Mention two ways in which the effects of friction can be minimised

Answer:

a)  v = 16.57 m / s, b)  a = 19.6 m / s², d)    N = 1.76 10³ N,     N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

          Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

          Em_f = K = ½ m v²

as they indicate that there is no friction

         Em₀ = em_f

         mgh = ½ m v²

         v = [tex]\sqrt{2gh}[/tex]

let's calculate

        v = [tex]\sqrt{2 \ 9.8 \ 14.0}[/tex]

         v = 16.57 m / s

b) the centripetal acceleration has the formula

           a = v² / r

           a = 16.57² / 14.0

           a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

               N-W = m a

               N - mg = m ar

               N = m (g + a)

let's calculate

               N = 60.0 (9.8 + 19.6)

               N = 1.76 10³ N

the relationship with weight is

              N / W = 1.76 10³/( 60 9.8)

              N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

Explanation:

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The Arrows on both sides that has equal length is wrong and the rest of the boxes are the correct

Answer:

option 5

Explanation:

because all u do is have to add them up

Answer:

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Explanation:

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Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

Answer: a= 2 m/s²

Explanation: acceleration = change of speed/ time = 20 m/s / 10 s

Answer:

At dawn your location on earth is pointed straight in the direction of the Earth's travel in its orbit.  Between midnight and dawn you are moving head-on through the location of the meteors in space, which means that you will, on average, observe more of them.

- public.nrao.edu

Explanation:

hope this helps

Answer:

2) a_y= -g  3) vₓ=constant v_y = v_{oy} - g t, 4)  vₓ = v₀ₓ - ax t

5)  changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

           v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

           vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

           x = v₀ₓ t

           y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

Answer:

The motion of an object is accelerated when its speed increases.

Answer:

Mass of Exoplanet =  0.58 kg

Explanation:

First, we will calculate the speed of the forest moon:

[tex]speed = v = \frac{Circumference}{time}\\[/tex]

circumference = 2πr = 2π(14441566 m) = 90739035.3 m

time = 6 days 10 hr = (6 days)(24 h/1 day)(3600 s/1 h) + (10 h)(3600 s/1 h)

time = 554400 s

Therefore,

[tex]v = \frac{90739035.3\ m}{554400\ s}\\\\v = 163.67\ m/s[/tex]

We know that the centripetal force on forest moon will be equal to the gravitational force given by Newton's Gravitational Law, as follows:

[tex]Centripetal\ Force = Gravitational\ Force\\\frac{m_{moon}v^2}{r} = \frac{Gm_{moon}m_{exoplanet}}{r^2}\\\\m_{exoplanet} = \frac{v^2r}{G}\\\\m_{exoplanet} = \frac{(163.67\ m/s)^2(14441566)}{6.67\ x\ 10^{-11}\ N.m^2/kg^2}[/tex]

Mass of Exoplanet =  0.58 kg

Answer:

A

Explanation:

The angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net external torque around that point:

d

→

L

d

t

=

0

or

→

L

=

→

l

1

+

→

l

2

+

⋯

+

→

l

N

=

constant

.

Note that the total angular momentum

→

L

is conserved. Any of the individual angular momenta can change as long as their sum remains constant. This law is analogous to linear momentum being conserved when the external force on a system is zero.

Answer:

his distance is too small (r = 0.01 mm), therefore the cut can be at any distance

Explanation:

For this exercise let's use the ampere law.

Let's use a cylinder as the circulating surface

          ∫ B. ds = μ₀ I

in this case the field is circular and ds is circular therefore the angle between them is zero and cos 0 = 1

          B 2π r =  μ₀ I

          r =  [tex]\frac{\mu_o I}{2\pi B}[/tex]

The field needed to demagnetize the card is B = 1000 gauss = 0.1 T

          r = [tex]\frac{4\pi 10^{-7} 5.0 }{2\pi \ 0.1}[/tex]

           r = 2 10⁻⁷  5.0/0.1

          r = 1  10⁻⁵ m

this distance is too small (r = 0.01 mm), therefore the cut can be at any distance

This problem involved half projectile.

initial velocity, vo = 20 m/s

time of flight, t = 7 s

(a) Simply use the formula to get the height, h:

h = vo*t - (1/2)gt^2

(b) To get the final vertical velocity or terminal velocity (vf), use the formula:

(vf)^2 - (vo)^2 = 2gh

(c) Use the formula find the horizontal distance traveled, R:

R = vo * cos(θ) * t

But since the angle involved with respect to horizontal is zero, and cos(0) = 1, we have

R = vo * t

Hope this helps~ `u`

Jai

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}[/tex]

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = [tex]k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}[/tex]

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = [tex]k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )[/tex]

            E_ {total} = [tex]k \frac{Q}{x_1^2}[/tex]   [tex]( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )[/tex]

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = [tex]-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}[/tex]

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}[/tex]

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = [tex]+ k \frac{Q}{(d-x_1)^2}[/tex]+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

5) Right side point

            E_ {total} = [tex]k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}[/tex]

             E_ {total} = [tex]- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )[/tex]- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

Answer:Air bags can leave you in even more injury, From the impact they give

You could end up with a broken nose,arm

concussion

Answer:

It’s behavioral which is learned via classical conditioning it’s when the neutral or the conditioned stimulus becomes something that causes fear to all organisms

A person acquire a fear of or aversion to something even though it has no negative effect on him or her due to the classical conditioning.

Any thing which has bad effects or something wrong can happen with the doing. The effect is the negative effect.

We assume that we do not fear, a dog that triggers a fear response, such as being bitten.

If we are exposed to the object or such situation we fear over and over without having any negative outcome. The  classical conditioning can truly help unlearn the fear.

Learn more about negative effect.

https://brainly.com/question/2026716

#SPJ2

Answer:

Acceleration = Δv/Δt or change in velocity over change in time

Explanation:

Answer:

x = 1, y = 1 and z = 0

Explanation:

Given equation;

[tex]P^x V^y T^z = constant[/tex]

Boyle's law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.

Mathematically the law is written as;

[tex]PV = constant[/tex]

From the given equation, the values of x, y and z that will match this law is calculated as follows;

[tex]P^1 V^1 T^0 = constant\\\\P^1 = P\\\\V^1 = V\\\\T^0 = 1\\\\P \times V \times 1 = PV = constant\\\\Thus, x = 1, \ y = 1 \ \ and \ z = 0[/tex]

Answer: A tsunami is a very long-wavelength wave of water that is generated by sudden displacement of the seafloor or disruption of any body of standing water.  Tsunami are sometimes called "seismic sea waves", although they can be generated by mechanisms other than earthquakes.  Tsunami have also been called "tidal waves", but this term should not be used because they are not in any way related to the tides of the Earth.  Because tsunami occur suddenly, often without warning, they are extremely dangerous to coastal communities. Ocean trenches are steep depressions in the deepest parts of the ocean [where old ocean crust from one tectonic plate is pushed beneath another plate, raising mountains, causing earthquakes, and forming volcanoes on the seafloor and on land.  

Explanation:

Answer:

OK we appreciate your concern.

Answer: kooi kooi

how r u and thanks for the free points :)

Answer:

1) 1.5 m/s^2

2) 4.5 N

Explanation:

From Newton's Second Law of motion, we know

[tex]F=m*a[/tex]

Which states that to calculate the force acting on an object, you multiply its mass and acceleration.

So, we know an object of mass 1.5 kg has an acceleration of 3 m/s^2, then

[tex]F=m*a=1.5*3=4.5[/tex]

A force of 4.5 N is acting on the object.

If a force of 4.5 N acts on a mass of 3kg we have

[tex]a=\frac{F}{m}=\frac{4.5}{3}=1.5[/tex]

So, it would give it an acceleration of 1.5 m/s^2.

Answer:

E =  1.64 x 10⁶ V/m

Explanation:

The electric field in the region between the plates can be given by the following formula:

[tex]E = \frac{\Delta V}{d}[/tex]

where,

E = Electric Field = ?

ΔV = Poetential Difference across the plates = 23 KV = 23000 V

d = distance between plates = 1.4 cm = 0.014 m

Therefore, using these values in the equation, we get:

[tex]E = \frac{23000\ V}{0.014\ m}[/tex]

E =  1.64 x 10⁶ V/m

Answer:

World War 1 was caused by entangled alliances, nationalism, imperialism, and major

advancements in military technology. Does the Treaty of Versaille address those issues?

Explain your answer using facts. (5 points)