A toy car can go 5 mph. How long would it take to go 12 miles?

The two aircrafts are skidded for 14.9 s.

To find the answer, we need to know about the Newton's equation of motion.

As per Newton's equation of motion

=> a= -1000/224.2= -4.5 m/s²

=> t= 20/4.5 = 4.4 s

Thus, we can conclude that the two aircrafts are skidded for 4.4 s.

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The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

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Answer:

[tex]\alpha=arccos[\frac{(a^2+b^2)(n-1)}{2ab(n+1)}].[/tex]

Explanation:

1) for A+B: a²+b²-2abcos(π-α);

2) for A-B: a²+b²-2abcos(α);

3) according to the condition (A+B):(A-B)=n, then

[tex]n=\frac{a^2+b^2-2abcos(\pi-a)}{a^2+b^2-2abcos(a)}; \ = > \ n=\frac{a^2+b^2+2abcos(a)}{a^2+b^2-2abcos(a)}; \ = > \ cos(\alpha)=\frac{(a^2+b^2)(n-1)}{2ab(n-1)}.[/tex]

according to definition of acceleration

a=v-u/t

t=v-u/a(equation 1)

according to the formula of average velocity

v+u/2*s/t

s=v+u/2*t(equation 2)

now putting the value of t in equation 2

s=v+u/2*v-u/a

s=v^2-u^2/2a

v^2=u^2+2as

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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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An example of an offensive action is B. kicking a ball into the goal

Something that is offensive upsets or embarrasses people because it is rude or insulting.

Offensive is an arranged campaign or plan of action, normally created by the military or planned to achieve some specific political aim or goal. A military plan to strike is an example of a military offensive. A plan to fight the war on medications is an instance of an offensive.

The four types of offensive operations are move to contact, attack, exploitation, and pursuit.

Leaders direct these offensive operations sequentially and in combination to generate highest combat power and destroy the enemy.

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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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The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool. Option A

The weed trimmer is a device that is used to trim the grasses on a lawn or a field. This device has a rotating shaft that does the actual trimming of the grasses.

The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool thus it does not affect the amount of time required to reach 9,000 rpm speed.

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B. The normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.

Fₙ = mg + mv²/r

where;

Fₙ = (2,000 x 9.8) + (2,000 x 18²)/24

Fₙ = 46,600 N

Fₙ =  4.7 x 10⁴ N

Thus, the normal force that is exerted on the coaster by the track at the lowest point is 4.7 x 10⁴ N.

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Answer:

NE BİLİM

Explanation:

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

"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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The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

v = √GM/r

where;

v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]

v = 5,916.36 m/s

Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

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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 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 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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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 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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A wave is a phenomenon that does not cause a permanent displacement in the particles of the medium through which it passes. And it transfers energy from one end of the medium to the other. Examples of waves include light waves, sound waves, water waves, x-rays, radiowaves, etc. Thus the required answers for each part of the question are:

a. The angle that the light wave would have is [tex]7.5^{o}[/tex].

b. The type of mirror that can be used is a plane mirror.

c. The image distance is 146.3 cm.

ii. The image formed by the mirror is a real image.

a. When a ray of light passes from one medium to another, then refraction occurs. The refraction depends on the refractive index of the medium considered.

Thus from Snell's law, we have:

refractive index, n, = [tex]\frac{Sin i}{Sin r}[/tex]

where: i is the angle of incidence, and r is the refracted angle.

Now given that n = 1.5, and i = 15.

Then;

n = [tex]\frac{Sin i}{Sin r}[/tex]

1.5 = [tex]\frac{Sin 15}{Sin r}[/tex]

Sin r = [tex]\frac{0.2588}{1.5}[/tex]

        = 0.17253

r = [tex]Sin^{-1}[/tex] 0.17253

  = 9.936

r ≅ [tex]10^{o}[/tex]

Since the light wave now moves from the glass into water, the determined refracted angle now becomes its angle of incidence in water. So that;

n = [tex]\frac{Sin i}{Sin r}[/tex]

1.33 = [tex]\frac{Sin 10}{Sin r}[/tex]

Sin r = [tex]\frac{0.17365}{1.33}[/tex]

       = 0.1306

r = [tex]Sin^{-1}[/tex] 0.1306

 = 7.504

r = [tex]7.5^{o}[/tex]

Therefore, the angle that the light wave would have is [tex]7.5^{o}[/tex].

b. The image formed would be the same size as that of the object. And also the same distance as that of the object to the pole of the lens.

The type of mirror that can be used is a plane mirror.

The ray diagram is attached to this answer.

c. From the mirror formula;

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]

where; f is the focal length of the mirror, u is the object's distance to the mirror, and v is the image's distance to the mirror.

Given; u = 65 cm, and f = 45 cm, then:

[tex]\frac{1}{45}[/tex] = [tex]\frac{1}{65}[/tex] + [tex]\frac{1}{v}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{1}{45}[/tex] -  [tex]\frac{1}{65}[/tex]

[tex]\frac{1}{v}[/tex] = [tex]\frac{20}{2925}[/tex]

v = [tex]\frac{2925}{20}[/tex]

v  = 146.25 cm

The image distance is 146.3 cm.

ii. The image formed by the mirror is a real image.

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The angle between the two beams inside the glass is mathematically given as

<(BR)= 0.563°

Generally, the equation for an angle for blue is  mathematically given as

[tex]< B= arcsin( sin\theta / i )[/tex]

Therefore

<B= arcsin( sin30 / 1.660 )

<B= 17.53°

For angle for red

<R= arcsin(sin30 / 1.610)

<R= 18.09°

For the angle in between

<(BR)= 18.093 - 17.530

<(BR)= 0.563°

In conclusion,  the angle between the two beams inside the glass

<(BR)= 0.563°

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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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Endarterectomy is used to reduce the risk of stroke and correct the stenosis in the artery.

The primary blood vessels that supply the brain with blood and oxygen are the carotid arteries.

The narrowing of these arteries is referred to as carotid artery disease. Carotid artery stenosis is another name for it. The main factor causing constriction is atherosclerosis.

This fat deposit reduces the blood flow to the brain which cause a stroke.

Carotid endarterectomy is a surgical treatment to remove plaque, an accumulation of fatty deposits that causes a carotid artery to become narrowed.

Hence, an Endarterectomy is used to reduce the risk of stroke, it will correct stenosis in the artery.

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SHM can be acquired by perpendicular projection of uniform circular motion of a particle on its diameter such a particle is called reference particle and its circular path is called reference circle.

A pendulum reaches its maximum velocity when the block is at its lowest point (the pendulum is vertical and pointing straight down). We can then use the term for conservation of energy to determine the maximum height of the block.

As the pendulum swings downward, gravity converts this potential energy into kinetic energy, so that at the bottom of the swing, the pendulum bob has zero potential energy, and its kinetic power, (1/2)mv2, equals the initial potential energy (mgh). (So the velocity, v, equals √(2gh).)

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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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The mass of the planet Mar, given the data from the question is 6.45×10²³ Kg

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by G

M = gr² / G

M = (3.724 × 3400000²) / 6.67×10¯¹¹

M = 6.45×10²³ Kg

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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 average kinetic energy of the gas's molecules directly relates to its temperature. Faster moving particles will more frequently and violently hit the container walls. Because of this, the force acting on the container's walls increases, increasing the pressure.

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When the volume of gas increased the KE of gases particles increases and hit the walls of container which increases the pressure on the walls .

What if gas will increase?

The gas expands in a closed container . The molecules strike and hit each other therefore the pressure at the walls increase.

According to ideal gas law , when volume is cnstant the Pressure and temperature both will increase.

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Answer:

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

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

Explanation:

Answer:

1. 13..6 grams per centimeters cubed.

2. Mass = 400kg

3. Volume = 200cm = 2m

Explanation:

1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.

2. [tex]density=\frac{mass}{volume}[/tex]

[tex]1000=\frac{mass}{2 * 1 * 0.2}[/tex]

[tex]1000*0.4=mass[/tex]

[tex]400kg = mass[/tex]

3. [tex]density=\frac{mass}{volume}[/tex]

[tex]0.6=\frac{120}{volume}[/tex]

[tex]volume=\frac{120}{0.6}[/tex]

[tex]volume= 200cm[/tex]

The magnitude of the tension in the string marked A and B is mathematically given as

A = 52.5 N

Generally, the equation for is mathematically given as

tan=3/8

negative x

[tex]B= tan\phi \\tan \phi=5/4[/tex]

negative x

C= tan=1/6

Hence, considering the Horizontal components

74.9cos(9.46) = A*cos(20.6) + B*cos(51.3)

A = 78.9 - 0.668B

Vertical components

74.9*sin(9.46) + Asin(20.6) = Bsin(51.3)

40.07 = 1.015B

B = 39.5 N

In conclusion, Sub the value of B is the equation of A

A = 78.9 - 0.668B

Sub

A = 78.9 - 0.668( 39.5 N)

A = 52.5 N

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