When traveling north from the United States into Canada, you'll see the North Star (Polaris) getting _________. lower in the sky higher in the sky dimmer brighter

Answer:

88 m ^2

Explanation:

Answer:

F = 32.28 N

Explanation:

For this exercise we must use the rotational equilibrium relation

          Σ τ = 0

In the initial configuration it is in equilibrium, for which all the torque and forces are compensated. By the time the payment lands on the bar, we assume that the counter-clockwise turns are positive.

          W_bird  L / 2 - F_left 0.595 - F_right 0.595 = 0

we assume that the magnitude of the forces applied by the hands is the same

          F_left = F_right = F

          W_bird L / 2 - 2 F 0.595 = 0

          F = [tex]\frac{m_{bird} \ g L} { 4 \ 0.595}[/tex]

we calculate

         F = 0.560 9.8 14.0 /2.38

         F = 32.28 N

Answer:

f=ma......10N=0.2a....=50m/s

Answer:

here,

force = mass× acceleration

10 = 0.2 kg × a

or, 10/0.2=a

or, a = 50km/h^2

is the required ans .

Answer:

MgO relative atomic mass is 40

Explanation:

Mg=24

O=16

Answer:

nothing

Explanation:

bocouse of darkness

Answer:

The orbital velocity an aircraft orbiting around a heavenly body is found as follows;

At the orbital velocity, [tex]F_G[/tex] = [tex]F_C[/tex]

Where;

[tex]F_G[/tex] = The gravitational force = [tex]\dfrac{G \cdot M \cdot m}{R_E^2}[/tex]

[tex]F_C[/tex] = The centripetal force = [tex]\dfrac{m \cdot v_0^2}{R_E}[/tex]

Therefore

[tex]v_0 = \sqrt{\dfrac{G \cdot M}{R_E} }[/tex]

Therefore, at the orbital velocity of the spacecraft, the centripetal force attracting the person away from the central region heavenly body is equal to the gravitational force pulling the person towards the center of the heavenly body (which was felt as her or his weight), and the person feels weightless while inside the orbiting spacecraft

Explanation:

Explanation:

electrical energy into heat energy

electrical , thermal

I think it is a education tips

Answer:

It depends upon SAE No. of oil. (SAE means Society of Automotive Engineers). However, it usually does protect against friction.

Explanation:

If we use very viscous oil, it does not reach all the parts. Very thin oil will flows away easily and gets wasted. Grease is used in such cases. It is generally used around ball-bearing. Normal grease or oil is never used where there is high pressure, high temperature and high speed. Special lubricants are used in such cases. In cold season the oil becomes thick and in hot season it becomes thin. Therefore selection of lubrication also depends on the season. It is always advisable to refer operating manual of the equipment before selecting the lubricant.

The intensity of friction depends on following factors:

i) The area involved in friction.

ii) The pressure applied on the surfaces.

Force = Pressure ´ Area Frictional force will increase, if the area of contact will increase or if pressure applied on the surface increased.

Answer:D

Explanation:

Answer:

Explanation:

Answer:

It was created during the French Revolution in 1799 and has enabled for the international exchange of scientific and technical information. Calculating with SI units is also a lot easier than using the English system.

Answer:

friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.

Answer:

Because it causes a lot of wear and tear in machine parts that move against each other. It erodes the surfaces and destroys their symmetries

Explanation:

Answer:

The postulates of the Kinetic Molecular Theory, KMT, are;

(1) In an ideal gas the molecules are in constant motion

(2) The collisions between molecules of gases are perfectly elastic

(3) The volume occupied by the molecules are negligible

(4) The temperature of the gas is directly proportional to its kinetic energy

(5) The intermolecular forces in the gas are negligible

According to the KMT, gaseous water vapor molecules are in constant motion and move at a speed that depends on their temperature. The intermolecular forces between the molecules are negligible and when they collide with the cold glass, they lose temperature to the glass, thereby reducing their temperature, kinetic energy and therefore, their speed is reduced.

The increasingly temperature of the water vapor coming in contact with the cold glass gives rise to reduced speed of the cooled gas molecules, thereby causing them to move closer together after having elastic collisions and to cluster with tiny particles in the air, to form tiny droplets

The rapid cooling on the cold glass surface causes the droplets to form rapidly on the cold glass surface which makes them visible as condensed water on the surface of the cold glass of water

Explanation:

Answer:

Option A.

Explanation:

We define the half time T as the time such that an initial quantity A reduces to its half.

So we can model the quantity as a function of time like:

P(t) = A*e^(-k*t)

Then for the half time, T, we will have:

P(T) = A/2 = A*e^(-k*T)

solving for k, we get:

A/2 = A*e^(-k*T)

1/2 = e^(-k*T)

ln(1/2) = ln( e^(-k*T)) = -k*T

-ln(1/2)/T = k

Here we know that the half time is T = 26.5s

if we input that in the above equation, we get:

-ln(1/2)/26.5s = k = 0.0262 s^-1

Then the correct option is A

Answer:

1) The kinetic energy of the object after it drops for 2.0 seconds is approximately 384.9 Joules

2) The potential energy of the object after it drops for 2.0 seconds is approximately 204 J

Explanation:

1) The given mass of the object, m = 2.0 kg

The height from which the object is dropped, h = 30 m

The kinetic energy of the object after it drops for 2.0 seconds = Required

Kinetic energy, K.E. = (1/2)·m·v²

Where;

v = The velocity of the object

The kinematic equation for finding the velocity of the object is presented as follows;

v = u + g·t

Where;

u = The initial velocity of the object = 0

g = The acceleration due to gravity of the object ≈ 9.81 m/s²

t = The time of motion of the object = 2.0 seconds

∴ The velocity after 2 seconds, v ≈ 0 + 9.81 m/s² × 2 s = 19.62 m/s

The kinetic energy, K.E. after 2 seconds as the object drops is given as follows;

[tex]K.E._{(after \ two \ seconds)}[/tex] = (1/2) × 2.0 kg × (19.62 m/s)² = 384.9444 J ≈ 384.9 J

2) The total energy, M.E. of the object at the top, h = 30 m, u = 0, is given as follows;

The total mechanical energy, M.E. = P.E. + K.E.

M.E. = m·g·h + (1/2)·m·u²

∴ M.E. = 2.0 kg × 9.81 m/s² × 30 m + 0 = 588.6 J

M.E. = 588.6 J

Given that the total mechanical energy, M.E., is constant, we have;

At 2.0 seconds, M.E. = 588.6 J , K.E. ≈ 384.9 J, P.E. = M.E. - K.E.

∴ P.E. = 588.9 J - 384.9 J ≈ 204 J

The potential energy after it drops 2.0 seconds, P.E. ≈ 204 J

Answer:

Explanation:

To find the horizontal component, the x component specifically, use the formula:

[tex]V_x=Fcos\theta[/tex] and for the vertical component, the y component, use the formula:

[tex]V_y=Fsin\theta[/tex]

where F is the magnitude of the force and theta is the angle in degrees.

For the x-component:

[tex]V_x=6cos250[/tex] so

[tex]V_x=-2.1[/tex] and depending upon whether this is a displacement vector or a velocity vector, the label would be meters/feet or m/s, respectively.

For the y-component:

[tex]V_y=6sin250[/tex] so

[tex]V_y=-5.6[/tex]

a.

Given,

height (h) = 100m

mass (m) = 5kg

acceleration due to gravitation (g) = 9.8ms^-2

Potential energy

= mgh

= 5kg × 9.8ms^-1 × 100m

= 4900 kgm²s^-2

= 4900 J

b.

Since the ball is not moving yet, its kinetic energy is 0.

c.

The total of potential and kinetic energy at every point of the journey is same, i.e., 4900 J.

Answer:

* Point charge outside the radius of the sphere r> R, the force in the two systems is the same

* Point charge inside the sphere  r <R,  therefore the force in the system with the insulating sphere is greater

Explanation:

To answer this question let's use the relation

          F = q E

with q being the point charge and E the electric field created by the sphere.

If we use Gauss's law

The electric field flux is proportional to the wax charge within the surface.

Let's analyze our situation.

* Point charge outside the radius of the sphere

          r> R

where R is the radius of the sphere and r the distance from the center of the sphere to the point charge

in this case the waxed charge for the insulating and conducting sphere is the same, therefore the force in the two systems is the same

* Point charge inside the sphere

           r <R

conductive sphere.

     As the charges are mobile, they are located on the surface of the sphere and there is no waxed charge within a Gaussian surface that passes through the point charge, therefore the electric field is zero and consequently the force

             F = 0

insulating sphere

      Charges cannot move therefore there is a fraction of charge within a surface that passes through the point charge, consequently the electric field is different from zero

            Fe> 0

for this second position the force on the conducting sphere is zero

therefore the force in the system with the insulating sphere is greater

Answer:

force=mass×acceleration

hence

acceleration is given by force÷mass

(500÷200)*5=12.5

C it reduces the amount of useful work done on objects move it up the ramp

All three cars experience the same impulse.

Impulse is equal to change in momentum.

Each car starts with the same amount of momentum and ends up with zero, so the magnitudes of all three changes are equal.

Answer:

A) How the size of a tree affect the bird species that prefer to live in it

Explanation:

I took the quiz

Answer:

One standard kilogram is define as the mass of platinium iridium cylinder having equal diameter and height kept at the particular condition of international bureo of weigh and measure sevre near paris.

Answer:

D. 2.8 hp

Explanation:

1 hp=746 watts

2088.8÷746= 2.8

Answer:

8392

Explanation:

d=s/t

Answer:

a) The velocity of the object as a function of time, v(t) is 2·b·t

b) The acceleration of the function of time, a(t) is 2·b

c) The time at which the object is at the flagpole is t = √(c/b)

Explanation:

The function that gives the position of the object north of the flagpole, x(t) is presented as follows;

x(t) = b·t² - c (b and c are constants)

a) The velocity of the object as a function of time, v(t), is derived as follows

v(t) = x'(t) = d(b·t² - c)/dt = 2·b·t

The velocity of the object as a function of time, v(t) = 2·b·t

b) The acceleration of the function of time, a(t) = v'(t) = d(2·b·t)/dt = 2·b

c) The time at which the object is at the flagpole is given by the x-intercept of the function, where x(t) = 0, as follows;

At the x-intercept, we have, x(t) = 0 and x(t) = b·t² - c

∴ 0 = b·t² - c, which gives

b·t² = c

t² = c/b

t = ±√(c/b), we reject the negative value to get;

The time at which the object is at the flagpole, t = √(c/b).

The current flowing through a resistor at a constant temperature is directly proportional to the potential difference across it. This is called Ohm's law

Answer:

The separation between the two stones is 36 m, when the second stone is approximately 10.9 m below the top of the cliff

Explanation:

The given parameters are;

The height of the cliff from which the stones are dropped, h = 60 m

The time at which the second stone is dropped = 1.6 seconds after the first

The distance below the top of the cliff when the distance between the two stones is 36 m = Required

We have;

The kinematic equation of motion that can be used is s = u·t - (1/2)·g·t²

For the first stone, we have, s₁ = u·t₁ - (1/2)·g·t₁²

For the second stone, we get; s₂ = u·t₂ - (1/2)·g·t₂²

t₁ = t₂ + 1.6

g = The acceleration due to gravity ≈ 9.81 m/s²

s = The distance below the cliff top

The initial velocity of the stones, u = 0

Let t represent the time from which the second stone is dropped at which the distance between the two stones is 36 m, we have;

s₁ = u·(t + 1.6) + (1/2)·g·(t + 1.6)²

s₂ = u·t + (1/2)·g·t²

u = 0

∴ s₁ - s₂ = 36 =  (1/2)·g·(t + 1.6)² - (1/2)·g·t²

2 × 36/(g) = (t + 1.6)² - t²  = t² + 3.2·t + 2.56 - t² = 3.2·t + 2.56

2 × 36/(9.81) = 3.2·t + 2.56

t = (2 × 36/(9.81) - 2.56)/3.2 =  ≈ 1.49 s

t ≈ 1.49 s

s₂ = (1/2)·g·t²

∴ s₂ = (1/2) × 9.81 × 1.49² ≈ 10.9

The distance below the top of the cliff of the second stone when the the separation between the two stones is 36 m, s₂ ≈ 10.9 m.

Answer:

Distance = 6.667 kilometres

Explanation:

Given the following data;

Speed = 20 km/h

Departure time = 7:00

Arrival time = 7:20

Time taken = 20 minutes

To calculate the distance travelled from home to school;

First of all, we would have to convert the value of time in minutes to hours.

Conversion:

60 minutes = 1 hour

20 minutes = X hours

Cross-multiplying, we have;

X = 20/60 = 1/3 hours

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 20 * 1/3

Distance = 20/3 =

Distance = 6.667 kilometres

Answer:

The answer is "4.97 m".

Explanation:

[tex]u = 1.23\ \frac{m}{s}\\\\[/tex]

[tex]H= 1.45 \ m\\\\[/tex]

[tex]\mu = 0.231\\\\[/tex]

The law of conservation tells us that heat energy at the top with kinetic energy at the top equals kinetic energy at the base.

[tex]mgh+\frac{1}{2}mu^2=\frac{1}{2}mv^2\\\\2gh +u^2 =v^2\\\\v=\sqrt{u^2+2gh}[/tex]

[tex]v=\sqrt{(1.23\ \frac{m}{s})^2+2(9.81 \frac{m}{s^2}) +(1.45\ m)[/tex]

   [tex]=\sqrt{1.5129+19.62 +1.45}\\\\=\sqrt{22.5829}\\\\=4.75\ \frac{m}{s}[/tex]

Friction force is given by the formula

[tex]f=-\mu mg \\\\ma= -\mu mg\\\\a=-\mu g\\\\[/tex]

[tex]= -(0.231) \ (9.81\ \frac{m}{s^2})\\\\=-2.26611 \ \frac{m}{s^2}[/tex]

Now by using an equation of motion as

[tex]v^2-u^2= 2as[/tex]

From the above the distance traveled is

[tex]S=\frac{v^2-u^2}{2a}[/tex]

[tex]S=\frac{(0)^2-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\[/tex]

   [tex]=\frac{-(4.75\ \frac{m}{s})^2}{2(-2.26611\ \frac{m}{s^2})}\\\\=\frac{-22.5625}{-4.53222}\\\\=4.97[/tex]

In other words, the distance from the bottom of the chute to the point where the toy zebra comes to rest is [tex]s = 4.97\ m[/tex]