m₂vₓ₁ ) ₓî = (m₁ + m₂) vₓ₂ ₓî

vₓ₂ = {(m₁ - m₂) / m₁ + m₂)} × vₓ₁

vₓ₂ = {(m₁ - m₂) / m₁ + m₂)} × v Cos Θ

t max = vₓ₁ / g

=  v Sin Θ / g

x =(D / 2) + vₓ₂t max

= (D / 2) { {(m₁ - m₂) / m₁ + m₂)} × v² Sin Cos Θ / g } ------ (1)

now,

D = 2 vₓ₁vₓ₂ / g

D / 2 = v² Sin Cos Θ / g

From equation (1) we get,

x = D/2 [ 1+{ (m₁ - m₂) / m₁ + m₂)}

Hence , D/2 [ 1+{ (m₁ - m₂) / m₁ + m₂)} is the place they will land.

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According to the work-energy theorem, twice the speed corresponds to 4 times the energy and 4 times the driving distance.

There are different forms of energy on earth. The sun is considered the elementary form of energy on earth. In physics, energy is considered a quantitative property that can be transferred from an object to do work. Therefore, we can define energy as the force required for any  physical activity. So we can define energy in simple words as follows Energy is the ability to do work  According to the laws of conservation of energy, "energy can neither be created nor destroyed, it can only be changed from one form to another". The SI unit of energy is the joule.

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The number of bright interference fringes in the central diffraction envelope is 3.

To determine the number of bright interference, we need to understand the equation of first minima in the diffraction pattern, and the equation of angular locations of the double slit interference pattern.

For the equation of first minima in the diffraction pattern is:

[tex]W[/tex]·[tex]Sin[/tex]θ = [tex]m_{1}[/tex]·λ

For the equation of angular locations of the double slit interference pattern is:

[tex]d[/tex]·[tex]Sin[/tex]θ = [tex]m_{2}[/tex]·λ..... (1)

Here, W is single slit width while d is slit separation

Next, we need to determine the number of bright interference fringes in the central diffraction envelope.

For the first minima, [tex]m_{1} = 1[/tex], then rewrite the equation (1) as follows.

=[tex]a[/tex]·[tex]Sin[/tex]θ = [tex]m_{1}[/tex]·λ

=[tex]a[/tex]·[tex]Sin[/tex]θ = [tex]1[/tex]·λ

=[tex]a[/tex]·[tex]Sin[/tex]θ = λ..... (2)

Then, from the equations (1) and (2)

[tex]=m_{2}=\frac{d}{w} \\=m_{2}=\frac{2w}{w}\\=m_{2}=2[/tex]

Therefore, there are 3 bright fingers, 1 at the centre and 2 in each side.

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If we keep oscilloscope plates close with potential difference 25 V , the  Charge on each plate will be 4.68 *10^-11 C.

The capacitance of a parallel plate capacitor is given by:

C = ∈₀A / d

where

∈₀ = 8.85 *10^-12 F/m is the vacuum permittivity

A is the area of each plate

d is their separation

For the capacitor in the problem:

A= 0.033m² = 0.0011m²

d = 5.2 mm = 0.0052 m

Substituting,

C = (8.85 *10^-12) (0.0011m)/ 0.0052 = 1.87 *10^-12 F

The capacity is related to the charge on the plate by:

Q=CV

where

V = 25.0 V is the potential difference between the plates

C = 1.87 *10^-12 F

Substituting,

Q = (25.0) (1.87 *10^-12) = 4.68 *10^-11 C

Charge on each plate = 4.68 *10^-11 C

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Answer:

relax

Explanation:

your answer is right there in front of you force = weight

area of feet = area

while your pressure = force per unit area i.e

force/area = 600/0.03

= 2 × 10⁵

The height of a hill from which the person throws a ball horizontally with a velocity of 12 m/s and the ball lands 10.1 m away from the hill is 3.47 m.

The height of a hill from which the person throws a ball horizontally with a velocity of 12 m/s and the ball lands 10.1 m away from the hill is calculated as follows;

The ball was projected horizontally from the top of a hill, therefore, there is no vertical component of the velocity of projection and the object is considered to be undergoing free fall under gravity from a height, h.

The height from which an object falls freely under gravity is calculated with the formula below:

H = gt²/2

where g is the acceleration due to gravity = 9.81 m/s²

t is time.

the time taken for the fall = horizontal distance / velocity

t = 10.1 / 12

t = 0.841 s

Therefore;

h = 9.81 * (0.841)² / 2

h = 3.47 m

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The Acceleration of the object = 6.4 m/s²

Mass of block (m) = 5 kg

Action force on block, (F₁) = 40 N

Frictional force opposing the motion (F₂) = 8 N

Acceleration of the object (a) = ?

⇒ Net force = Action force on block - Opposing friction force

⇒ F = F₁ - F₂

⇒ F = 40 - 8

⇒ F = 32 N

Net force of the block (F) = 32 N

Mass of block (m) = 5 kg

F is the Force in N.

m is the Mass in kg.

a is the Acceleration in m/s².

F = ma

⇛ a = F/m

⇛ a = 32/5

⇛ a = 6.4 m/s²

An electron degenerate object is a white dwarf, whereas a neutron star is a neutron degenerate object. A white dwarf is less compact and has a larger radius than a neutron star.

A white dwarf would collapse into a denser entity known as a neutron star if it reached the Chandrasekhar limit and nuclear reactions were to stop. This would happen if nuclear reactions continued to occur.

White dwarf stars are prevented from collapsing by the fact that electrons are fermions, while neutron stars are prevented from collapsing by the fact that electrons are fermions.

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The magnification of this telescope is 12.

Solution:

F = 1200 mm = 120 cm

f = 32mm, 25mm, 14 mm, 10mm

M = F/f 120/ = 3.75

M = 120/25 =  4.8

M = 120/14 = 8.57

M = 120/10 = 12.

Magnification ratio means that the ratio of subject size on the sensor plane is greater than or equal to the actual size of the subject. This allows macro lenses to capture very sharp close-ups of insects, etc. The magnification produced by a lens is equal to the ratio of image distance to object distance. Total magnification In a compound microscope the total magnification is the product of the objective and ocular lenses.

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The average flow rate is 3.611 cm^3/s

The speed V is = 104Km/h

The velocity of vehicle z is = 8km/l

Qavg = V/Z

         = 104/8 =13

         = 13 (1000)/1(3600)

Qavg = 3.611 cm^3/s

The physical parameters flow rate and velocity are linked yet very distinct. Consider a river's flow rate to help you understand the difference. The flow rate of the river increases as water velocity increases. However, the size of the river also affects the flow rate. The Amazon River in Brazil, for instance, carries much more water than a swift alpine stream. When A is the cross-sectional area and v is the average velocity, the flow rate Q and velocity v are precisely related. This equation seems to make sense. According to the relationship, the size of a river, pipe, or other body of water as well as the magnitude of the average velocity (hereinafter referred to as the speed) directly affect the flow rate.

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Thermonuclear fusion occurs faster in massive stars so larger stars use up all their fuel in less time.

The more mass a star has the faster it will exhaust its fuel supply and the shorter its lifespan. The most massive stars could burn up and explode in a supernova after just a few million years of nuclear fusion. Massive stars are the largest, hottest, and brightest main-sequence stars and are blue, blue-white, or white in color.

Massive stars run out of hydrogen fuel very quickly and therefore have short lives. This is because the more massive the star the greater the fuel consumption. Even if a high-mass star has more fuel it uses it up very quickly so it does not live as long as a low-mass star. A main-sequence star's mass determines the fundamental properties of its luminosity surface temperature radius and lifetime.

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The GPE that the barbells have at max height is ,

3680.h j=11,000 units

where h= maximum height the barbells were lifted

work done by the championship lifter ,W = 11,000 units

weight of the barbells, N = 3680 N

The gravitational potential energy, P.E., the barbells had at their maximum height of lift is given as follows;

P.E. = m × g × h

Where;

m = The mass of the barbells;

g = The acceleration due to gravity = 9.8 m/s^2

h = The maximum height to which the barbells are lifted

m × g = The weight of the barbells = 3680 N

∴ P.E. = 3680 N × h = 3680·h J

we know the law of conservation of energy, according to this the work done by the weight lifter is equals to the maximum gravitational potential energy gained by the barbell is equal to energy at maximium height i.e P.E

therefore, GPE = 3680.h j = W = 11,000j is your answer.

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The direction of the polarized light after passing the first and the second filter are the same, however the intensity of the light after passing the first and second polarizer are half and fourth of its original intensity, respectively.

Light polarization is a filter of electromagnetic waves such that it propagates into one transmission axis only.

When light passes through a polarizer, its intensity will decrease by half.

Suppose the intensity of unpolarized light is  I₀,  after passing the first polarizer, its intensity becomes:

I₁ = 1/2 . I₀

After passing the second polarizer, the intensity will further decrease by half, or:

I₂ = 1/2 . I₁

I₂ = 1/2 .1/2 . I₀ = 1/4. I₀

Since both polarizers are vertical filters, then the direction of the polarized light after passing the first and the second filter are the same.

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In both the cases gravitational potential energy is same.

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration can be assumed to be constant at about 9.8 m/s².

Given two cases height of the top is 3 ft so the potential energy is same as potential energy is dependent on height of the ball.

In both the cases gravitational potential energy is same.

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Compared to the orbital speeds of the planets in our solar system, the Milky Way Galaxy's rotation curve is far more flat and steady. While there is less mass as we get farther from the sun in our solar system.

The orbital speed of an astronomical body or object (such as a planet, moon, artificial satellite, spacecraft, or star) in gravitational bound systems is its speed relative to the most massive body's centre of mass, or the barycenter, if one body is significantly more massive than the other bodies in the system put together.

The phrase can be used to describe either the mean orbital speed (i.e., the speed throughout the course of an orbit) or its instantaneous speed at a specific location in the orbit. For objects in closed orbits, the maximum (instantaneous) speed occurs at periapsis (perigee, perihelion, etc.), whereas the smallest speed occurs at apoapsis (apogee, aphelion, etc.). When two-body systems are perfect, things An astronomical body or object (such as a planet, moon, man-made satellite, spacecraft, or star) moves through space at a certain speed called its orbital speed in gravitational bound systems.

When a system closely resembles a two-body system, the object's specific orbital energy, also known as "total energy," and its distance from the central body can be used to calculate the object's instantaneous orbital speed at a specific point in the orbit. Independent of position, the specific orbital energy is constant.

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Answer:

1.25 cm³

Explanation:

m = 7.0 g

d = 5.6 g/cm³

v = ?

Using,

[tex] d \: = \frac{m}{v} [/tex]

[tex]v = \frac{m}{d} [/tex]

[tex]v = \frac{7}{5.6} [/tex]

[tex]v = 1.25 \: c {m}^{3} [/tex]

Answer:

2. The object must have zero average acceleration over the interval.

Explanation:

Notes:

-Velocity can have zero change if it is either at rest or moving at a constant velocity. This also means there is zero acceleration (acceleration is any change in velocity).

-Velocity is also speed and direction, so if it changes direction (for example: moving backward or in a circle) it is not constant even if it has a constant speed. It also means the object is accelerating.

1: Nothing can be determined without additional information.

Incorrect, because it is solvable by the process of elimination.

2: The object must have zero average acceleration over the interval.

Correct, because zero acceleration equals zero change in velocity.

3: The object must be changing position.

Incorrect, because you don't know whether or not the object is accelerating.

4:  The object must have zero average velocity over the interval.

Incorrect, because this implies it is at rest. To have zero change, it can be at rest OR moving at a constant velocity.

5:  The object must have constant velocity over the interval.

Incorrect, because zero change in velocity can mean either be at rest OR moving at a constant velocity.

6: The object must begin and end at the same position.

Incorrect, because to begin and end at the same position implies that it was at rest or it changed direction while moving to end at the same position. (see Notes above for explanation)

7: The object must have constant acceleration over the interval.

Incorrect, because you can't have zero change in velocity if there is acceleration (Acceleration is any change in velocity).

8: The object must be at rest.

Incorrect, because to have zero change, it can be at rest OR moving at a constant velocity.

When an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about  the amount of energy the star gives off each second.

Luminosity and apparent brightness are about brightness, but from a different point of view. The difference between luminosity and apparent brightness is that luminosity tells us exactly how bright a star really is while apparent brightness only tells us its brightness seen from Earth.

L / Lsun = ( d / dsun )² = b / bsun

L = Luminosity of a star

Lsun = Luminosity of sun

d = Distance of star

dsun = Distance of sun

b = Brightness of star

bsun = Brightness of sun

Therefore, when an astronomer rambles on and on about the luminosity of a star she is studying, she is talking about  the amount of energy the star gives off each second.

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The Schwarzschild Radius Rs of this black hole is 5.69 x [tex]10^{10}[/tex] m

The Schwarzschild radius or gravitational radius is a physical parameter in the Schwarzschild solution of Einstein's field equations that corresponds to the radius defining the event horizon of a Schwarzschild black hole.

Let's assume that the ring shaped disk is circular and thus, to find the mass of the object at the center of the milky way, we can use the circular orbit equation;

v = √(GM/r)

where;

G is the gravitational constant and has a value of 6.67 x 10^(-11) Nm²/kg²

R is the radius of the orbit,

M is the mass of the larger object

v = velocity = 190km/s = 190000 m/s

Also

1 light year = 9.4605 x 10^(15) m

Thus,7.5 light years = 7.5 x 9.4605 x 10^(15) m

so, M = v²r/G

Putting all the values in the equation,

M = [190000² x 7.5 x 9.4605 x 10^(15)]/6.67 x 10^(-11)

M = 3.84 x 10^(37) kg

Now, mass of the sun generally has a value of 1.9891 x 10^(30) kg

Thus, value of mass of object in solar masses = 3.84 x 10^(37)/1.9891 x 10^(30) = 1.931 x 10^(7) solar masses

The formula for Schwarzschild radius is given as;

Rs = 2GM/c²

Where c is speed of light = 3 x 10^(8) m/s

Thus,

Rs = 2 x 6.67 x 10^(-11) x 3.84 x 10^(37)/(3 x 10^(8))² = 5.69 x [tex]10^{10}[/tex] m

The Schwarzschild radius Rs of this black hole is 5.69 x [tex]10^{10}[/tex] m

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She might be barely lower than you however shifting upward in the direction of you.

Oscillation is the repetitive or periodic variation, generally in time, of a few measures approximately a critical price or among or greater distinct states. familiar examples of oscillation consist of a swinging pendulum and alternating modern-day.

A pendulum is a weight suspended from a pivot so that it can swing freely. whilst a pendulum is displaced sideways from its resting, equilibrium position, it's miles issue to a restoring force due to gravity so one can accelerate it returned toward the equilibrium function.

Oscillation is defined as the method of repeating versions of any quantity or degree about its equilibrium price in time. Oscillation can also be defined as a periodic variant of a rely on among values or approximately its crucial price.

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Answer:

false is the answer . in my point of view

The car would be moving with a velocity of 21 m/s at the end of the five seconds.

We know that the first equation of motion is ⇒ v = u + at (i)

Here, the car moves with an initial velocity, u =  8.5 m/s (ii)

The car accelerates at a constant rate, a = 2.5 m/s² (iii)

The total time during this process, t = 5 seconds (iv)

Putting the values of the initial velocity, acceleration, and time (ii, iii, and iv) in equation (i), we get ⇒ v = 8.5 + (2.5)(5)

v = 21 m/s

Therefore, the car moves with a final velocity of 21 m/s at the end of the five seconds.

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Mass of the heaviest book is 0.733 kg.

Heaviest book has weight

= (6 + 6) * 0.6 N

= 7.2 N

If m = mass of the heaviest book in kg

m = 7.2/9.81 kg

   = 0.733 kg.

Children often bring a well-established "life-view" of friction with them because of their experiences with slippery surfaces like frozen ponds (low friction) and "gripping" surfaces like deep pile carpets (high friction) and the effects they have on movement. This way of existence needs to be expanded and understood in the perspective of science.

The force that modifies movement as a result of surface/surface interaction is known scientifically as friction (all changes of movement require the action of a force). When using diagrams to portray forces in action, the direction of the frictional force should be depicted as being opposed to that of the movement.

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The acceleration of the box is 3.4m/s² and if the box starts from rest at constant force for 3 seconds, then the speed of the box will be 10.2m/sec

The frictional force (f) = μN

where μ= coefficient of kinetic friction

N = Normal force = mg

The given value for μ = 0.10

∴ f = 0.10 × 45 × 10

f = 45 N

Hence F net = applied force - frictional force

F net = 200 - 45

F net = 155N

Thus, the acceleration of box (a) will be, F = ma

where m = mass of box =45kg

a = F /m

a = 155/45

a = 3.4 m/s²

a) Hence acceleration of the box pulled horizontally is 3.4m/s²

The speed of the box pulled, if initially the box was at rest and constant force is applied for 3 seconds,

v = u + at

v = 0 + 3.4 × 3

v = 10.2 m/s

b) Hence, the speed of the box at constant force is 10.2m/s

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