Aster walks first 70 m in the direction 37° north of east, and then walks 82 m in the

direction 20° south of east, and finally walks 28 m in the direction 30° west of north.(2pt)

a) How far and at what angle is the Aster's final position from her initial position?

b) In what direction would she has to head to return to her initial position?

C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.

As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.

Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

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Answer:

11,890

Explanation:

First we need to know what is considered a significant figure.

A significant figure is a value that is not a zero at the start OR end of a value.

Which means, the 0 in the value of 90 or 0.363 are not considered a significant figure.

The 0 in the value of 3056 is considered a significant figure.

So from the table, we can deduce:

0.275 has 3 significant figures

750 has 2 significant figures

[tex]10.4 \times {10}^{5} = 1040000[/tex]

has 3 significant figures.

11,890 has 4 significant figures.

320,050 has 5 significant figures.

So from the above, we can already see the answer.

The range of horizontal projectile motion is 3737.5 m.

A projectile's horizontal velocity is constant (a never changing value), Gravity causes a downward vertical acceleration with a magnitude of 9.8 m/s/s. A projectile's vertical velocity fluctuates by 9.8 m/s per second, A projectile's vertical motion is unrelated to its horizontal motion.  we know that sinθ is maximum at 90°. Therefore horizontal range will maximum at 45°.

To calculate the range of horizontal projectile motion we use;

Δx = vₓ t

where , Δx = Range

             vₓ  = Velocity

              t   = Time

Initial horizontal velocity, vₓ = 650 m/s

Time = 5.75 sec

                Δx = 650 × 5.75

                     = 3737.5 m

Therefore, the range of horizontal projectile motion is 3737.5 m.

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The speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.

The speed of the roller coaster at any position is calculated by applying the principle of conservation of energy.

K.E = P.E

¹/₂mv² = mgh

v = √2gh

where;

v(2) = √[(2 x 9.8)(39 - 0)]

v(2) = 28 m/s

v(3) = √[(2 x 9.8)(39 -  26)]

v(3) = 16 m/s

v(4) = √[(2 x 9.8)(39 - 14)]

v(4) = 22 m/s

Thus, the speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.

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The weight of the automobile is 17,533.6 N.

The weight of the automobile is calculated as follows;

P = F / A

F = (W/4)

P = (W/4) / A

P = W/4A

W = 4AP

where;

W = (4)(217 x 10⁻⁴ m²)(2.02 x 10⁵ Pa)

W = 17,533.6 N

Thus, the weight of the automobile is 17,533.6 N.

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"V = IR" is the equation that relates current to voltage.

Relationship between current and voltage:

                                           V = IR or,

                                           I = V/R,

        the amount of current (I) flowing through a circuit is inversely    

        related to the amount of resistance (r) and directly proportional to  

        the voltage (v).

Therefore the correct answer is option C i.e., "V = IR".

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Answer:

NE BİLİM

Explanation:

PUAN İÇİN YAZIYORUM SEN BENİ GÖRMEZDEN GEL ANLDIN MI DOSTUM bu arada Suriye >>>>>

(a) The weight of the drop is 1.54 x 10⁻²⁵ N,

(b) The the charge on the drop, in terms of e is 5 x 10⁻¹²e and

(c) The excess electrons is 5 x 10⁻¹² electron.

The weight of the drop is calculated as follows;

Volume of the drop; V = ⁴/₃πr³

V = ⁴/₃π(1.64 x 10⁻⁶)³ = 1.845 x 10⁻¹⁷ m³

mass = density x volume

mass =  0.851 g/cm³ x  1.845 x 10⁻²³ cm³ = 1.572 x 10⁻²³ g =  1.57 x 10⁻²⁶ kg.

Weight = 1.57 x 10⁻²⁶ kg x 9.8 m/s² = 1.54 x 10⁻²⁵ N.

q = F/E

q = (1.54 x 10⁻²⁵ N)/(1.92 x 10⁵)

q = 8.01 x 10⁻³¹ C

1.6 x 10⁻¹⁹ C = 1e

8.01 x 10⁻³¹ C = ?

= 5 x 10⁻¹²e

1.6 x 10⁻¹⁹ C ------- 1 electron

8.01 x 10⁻³¹ C ------- ?

= 5 x 10⁻¹² electron

Thus, the weight of the drop is 1.54 x 10⁻²⁵ N, the the charge on the drop, in terms of e is 5 x 10⁻¹²e and the excess electrons is 5 x 10⁻¹² electron.

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The magnitude of the static friction force = 123 N

The coefficient of static friction = 0.31

Static friction is the frictional force that must be overcome in other for a body to that moving over another.

Since the crate does not move, the magnitude of the static frictional force is equal to the applied force.

The magnitude of the static frictional force = 123 N

The coefficient of static friction = frictional force/normal reaction

The coefficient of static friction = 123/(40* 9.8)

The coefficient of static friction = 0.31

In conclusion, the static frictional force on the on the crate is equal to the applied force on the crate.

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The total resistance in the circuit, R is 4 Ω .

The meters connected to the 12 Ω resistance are:

If the switch is opened, only the 4 Ω and 2 Ω resistances will stop working.

The resistances in the circuit are connected both in series and in parallel

The resistances in series are the 4 Ω and the 2 Ω resistances.

Equivalent resistance = 4 + 2 = 6 Ω

The 4 Ω and the 2 Ω resistances are then connected in parallel with the 12 Ω resistance.

Total resistance, R is calculated as follows:

1/R = 1/12 + 1/6

1/R = 3/12

R = 12/3

R = 4Ω

The meters connected to the 12 Ω resistance are:

If the switch is opened, only the 4 Ω and 2 Ω resistances will stop working because the circuit connecting them to the cell is broken whereas the circuit to the 12 Ω resistance is continuous.

In conclusion, resistances can be connected in parallel or in series.

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The solution for the questions below is mathematically given as

(A)

Generally, the equation for the flux is  mathematically given as

[tex]\Phi _{B}=B.A[/tex]

[tex]\Phi _{B}=B.A\\\\\Phi _{B}=B(40*30*10^{-4}) ,[/tex]

So, the magnetic flux through the coil is constant.

From faradays law,

[tex]\varepsilon =-\frac{\mathrm{d} \Phi _{B}}{\mathrm{d} t}[/tex]

---(1) for the induced emf. Since magnetic flux is constant, LHS. of (1) =0

induced emf=0

(B)

Let x be the length of the coil's magnetic field area.

[tex]then, \Phi _{B}=B.A=B(40*x*10^{-4}) \\\\\frac{\mathrm{d}\Phi _{B} }{\mathrm{d} t}=40\\B*10^{-4}*\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]

induced emf=0.0132V

(C)

In conclusion, Therefore, there is no variation in the magnetic flux across the coil when magnetic flux=0.

induced emf=0

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The mechanical energy lost due to friction is 360 J.

The distance the runner slides is, s = 0.655 m

The mechanical energy lost due to friction acting on the runner is equal to the change in kinetic energy.

Change in Kinetic energy = 1/2m(v -u)²

Change in Kinetic energy = 80 * (0 - 3.0)²/2

Change in Kinetic energy = 360 J

Mechanical energy lost = 360 J

Distance he slide is determined using the formula below as follows:

Acceleration of runner = coefficient of friction * acceleration due to gravity

acceleration, a = 0.7 * 9.8 = 6.86

v² = u² - 2as

s = v² + u²/2a

s = 0 + 3³/2 * 6.86

s = 0.655 m

In conclusion, the mechanical energy lost due to friction is the loss in kinetic energy.

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The spring constant, k, in newtons per meter is 1,955.9 N/m.

h = v²sin²θ/2g

v = √[(2gh)/sin²θ]

v = √[(2 x 9.8 x 4.4)/ (sin 39)²]

v = 14.76 m/s

E = K.E + P.E

E = ¹/₂m(v cosθ)²  +  mgh

E =  ¹/₂(1.1)(14.76 cos39)²  +  (1.1 x 9.8 x 4.4)

E = 119.8 J

E = ¹/₂kx²

k = 2E/x²

k = (2 x 119.8)/(0.35²)

k = 1,955.9 N/m

Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.

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There is a 446N force in the horizontal massless cable CD. The vertical component of the pivot A's force on the aluminum pole, which is 328.3N. The force that the pivot A applies to the aluminum pole has a horizontal component of 446N.

We need to be aware of the force in order to discover the solution.

where, H=3.8m, l=7.6m, m=10kg, M=23.5 kg and alpha= 37 degree,

                      [tex]F_V-mg-Mg=0\\F_v=328.2N\\[/tex]

                  [tex]F_H=T=446N[/tex]

In light of this, we may say that the tension in the horizontal massless cable CD and the horizontal component of the force applied by the pivot A to the aluminum pole are identical and each exert 466N of force, whereas the vertical component of that force is 328.3N.

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The change in the gravitational potential energy of the pole vaulter is 2,156 J.

The change in the gravitational potential energy of the pole vaulter is calculated as follows;

ΔP.E = mg(Δh)

where;

ΔP.E = (55)(9.8)(4)

ΔP.E = 2,156 J

Thus, the change in the gravitational potential energy of the pole vaulter is 2,156 J.

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Answer:

Approximately [tex]25\; {\rm N}[/tex] assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].

Explanation:

This mass is in a circular motion of radius [tex]r[/tex]. Hence, when the velocity of the mass is [tex]v[/tex], the acceleration of this mass should be [tex](v^{2} / r)[/tex]. The net force on this mass should be [tex](\text{net force}) = (m\, v^{2}) / r[/tex] towards the center of the circle.

When this [tex]m = 0.80\; {\rm kg}[/tex] mass is at the top of this circle, both gravitational pull and the force of the string (tension) point downwards. Hence, the net force on this mass would be:

[tex](\text{net force}) = (\text{weight}) + (\text{tension})[/tex].

Thus:

[tex]\begin{aligned} (\text{tension}) &= (\text{net force}) -(\text{weight})\\ &= \frac{m\, v^{2}}{r} - m\, g \\ &= m\, \left(\frac{v^{2}}{r} - g\right) \\ &= 0.80\; {\rm kg}\times \left(\frac{(9.0\; {\rm m\cdot s^{-1}})^{2}}{2.0\; {\rm m}} - 9.81\; {\rm m\cdot s^{-2}}\right) \\ &\approx 25\; {\rm kg \cdot m\cdot s^{-2}} \\ &= 25\; {\rm N}\end{aligned}[/tex].

An astronaut's spacesuit is considerably more than just a pair of garments they put on during spacewalks. An entire spacesuit is actually a single-person spacecraft. The Extravehicular Mobility Unit, or EMU, is the official name for the spacesuit used on the International Space Station and Space Shuttle. "Extravehicular" refers to something that is outside of a vehicle or spaceship. To be "mobile" implies to be able to move around while wearing an astronaut suit. The astronaut is shielded by the spacesuit from the perils of being outdoors in space.

The human body cannot survive in space's hostile atmosphere without protection. A spacesuit must shield the astronaut from the vacuum of space, the drastic temperature changes in space, and, if at all possible, it must lessen the astronaut's exposure to radiation. Therefore, the suit's primary function is to act as a pressure vessel. It must keep an air environment close to the skin constant, deliver a constant supply of clean air to the lungs, and expel stale, carbon dioxide-rich air. Every 90 minutes or so, an astronaut in low Earth orbit (LEO) experiences day and night. They can fast chill to -250 F during the night and reach 250 F while the sun is shining on them (-156 C). The body's normal temperature of 98.6 F must be maintained by the suit (37 C).

To do all this, the current NASA suit has 14 layers. The liquid ventilation and cooling garment is the first three layers. It resembles a body-hugging spandex garment that has tubes inside that carry cool water across the body to dissipate extra heat. Air below the next layer cannot escape since it is a pressure vessel made of nylon coated with urethane. The following layer resembles a tent since it is constructed of Dacron. Its goal is to exert pressure on the pressure garment so that it keeps its form and doesn't expand. Neoprene coasted nylon ripstop is the following layer. It is quite resilient to tears. Seven layers of mylar film and foil blanket are used to cover it. These restrict warmth transfer into or out of the suit by acting like a thermos. Due to its several layers, it also provides defense against very small micrometeroids, which drain energy as they pierce and break each layer. The top layer is made of orthofabric, a goretex, nomex, and kevlar mixture. It is very impervious to tearing and thermally reflecting to help regulate temperature. The suits are made by ILC Dover. Incorporating materials that minimize radiation penetration through the suit is something they are always working on.

Answer:

overloading can be avoided if two many appliances are not connected to a single socket short circuiting is a name given to a situation in which they live and the natural voice accidentally coming contact

The magnitude of the acceleration of block m2  is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.

The acceleration of the blocks is calculated from the net force on the blocks.

∑F = ma

a = ∑F/m

a = (F) / (m₁ + m₂)

where;

a = (0.03)/(0.04 + 0.02)

a = 0.09 m/s²

T = m₂a

T = (0.02 x 0.09) = 1.8 x 10⁻³ N

Thus, the magnitude of the acceleration of block m2  is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.

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DNA sequencing is the technique which allows multiple peoples DNA to be compared to look for similarities and differences in order to solve crime and is denoted as option C.

This is referred to as the deoxyribonucleic acid and contains the genetic components of organisms and is also located in the nucleus. On the other hand, DNA sequencing refers to the process of determining the nucleic acid sequence.

Each organisms has a unique DNA sequence which is why it is used to identify individuals through the use of bodily fluids such as saliva, blood etc.

It is used to compare the similarities and differences in order to solve crime thereby making it the most appropriate choice.

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Answer:

Explanation:

An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa

A. The magnitude of the spring force (in N) acting upon the object is 15.9 N

B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²

C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).

F = Ke

F = 106 × 0.15

F = 15.9 N

F = ma

Divide both sides by m

a = F / m

a = 15.9 / 0.52

a = 30.58 m/s²

Considering the diagram, we can see that the spring was pulled away from the equilibrium point.

Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.

Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.

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(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.

(b) The net work done on the crate while it is on the rough surface is -16.04 J.

(c) The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.

F(net) = F - μFf

F(net) = 299 - 0.351(92 x 9.8)

F(net) = -24.67 N

W = F(net) x L

W = -24.67 x 0.65

W = - 16.04 J

a = F(net)/m

a = -24.67/92

a = - 0.268 m/s²

v² = u² + 2as

v² = 0.88² + 2(-0.268)(0.65)

v² = 0.426

v = √0.426

v = 0.65 m/s

Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 24.67 N, opposite as the motion of the crate.

The net work done on the crate while it is on the rough surface is -16.04 J.

The speed of the crate when it reaches the end of the rough surface is 0.65 m/s.

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The maximum height h of the curved path is 7.38 m.

Apply the following kinematic equation;

v² = u² - 2gh

where;

(u² - v²)/2g = h

(39² - 37.1²)/(2 x 9.8) = h

7.38 m = h

Thus, the maximum height h of the curved path is 7.38 m.

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Answer:

No

Explanation:

Plastic bags are not magnetic materials, only magnetic materials (such as iron) can be attracted by the magnet.

Hope this helps.

Answer:

Projectile motion formula or equations derived (In Tabular format)Motion Path equation:

y = (tanθ) x – (1/2) g . x2/(V0 cosθ)2

Explanation:

The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.

The work done is given by the use of the formula;

W = F * x

Where;

F = force applied

x = distance covered

W = 150 N *  4.95 m = 742.5 J

Now;

The coefficient of kinetic friction is given by;

μ = F/mg

μ = 150/ 33 * 9.8

μ = 0.46

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The change in temperature is 20 kelvin


What is Temperature?


Temperature can simply be described as how hot or how cold an object is at a particular period in time. The unit of temperature is Kelvin.

The formula for calculating change in temperature is

Final temperature - Initial temperature

Final temperature = 300 kelvin
Initial temperature= 280 kelvin

Change in temperature= 300-280

= 20 kelvin

Thus the change in temperature for this process is 20 kelvin


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The upward force exerted on the board by the support is 530.8 N.

The upward force exerted on the board by the support is calculated as follows;

F(up) = 52.8 N  +  206.0 N  +  272.0 N

F(up) = 530.8 N

Thus, the upward force exerted on the board by the support is 530.8 N.

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The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

The magnitude of e.m.f induced in the loop is calculated as follows;

emf = dФ/dt

Ф = 6t² + 7t

dФ/dt = 12t + 7

at t = 2 seconds

emf = dФ/dt = 12(2) + 7 = 31 V

Thus,  the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

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