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Answer:

warm front .,

Explanation:

...........

Answer:

the correct answer is D

Explanation:

For this exercise we must remember that the poles of the same if not repel each other and the poles of different signs attract.

With this, let's analyze the situation presented.

The two magnets are perpendicular, with the second magnet to the right of the first.

We have two cases:

* first magnet with the north pole to the right

  If the north of the second magnet is upwards there is a repulsion and the south pole of the second magnet there is an attraction with the north pole of the first magnet, so it would have a force that has to rotate the second magnet.

The force with the south pole of the first magnet that is at a greater distance is less, so the resultant of force is determined by the nearest poles.

If the poles of the second magnet are reversed, that is, the South pole up and the North pole down, the same result is obtained, but with a twist in the opposite direction.

* first magnet with the south pole to the right

    Repels with the south pole of the second magnet and attracts with the north of the second magnet

therefore in both possibilities the second magnet acquires a rotational movement

Consequently the correct answer is D

Answer:

B

Explanation:

The crest of all three waves are of equal height

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh   (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT   (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C    (3)

ΔT = 5.39x10⁹ / 4200 * 75000

Hope this helps

Answer:

The height of the hill is 0.46 m.

Explanation:

Given;

mass of the child and sled, m = 50 kg

initial velocity of the sled, u = 0

final velocity of the sled, v = 3 m/s

The height of the high is calculated from the law of conservation of energy;

P.E at top = K.E at bottom

mgh = ¹/₂mv²

gh = ¹/₂v²

[tex]h = \frac{v^2}{2g} \\\\h = \frac{3^2}{2\times 9.8} \\\\h = 0.46 \ m[/tex]

Therefore, the height of the hill is 0.46 m.

Answer:

A. v = √2gh

B. No! The final velocity does not depend on the mass of the car.

C. Yes! the final velocity depends on the steepness of the hill

D. 3.28 m/s

Explanation:

A. Determination of the final velocity.

½mv² = mgh

Cancel out m

½v² = gh

Cross multiply

v² = 2gh

Take the square root of both side

v = √2gh

B. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is no mass (m) in the formula.

Thus, the final velocity does not depend on the mass of the car.

C. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is height (h) in the formula.

Thus, the final velocity depends on the steepness of the hill

D. Determination of the final velocity.

Height (h) = 0.55 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = √2gh

v = √(2 × 9.8 × 0.55)

v = √10.78

v = 3.28 m/s

Answer:

The cutoff wavelength is the minimum wavelength in which a particular fiber still acts as a single mode fiber. Above the cutoff wavelength, the fiber will only allow the LP01 mode to propagate through the fiber (fiber is a single mode fiber at this wavelength).

Explanation:

Answer:

Ingenstion

Explanation:

The most common tapeworms that infect cats worldwide are Dipylidium caninum and Taenia taeniaeformis. Dipylidium caninum is transmitted to cats by fleas. The immature fleas larvae ingest the eggs of the worm, but infection is then passed on to a cat when it swallows an infected flea during grooming.

Answer:

The most common worms that housecats get are tapeworms. The housecats always lick their fur to clean them. The tapeworms love where it is damp. After the housecats clean their fur, they will play in the house sometimes they will roll on the floor. Then they get tapeworm eggs, after the eggs hatch the tapeworm will live on the fur of the cats.

Explanation:

Answer:

 ρ= 1.378 10⁴ Ω / m

Explanation:

Let's use ohm's law

          V = i R

           R = V / i

let's calculate

           R = 9.0 / 230 10⁻⁶

           R = 3.9 10⁴ Ω

now we can use the definite of resistance

           R = ρ [tex]\frac{L}{A}[/tex]

the area of ​​a circle is

           A = π r² = π (d/2)²

           ρ = R A / L

           ρ = π R [tex]\frac{d^2}{4L}[/tex]

let's calculate

          ρ = π 3.9 10⁴  [tex]\frac{(1.5 \ 10^{-3}^2 }{4 \ 5 \ 10^{-2}}[/tex]

          ρ= 1.378 10⁴ Ω / m

Answer:

Explanation:

a ) The volume of blood flowing per second throughout the vessel is constant .

a₁ v₁ = a₂ v₂

a₁ and a₂ are cross sectional area at two places of vessel and v₁ and v₂ are velocity of blood at these places .

2A x v₁ = A x .40

v₁ = .20 m /s

b )

Let normal pressure be P₁ when cross sectional area is 2A and at cross sectional area A , pressure is P₂

Applying Bernoulli's theorem

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

P₁ - P₂ = 1/2  ρ(v₂² - v₁² )

= .5 x 1060 ( .4² - .2² )

= 63.6 Pa .

If a boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water, then the horizontal distance traveled by the boy would be 2.58 meters.

There are three equations of motion given by Newton,

v = u + at

S = ut + 1/2 × a × t²

v² - u² = 2 × a × s

As given in the problem if a boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water,

3 = ut + 1/2 × a × t²

3 = 0 + 0.5 × 9.8 × t²

t = 3 / 4.9

t = 0.7824

The horizontal distance traveled by the boy = 3.3 ×  0.7824

                                                                         = 2.58 meters

Thus, the horizontal distance traveled by the boy would be 2.58 meters.

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Answer:

A) t = 1.249 10⁻² s, B)  f = 80 Hz, C) f = 20 Hz,

D)  slowly advancing an angle of approximately    Δθ = 0.05 rad each flash

E) In each flash it seems to go backward an angle of Δθ = -0.05 rad

Explanation:

A) To make it appear that the blades are immobile, it implies that every time the light turns on, a blade should be in the same position, therefore, as we have 4 blades, they must rotate an angle of 2π/4,  

         θ = π / 2  

         θ = 1.57 rad  

taking the angle let's use the endowment kinematics relations  

          θ = w₀ t + ½ α t²  

in general the fans rotate at constant speed α= 0  

         θ = w₀ t  

         t = θ / w₀  

let's reduce the magnitudes to the SI system  

        w₀ = 20 rev / s (2π rad / 1rev) = 125.66 rad / s  

let's calculate  

        t = 1.57 / 125.66  

        t = 1.249 10⁻² s  

B) the fastest speed for the blades to rotate is when one blade of a complete turn , we use the relationship between the fecuance and the period  

        f = 1 / T  

        f = 1 / 1.25 10⁻²  

       f = 80 Hz

C) we have two possibilities:  

* a yellow dot is placed on each sheet  

In this case the angular velocity of the blade is the same at all points, therefore the results obtained should not change

* a yellow dot is placed on a single sheet.  

Here for the point to remain fixed the angle of rotation must be

       θ= 2π rad  

the time is  

       t = 2π / 125.66  

       t = 5 10⁻² s  

the maximum frequency is  

      f = 1/5 10⁻²  

      f = 20 Hz

D) The copy strobe rotates at f = 19 Hz, the time between each flash is  

      t = 1/19  

      t = 5.26 10⁻² s  

this time is higher, so the angle turned is large  

       θ = w t  

       θ = 125.66 5.26 10⁻²  

       θ = 6.61 rad  

the relationship between this angle and the angle of a circle is  

θ = 1,052

We can see that it is this time the blade rotates 1 complete turns, for this the position of the blade changes us, for the other 0.052 rad the blade rotates a little more than the circumference therefore it seems that it is slowly advancing an angle of approximately  

         Δθ = 0.05 rad each flash  

E) in this case changes the flash speed  

       t = 1/21  

       t = 4.76 10⁻² s  

the angle rotated is  

      θ = 125.66 4.76 10⁻²  

      θ = 5.984 rad  

      θ / 2π = 0.95  

in that case, the blade did not complete the turn, therefore in each flash it seems to go backward an angle

Δθ = -0.05 rad

This question is incomplete, the missing image is uploaded along this answer below;

Answer:

a) the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) the time at which the contact occur is 8 seconds

Explanation:

Given the data in the question;

first we convert the given angular velocity to rad/s

angular velocity = 240 rpm = ((240/60) × 2π ) = 8π rad/s

so

ωA = 8π rad/s

next we determine angular acceleration at point A; so

ωA = at

8π rad/s = at -------let this be equation

thus, angular acceleration of disk A is ωA and rotates in counter clockwise direction.

Next we determine the velocity of point C;

Vc = rA × ωA

where Vc is velocity at point C, rA is radius of A ( 150/1000)m,  { from the diagram }

so we substitute

Vc = 0.15m × 8π

Vc = 1.2π m/s

for angular velocity at point B;

Vc = rB × ωB

where rB is the radius of B ( 200/1000)m

we substitute

1.2π = 0.2 × ωB

ωB = 1.2π / 0.2

ωB = 6π rad/s

Thus, the wheel B rotates with an angular velocity of 6π rad/s in clock wise direction.

Now,

a) Determine the required angular acceleration magnitude α

we find the the angular acceleration of disk B after 2 seconds, using the expression;

ωB = at

where angular acceleration is a and t is time ( t - 2)

we substitute

ωB = at

6π = a( t - 2) -------- let this be equation 2

now, lets substract equation 1 form equation 2

(6π = a( t - 2)) - (8π = at)

(6π = at - 2a) - ( 8π = at)

-2π = 0 + -2a

2π = 2a

a = 2π/2

a = π rad/s² or 3.14 rad/s²

Therefore, the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) determine the time at which the contact occurs;

from equation 1

8π = at

we substitute in the value of a

8π = π × t

t = 8π / π

t = 8 seconds

Therefore, the time at which the contact occur is 8 seconds

Answer:

A. Coefficient of kinetic friction, μ = 0.34

B. The box moves a distance of 9.64 m before coming to a stop

C. The heavier box will stop after 2.1 seconds

Explanation:

a. The coefficient of kinetic or sliding friction is given as: μ = F/R

where applied force, F = m × (∆v)/t

∆v = v - u

∆v = 0 - 8 m/s = -8 m/s; t = 2.4 s

R = normal reaction = m×g

where g = 9.8 m/s²

Substituting in the kinetic friction formula; μ = m∆v/t ÷ 1/m×g

μ = ∆v/g×t

μ = 8 / 9.8 × 2.4

μ = 0.34

b. Using the equation v² = u² + 2as to calculate the distance travelled by the box

where v = 0 m/s; u = 8.0 m/s; a = ? s = ?

From F = ma = μR

a = μR/m = (μ × m × g)/m

a = μg

a = 0.34 × 9.8

a = 3.32 m/s²

This is negative acceleration or deceleration

Substituting in the equation of motion

8² + 2 × -3.32 × s = 0

-6.64s = -64

s = 9.64 m

Therefore, the box moves a distance of 9.64 m before coming to a stop

c. The coefficient of friction is independent of mass.

Using the formula in (a): μ = ∆v/g×t

t = ∆v/μg

t = 7/0.34 × 9.8

t = 2.10 s

Therefore, the heavier box will stop after 2.1 seconds

The  coefficient of kinetic friction of the box is 0.34.

The distance traveled by the box is 9.6 m.

The time taken for the heavier box to stop is 2.1 s.

The  coefficient of kinetic friction of the box is calculated as follows;

[tex]\mu mg = ma\\\\\mu g = a\\\\\mu g = \frac{v}{t} \\\\\mu = \frac{v}{gt} \\\\\mu = \frac{8}{9.8 \times 2.4} \\\\\mu = 0.34[/tex]

The distance traveled by the box is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\2as = -u^2\\\\s = \frac{-u^2}{2a} \\\\s = \frac{-u^2}{2\mu g} \\\\s = \frac{-(8)^2}{2\times 0.34 \times 9.8} \\\\s = -9.6 \ m\\\\|s| = 9.6 \ m[/tex]

The time taken for the heavier box to stop is calculated as;

[tex]\mu = \frac{v}{gt} \\\\t = \frac{v}{\mu g} \\\\t = \frac{7}{0.34 \times 9.8}\\\\ t = 2.1 \ s[/tex]

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Answer:

47947.52 J.

Explanation:

From the question,

Amount of heat given of (Q) = mc(t₁–t₂).................... Equation 1

Where m = mass of water, c = specific heat capacity of water, t₁ = initial temperature, t₂ = final temperature.

Given, m = 640 g = 640 g, c = 4.816 J/g°C, t₁ = 76 °C, t₂ = 28 °c.

Substitute these values into equation 1 above

Q = 640×4.816(48)

Q = 147947.52 J.

Hence the amount of heat given off is 47947.52 J.

Answer:

Check Newton's Rings:

d = height of air film

s = distance from center to ring being considered

R = radius of circle considered

The approximate formula is:

d = s^2 / (2 R)   or s = (2 R d)^1/2

If we just use 4000 mi for R and 1/4 mi for d the height

we get s = (2 * 4000 * 1/4)^1/2 = 2000^1/2 mi = 45 mi

Answer: B. The elevator is not an inertial frame of reference.

F. The elevator must be accelerating.

Explanation:

Since we've been given the information that the tension in the string is measured to be exactly equal to the force due to gravity on the load and that no other forces are acting on the load, the statements about the elevator that are correct include:

• The elevator is not an inertial frame of reference.

• The elevator must be accelerating.

If the elevator isn't accelerating, the tension in the string can't be equal to the force of gravity.

Answer:b

Explanation:

Guess

Answer: d = st

Explanation:

We know that the distance is equal to the rate (speed) times the time

d = st

According to the ideal gas law, pressure will rise as a gas's temperature rises. There is a limit to how much the tire can expand before the rubber gives in to the pressure build-up.

For every 10 degrees that the temperature drops, the inflation pressure in tires typically decreases by 1 to 2 psi. Moreover, as the tire pressure heats up during the first 15 to 20 minutes of driving, it will increase by one psi every five minutes.      

The ideal gas law states that pressure will increase as a gas's temperature increases. Before the rubber gives in to the pressure build up, the tire can only expand so far.

Therefore, The pressure in your tires will increase due to the increased particle movement in hot air, which will cause the centre of the tread to bow out and wear out first. Increasing the demand for new tires.

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Answer:

the  frequency of the oscillation is 2.5 Hz.

Explanation:

Given;

time to complete the oscillation, t = 0.4 s

number of oscillations, n = 1

The frequency of the oscillation is calculated as;

[tex]F = \frac{n}{t} \\\\F = \frac{1}{0.4} \\\\F = 2.5 \ Hz[/tex]

Therefore, the  frequency of the oscillation is 2.5 Hz.

Answer:

The warm seas create a large humid air mass. The warm air rises and forms a low pressure cell, known as a tropical depression.

Explanation:

Hurricanes arise in the tropical latitudes (between 10 degrees and 25 degrees N) in summer and autumn when sea surface temperature are 28 degrees C (82 degrees F) or higher.

Answer:

air

Explanation:

Answer:

is there a. b.  c  or d?

Explanation:

Answer:

d. 110 MW

Explanation:

The computation of the power loss is given below:

= I^2 × R

= 1000^2 × 110

= 110 × 10^6 W

= 110 MW

Here I represent the current through the resistance

ANd, R represent the resistance

So, the correct option is d

Answer:

Depending, on how much it's push against together.

Explanation:

Since, the two objects are getting in contact. But, if it's a type of item/thing there's a different frictions, but I know it's normal friction when two objects comes in contact. But, its depending on how much you push it against the two items.

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

Answer:

Displacement = 30 miles due east.

Explanation:

Let the distance due west be A

Let the distance due east be B

Given the following data;

A = 50 miles

B = 80 miles

To find the displacement;

Displacement can be defined as the change in the position of a body or an object. It is a vector quantity because it has both magnitude and direction.

Thus, the displacement would be calculated by subtracting distance A from distance B because the rider rode in opposite directions.

Displacement = B - A

Displacement = 80 - 50

Displacement = 30 miles due east.

Answer:

10 mm

Explanation:

We'll begin by calculating the spring constant of the spring. This can be obtained as follow:

Extention (e) = 5 mm

Force (F) = 125 N

Spring constant (K) =?

F = Ke

125 = K × 5

Divide both side by 5

K = 125 / 5

K = 25 N/mm

Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:

Force (F) = 250 N

Spring constant (K) = 25 N/mm

Extention (e) =?

F = Ke

250 = 25 × e

Divide both side by 25

e = 250 / 25

e = 10 mm

Thus, the spring will stretch 10 mm when a 250 N force is applied.

Answer:

The first choice, one block per minute.