A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizontal. What is the horizontal force exerted by the ground on the ladder when an 81 kg object is 4.0 m from the bottom?

Answer: 3 seconds

Explanation:

Since they provide the same impulse,

Impulse= Ft

F1 = 6N

t1 = 2s

F2= 4N

t2= ?

F1= Force of first engine

t1= time elapsed by first engine

F2= force of second engine

t2= time elapsed by second engine

F1t1 = F2t2

6 × 2 = 4t2

t2= 12/4

t2 = 3 seconds

This second engine fires for 3 s

[tex]\texttt{ }[/tex]

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

thrust of first engine = F₁ = 6 N

elapsed time of first engine = t₁ = 2 s

thrust of second engine = F₂ = 4 N

Asked:

elapsed time of second engine = t₂ = 2 s

Solution:

We will use Newton's Law of Motion to solve this problem as follows:

[tex]\Sigma F = ma[/tex]

[tex]\Sigma F = m \Delta v \div t[/tex]

[tex]\Sigma F = I \div t[/tex]

[tex]\boxed {I = \Sigma F \times t}[/tex] → Impulse Formula

[tex]\texttt{ }[/tex]

Two rocket engines provide the same impulse :

[tex]I_1 = I_2[/tex]

[tex]\Sigma F_1 \times t_1 = \Sigma F_2 \times t_2[/tex]

[tex]6 \times 2 = 4 \times t_2[/tex]

[tex]12 = 4 \times t_2[/tex]

[tex]t_2 = 12 \div 4[/tex]

[tex]\boxed {t_2 = 3 \texttt{ s}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Dynamics

Answer:

Option-(D): 10¹¹ waves per second.

Explanation:

The electromagnetic waves is such form of a energy transfer or wave propagation through any space with or without having any medium(particles).As, the medium or particles inside a space are able to transfer the amount of energy from the origin towards the receiver, which makes it very easy for the wave propagation through a medium.Now, the electromagnetic waves are generated from the sensor on a traffic light when the frequency,f level of the wave generation is about 10¹¹ Hertz(Hz).

Answer:

answer is A

Explanation:

Answer:

autonomic specificity

Explanation:

The hypothesis which states that the different emotions in our body involve different and unique physiological profiles is called autonomic specificity hypothesis.

Viewed from this hypothesis perspective, emotions in the human body can be seen as time-tested solutions to timeless problems and challenges. With emotions, evolution in the human body has provided the human with  at least one generalized response to tackle these problems.

Answer:

B. False

Explanation:

Acceleration is the slope of a velocity vs. time graph.

Displacement is the area under a velocity vs. time graph.

Answer:

A torque of 102.5375 Nm must be exerted by the fireman

Explanation:

Given:

The rate of water flow = 6.31 kg/s

The speed of nozzle  = 12.5 m/s

Now, from the Newton's second law we have  

The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction

thus we have,

F = 6.31 kg/s x 12.5m/s

or

F = 78.875 N  

Now,

The torque (T) exerted by water force about the fireman's will be

T = (F x d)

or

T = 78.875 N x 1.30 m

T = 102.5375 Nm

hence,

A torque of 102.5375 Nm must be exerted by the fireman

The torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.

Newtons second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.

It can be given as,

[tex]F=ma[/tex]

Here, (m) is the mass of the body and (a) is the acceleration.

For the flow of water, the second law of motion can be given as,

[tex]F=Q\times v[/tex]

As he rate of water flow is 6.31 kg/s, and the nozzle speed is 12.5 m/s. Thus the force of this can be given as,

[tex]F=6.31\times12.5\\F=78.875\rm N[/tex]

The torque for the fireman exert on the hose is equal to the product of force applied and the distance traveled. Therefore the value of torque is,

[tex]\tau=78.875\times1.30\\\tau=102.5375\rm Nm[/tex]

Thus, the torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.

Learn more about the Newtons second law of motion here;

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Answer:

Voltage-gated K+ channels

The correct answer is that potassium ion channels mediate the falling phase of an action potential.

During the falling phase of an action potential, the membrane potential of the neuron must return to its resting state after being depolarized. This is primarily achieved through the efflux of potassium ions (K^+) out of the cell. The opening of voltage-gated potassium channels allows for this efflux, which repolarizes the membrane potential back towards the resting membrane potential.

 Here is the sequence of events during an action potential:

 1. Resting potential: The neuron is at its resting membrane potential ,typically around -70 mV, due to the concentration gradients of ions across the membrane and the selective permeability of the membrane to potassium ions via leak channels.

 2. Rising phase (Depolarization): When a stimulus reaches the threshold level, voltage-gated sodium channels open, allowing sodium ions (Na^+) to rush into the cell. This influx of positive charges depolarizes the membrane potential towards +30 mV.

3. Peak of the action potential: The membrane potential reaches its peak when the sodium channels become inactivated, stopping the influx of sodium ions.

4. Falling phase (Repolarization): Voltage-gated potassium channels open, and potassium ions move out of the cell, restoring the membrane potential towards the resting potential. The movement of potassium ions out of the cell is slower than the initial sodium influx, which is why the falling phase is slower than the rising phase.

5. Overshoot: Sometimes, the membrane potential temporarily overshoots the resting potential, becoming more negative than at rest, due to the continued efflux of potassium ion.

6. Return to resting potential: The potassium channels close, and the sodium-potassium pump restores the original ion distribution across the membrane, bringing the membrane potential back to the resting potential.

In summary, the falling phase of an action potential is mediated by the opening of voltage-gated potassium channels, which allows for the efflux of potassium ions, thereby repolarizing the neuron's membrane potential."

Answer: If it has ions, it is an electrolyte

Explanation:

Let's start by explaining that electrolytes are compounds that contain charged particles or ions, which can be cations (positive ions) or anions (negative ions).

So, it is this composition that makes an electrolytic material conduct electricity.

In this sense, the way to identify if a material is an electrolyte or not, is knowing whether it is composed of ions or not.

Applying the work-energy theorem, which states that the work done by nonconservative forces equates to the change in kinetic plus potential energy, the skateboarder's change in potential energy results in -18.4J, and the absolute change in height becomes 0.032 m.

The student is asking about the concept of conservation of energy in physics, particularly in the context of the work-energy theorem. This theorem states that work done by nonconservative forces (like friction) is equal to the change in the kinetic and potential energy of the system. According to the given problem, the initial kinetic energy of a 55.6-kg skateboarder is KE0 = 0.5 * m * v², and after performing 80.4J of work on himself, while friction does -244J of work on him, the final kinetic energy becomes KEf = m * v² / 2 where 'v' is the final speed of 7.24 m/s.

(a) Considering that there are no conservative forces, the work-energy theorem states that Wnc = ΔKE + ΔPE. From here, we can calculate that ΔPE or PEf - PE0 results in -18.4J.

(b)The change in vertical height can be obtained by Δh = ΔPE / (m*g), where 'g' is the gravitational constant. Thus, Δh =-18.4J divided by (55.6 * 9.8), which results in -0.032 m, but since we are asked for the absolute value, the height change is 0.032 m.

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Answer:

1.034 m above the floor

Explanation:

The location of center of body for a compound body, when the weights are given is calculated as:

[tex]\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}[/tex]

where,

[tex]\bar x[/tex] is the center of gravity of the entire body

W = weight of the individual body

x = center of gravity of the individual body

Thus on substituting the values we get,

[tex]\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}[/tex]

or

[tex]\bar x = \frac{691.83}{669}[/tex]

or

[tex]\bar x =1.034m[/tex]

Hence, the center of gravity of the entire body lies 1.034 m above the floor

Answer:

50°

Explanation:

It is given that the both scales are the same at 18˚

Now,

at 30˚ on the Cantor scale, the temperature is 30˚ − 18˚ = 12˚ above

the 18˚ mark.

also, it is given that with every 3˚ increase in the temperature on the Cantor scale is there is an 8˚ increase on the Frobenius scale,

mathematically, we can write it as (by unitary method)

3˚ increase in the temperature on the Cantor =  8˚ increase on the Frobenius scale

or

1˚ increase in the temperature on the Cantor =  (8/3)˚ increase on the Frobenius scale

thus, for x˚ increase in the temperature on the Cantor =  ((8/3)˚ × x) increase on the Frobenius scale

hence, for 12° increase we have

12˚ increase in the temperature on the Cantor =  ((8/3)˚ × 12) increase on the Frobenius scale

or

12˚ increase in the temperature on the Cantor =  32° increase on the Frobenius scale

hence, the final reading on the Frobenius scale will be, 18˚ + 32˚ = 50˚.

Answer:given below

Explanation:

Cart is pulled 14 cm from mean position

and its position is given by

[tex]x=14cos\left ( \pi t\right )[/tex]

therefore its velocity is acceleration is given by

[tex]v=-14\pi sin\left ( \pi t\right )[/tex]

[tex]a=-14\pi ^2cos\left ( \pi t\right )[/tex]

[tex]\left ( a\right ) cart\ max.\ speed\ is[/tex]

[tex]v_{max}=14\pi at\ t=0.5sec[/tex]

and its position is x=0

acceleration at t=0.5sec

a=0

[tex]\left ( b\right )[/tex]

[tex]a_{max}=14\pi ^2 cm/s^2[/tex]

at t=0,1,2 sec

at t=0

x=14 cm

v at t=0

v=0 cm/s

for t=1 sec

x=-14 cm i.e. 14 cm behind mean position

v=0 m/s

The cart's maximum speed is 14π m/s and occurs at the equilibrium point in its oscillation at integer multiples of the half period. At these moments, the magnitude of the acceleration is 14π² m/s². The magnitude of the acceleration is greatest when the cart is at the extreme points of its oscillation, where its speed is zero.

The cart, undergoing simple harmonic motion, has a maximum speed when it passes through equilibrium - the midpoint of its oscillating path. From the equation for the straightforward harmonic motion, we know that the speed v of the cart is given by the derivative of s(t) = 14cos(πt). The derivative of this function is v(t) = -14πsin(πt), and the maximum speed occurs when sin(πt) = ±1, which gives a maximum speed of ±14π m/s.

The maximum speed is attained at integer multiples of the half period (1/2, 3/2, 5/2 seconds, and so on). At this moment, the cart is at the equilibrium position, s=0. Acceleration a(t) is given by the second derivative of the position function s(t), which is a(t) = -14π²cos(πt). The magnitude of the acceleration at the time of maximum speed is 14π² m/s².

In the case of the maximum magnitude acceleration, this is attained at the turning points of the oscillation when cos(πt) = ±1. At these moments, the speed of the cart is zero.

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Answer:

C. Interference from the sun causes data to be collected inaccurately.

Explanation:

Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.

Answer:

7 mph

Explanation:

Let the speed of river is c mph.

The effective downstream speed = 14 + c

The effective upstream speed = 14 - c

Total time = 4 hrs

Total distance = 42 miles

Time for upstream + time for downstream = 4 hrs

21 / (14 + c) + 21 / (14 - c) = 4

21 (14 - c + 14 + c) = 4 (196 - c^2)

21 x 28 = 4 (196 - c^2)

c^2 = 196 -147 = 49

c = 7 mph

The speed of the river is 7 miles/h.

Speed of Bob in still water is 14 miles/h.

Speed of Bob downstream = (14 + x) miles/h

Speed of Bob upstream = (14 - x) miles/h

here x = speed of the river.

The total distance downstream and back is 42 miles. Hence,

Distance upstream = Distance downstream = 21 miles

Given that total time of journey upstream and downstream is 4 hours. Then, using the formula:

[tex]t = \frac{distance}{speed}[/tex], we get:

[tex]\frac{21 \hspace{0.5mm} miles} {(14 + x) \hspace{0.5mm} miles/h} + \frac{21 \hspace{0.5mm} miles}{(14 - x)\hspace{0.5mm}miles/h} =4 \hspace{0.5mm} hours[/tex]

or, [tex]\frac{2 \times 21 \times 14 \hspace{0.5 mm} h}{14^{2} -x^{2} } = 4 \hspace{0.5mm} hours[/tex]

or, 21 × 7 h = 14² - x²

or, x² = 49

or, x = 7 miles/h

Answer:

[tex]\lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

Explanation:

the velocity as a function of time is

[tex]v(t)=\frac{mg}{c}(1-e^{\frac{-ct}{m}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})[/tex]

[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-\frac{1}{\infty })\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-0)\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]

The limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance considered, is the terminal velocity mg/c. The exponential term in the equation approaches zero as time becomes very large, leading to the object reaching a constant terminal velocity.

To calculate the limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance taken into account, we use the provided equation v = mg/c (1 − e^{-ct/m}), where g is the acceleration due to gravity, which averages 9.80 m/s², and c is a positive constant representing air resistance. The limit represents the object's terminal velocity, which is the constant speed an object reaches when the force of gravity is balanced by the drag force of air resistance.

As t → ∞ (increases towards infinity), the exponential term e^{-ct/m} approaches zero. Hence, the limit of the speed v becomes:

∑ lim t→∞ v = lim t→∞ mg/c (1 − e^{-ct/m}) = mg/c.

The value mg/c is known as the terminal velocity, which is the maximum speed that the object will reach as it continues to fall.

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Answer: 768000km

Explanation:

Velocity is given by the relation between the distance [tex]d[/tex] and the time it takes to travel that distance [tex]t[/tex]:

[tex]V=\frac{d}{t}[/tex]   (1)

In this problem we are told the time it takes for radio wave to travel from the Earth to the Moon and back is the "echo":

[tex]t=2.56s[/tex]  (2)

In addition, we know radio waves are electromagnetic waves (light), and its velocity is:

[tex]V=3(10)^{8}m/s[/tex]   (3)

Substituting (2) and (3) in (1):

[tex]3(10)^{8}m/s=\frac{d}{2.56s}[/tex]   (4)

And finding [tex]d[/tex]:

[tex]d=(3(10)^{8}m/s)(2.56s)[/tex]   (5)

Finally we can obtain the distance:

[tex]d=768000000m=768000km[/tex]  

Answer:

384,000 km

Explanation:

Given

Velocity of radio wave [tex]v = 3.00 \times 10^{8}m/s[/tex]

Duration of echo T = 2.56 s

Solution

Time taken for the radio wave to travel to moon and to travel back to earth as it was picked up by the astronaut's microphone is 2.56 s

Since any time delays in the electronic equipment can be ignored

time taken for the radio wave to reach moon

[tex]t = \frac{T}{2}\\\\t = \frac{2.56}{2} \\\\t = 1.28 s[/tex]

[tex]v = \frac{d}{t}\\\\d = vt\\\\d = 3 \times 10^8 \times 1.28\\\\d = 3.84 \times 10^8 m\\\\d = 384,000 km[/tex]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

[tex]E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2[/tex]

[tex]E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2[/tex]

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

Answer:

F = 32.2m

Explanation:

The force of gravity on an object is given by:

[tex]F=mg[/tex]

where

m is the mass of the object

g is the acceleration due to gravity

Here we have:

- An object of mass m

- The acceleration of gravity is expressed as [tex]g=32.2 ft/s^2[/tex]

Therefore, substituting into the formula above, we find that the force of gravity on the object is

[tex]F=m\cdot 32.2 = 32.2m[/tex]

Answer:

F = 32.2m

Explanation:

Answer:

   V₂  =  225 m^{3}

so no exact option is match with the calculated answer.

Explanation:

According to Boyle's Law, we have

                                     P_1 V_1  =  P_2 V_2    ----------- (1)

Data Given;

                  P_1  =  1.0 atm

                  V_1  =  180 m³

                  P_2  =  0.80 atm

                  V_2  =  ?

Solving equation 1 for V₂,

                  [tex]V_2= \frac{P_1 V_1}{P_1}[/tex]

Putting values,

                   [tex]V_2 = \frac{1.0*180}{0.80}[/tex]

                  [tex]V_2 = 225 m^{3}[/tex]

so no exact option is match with the calculated answer.

Answer:

E

Explanation:

Answer:

i = 7.777 × 10⁻⁴ A = 0.77 mA

Explanation:

Given:

loop radius, r = 3.0 cm = 0.03 m

Area, A = π x r² = π x 0.03² = 0.0028 m²

Magnetic Field, B = 0.75 T

Loop resistance, R = 18 Ω

time, t = 0.15 seconds

Now,

the induced emf is given as:

EMF = [tex]-\frac{BA}{t}[/tex]

also

EMF = i x R

Where, i is the current flowing

equating both the formulas for EMF, we get

[tex]{i}{R}=-\frac{BA}{t}[/tex]

or

[tex]{i}=-\frac{BA}{tR}[/tex]

substituting the values in the above equation we get

[tex]{i}=-\frac{0.75\times 0.0028}{0.15\times 18}[/tex]

or

the magnitude of the current, i = 7.777 × 10⁻⁴ A = 0.77 mA

The electric current produced in a loop due to a changing magnetic field can be calculated using Faraday's law of electromagnetic induction. Based on the given data, the magnitude of the induced current in the loop is approximately 1.18 mA.

The scenario described in the question depicts a situation where a magnetic field decreases uniformly in magnitude, thus inducing an electric current within a circular loop according to Faraday's law of electromagnetic induction.

The electric current produced in the loop as a result of the change in the magnetic field can be calculated using the expression, I = ΔΦ/ Δt * R, where ΔΦ is the change in magnetic flux, Δt is the time interval, and R is the resistance of the loop.

In this case, the initial magnetic flux, Φ1 = B1 * A, where B1 is the initial magnetic field and A is the area of the loop. Since the magnetic field decreases to zero, the final magnetic flux, Φ2, is zero. So, ΔΦ = Φ2 - Φ1 = - B1 * A. Here, A = πr², where r is the radius of the loop.

So, I = - [ (B1* πr²) / ( Δt * R ) ]. Substituting the given values into the equation, we get I = (0.75 T * π * (0.03 m)²) / ( 0.15 s * 18 Ω ) = 0.00118 A or 1.18 mA.

So, the magnitude of the induced current is approximately 1.18 mA.

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Answer:

Required mass of sand is 20 kg

Explanation:

Given:

Mass of the plank = 25 kg

Distance of the Center of gravity of the Plank from the fulcrum = [tex]\frac{2}{2}-0.50 = 0.5m[/tex]

Distance of the Center of gravity of the sand box from the fulcrum = [tex]\frac{2}{2}-\frac{0.75}{2}= 0.625m[/tex]

Balancing the torque due to the plank and the sand box with respect to the fulcrum

Torque = Force × perpendicular distance

thus, we get

(25 × g) × 0.5 = weight of sand × 0.625

where, g is the acceleration due to gravity

or

(25 × g) × 0.5 = (mass of sand × g) × 0.625

or

mass of sand = 20 kg

Hence, the required mass of the sand is 20 kg

Answer:

20 Kg mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed horizontally on a fulcrum placed just below its midpoint.

Explanation:

Use the second condition of equilibrium.

[tex]$\sum} \tau=0$[/tex]

[tex]MgL-$M g x_{c m}=0$[/tex]

[tex]$M=\frac{m x_{c m}}{L}[/tex]

[tex]=\frac{25(0.50)}{0.625}[/tex]

[tex]=20 \mathrm{~kg}$[/tex]

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Answer:

Boat speed = 8 miles/hr

Explanation:

Let the speed of the boat be U

Let the speed of the current be V.

Therefore, the downstream speed is U+V

and the upstream speed is U-V

Now we know that Speed = distance / time

Therefore, downstream speed, U+V = 63 / 7

                                                    U+V = 9 miles/hr   ------(1)

                  Upstream speed, U-V = 63 / 9

                                                U-V = 7 miles/hr      --------(2)

Therefore subtracting (2) from (1), we get

( U+V) - ( U-V ) = 9-7

2V = 2

V = 1 miles/hr

Therefore the speed of the current is V = 1 mile/hr

Now from (1) we get

U+V = 9

U+1 = 9

U = 8

Therefore, the speed of the boat is U = 8 miles/hr

Answer:

Part a)

2)  >  1)  >  3)

Part b)

1)  =  2)  =  3)

Explanation:

Due to charge induction the magnitude of charge on the inner surface of the outer shell is having same charge as that of the small sphere inside but the sign of charge must be opposite.

So here we can say

1)+ 4q, 0

so inner surface has charge - 4q and outer surface charge is +4q

2) -6q , +10q

so inner surface charge is +6q, outer surface charge is +4q

3) +16q , -12q

so inner surface charge is -16q, outer surface charge is +4q

Part a)

situations in which inner surface charge is arranged in decreasing order is given as

2)  >  1)  >  3)

Part b)

Situations in which outer surface charge is arranged in decreasing order is given as

1)  =  2)  =  3)

Situation (1) has the most positive charge on the inner surface of the shell, while situation (2) has the most positive charge on the outer surface.

To rank the situations according to their charge on the inner and outer surfaces of the metallic spherical shell, we need to consider the net charges on the small charged ball and the shell. Let's analyze each situation:

Therefore, ranking the situations according to the charge on the inner surface would be: (1), (3), (2) - from most positive to least positive. For the outer surface, the ranking would be: (2), (1), (3) - from most positive to least positive.

In a Northern Hemisphere extratropical cyclone, the surface winds blow counterclockwise and inward due to the Coriolis effect, a result of the Earth's rotation.

Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow counterclockwise and inward about the center. This is due to the Coriolis effect, a phenomenon caused by the Earth's rotation which influences the direction that wind travels across the globe. It results in winds in the Northern Hemisphere curving to the right, thus causing cyclone winds to move counterclockwise.

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In a Northern Hemisphere extratropical cyclone, the surface winds blow in a counterclockwise and inward direction. This pattern is influenced by the Coriolis force, which causes winds to be deflected to the right, leading to this counterclockwise rotation. The phenomenon also aligns with the fact that air is attracted towards low-pressure centers like cyclones.

In the Northern Hemisphere, when viewed from above, surface winds around an extratropical cyclone flow in a counterclockwise and inward direction. This phenomenon is influenced by the Coriolis force, which causes winds to be deflected to the right in the Northern Hemisphere. This deflection results in a counterclockwise rotation. The Coriolis force similarly influences tropical cyclones, causing them to display the same pattern of rotation.

A key point to note is that this rotation is associated with areas of low pressure, such as the center of these cyclone systems. This low-pressure center attracts air, causing winds to flow inward. As the air rises within these low-pressure zones, it cools and forms clouds, making such cyclonic weather patterns visible from space.

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Answer:

intensity.

Explanation:

when the light collected by the lens is focused into a small spot it tends to increase the intensity of the light.

as different path of light with different intensity combines from passing through the lens it tends to make the light path and intensity coherent and after being coherent there intensity increases.

Answer:

Kinetic Energy

Explanation:

Heat energy is another name for thermal energy. Kinetic energy is the energy of a moving object. As thermal energy comes from moving particles, it is a form of kinetic energy.

Answer:

B,A,C

Explanation:

Hope this helps!!!

Answer:

B, A, C

Explanation:

Momentum:

[tex]P = m*v[/tex]

[tex]P = momentum[/tex]

[tex]m = mass[/tex]

[tex]v = velocity[/tex]

Momentum of car A

[tex]P_{A} = 1500kg*10m/s = 15,000kg*m/s[/tex]

Momentum of car B

[tex]P_{B} = 1500kg*25m/s = 37,500kg*m/s[/tex]

Momentum of car C

[tex]P_{C} = 1000kg*10m/s = 10,000kg*m/s[/tex]

In decreasing order:

B, A, C

Answer:

a) Increase by a factor of more than 2

Explanation:

The compression process is an adiabatic one. The opposite of this kind of process is the isothermal process, where the heat flow makes the temperature to be constant.

Let's consider this kind of process by means of an equation of state; for example, the ideal gas law:

[tex]P=\frac{RT}{v}[/tex]

Be [tex]v_{0}[/tex] the initial volume. And [tex]v_{f}=2v_{0}[/tex] the final volume.

If we consider the ideal gas law, it is evident that if the temperature remains constant (isothermal process), the pressure increases by a factor of 2; but in an adiabatic process the temperature of a gas tends to increase its temperature, so the pressure will be a higher than the resultant for the isothermal process.

Answer:

Ffs = 251 N

Fn = 691 N

Explanation:

Take the y direction to be normal to the ramp and the x direction to be parallel to the ramp.

The angle of the ramp is 20°, so the angle that the weight vector makes with the normal is also 20°.  Therefore:

Fgx = Fg sin 20°

Fgy = Fg cos 20°

Sum of the forces in the x direction (parallel to the ramp):

∑F = ma

Ffs − Fgx = 0

Ffs = Fgx

Ffs = Fg sin 20°

Ffs = 735 sin 20°

Ffs ≈ 251

Sum of the forces in the y direction (normal to the ramp):

∑F = ma

Fn − Fgy = 0

Fn = Fgy

Fn = Fg cos 20°

Fn = 735 cos 20°

Fn ≈ 691

To answer that question we need to apply equations of movement ( from Newton´s laws )

In equilibrium:

∑F  = 0         or    ∑Fx = 0     ;  ∑Fy = 0

Solution is:

a) F(sf) = 251 [N]

b) Fn = 691 [N]

From the attached drawings we can see:  ( Body free diagram)

∑ Fₓ  = F(sf)  - Pₓ  = 0           where P = m×g  = 735 [N] ( the weigth)

and Pₓ = P× cos20°

Then     F(sf)  = Pₓ × sin 20°

F(sf) = m×g×cos20°  =  735× 0.34202 [N]

F(sf) = 251.3847

F(sf) = 251 [N]

rounding to the nearest number

F(sf) = 691 [N]

∑ Fy = 0

∑ Fy = Fn - Py  = 0                     Py = P×cos20°      Py = m×g×cos20°

Py = 735×0.939693 [N]     Py =   [N]

Fn = Py = 690.674 [N]

rounding to the nearest whole number

Fn = 691 [N]

Related question  Brainly.com/question/ 11544804

Answer:

Speed of A = 891 km/h

Speed of B = 804 km/h

Explanation:

Let the speed of aeroplane B is v, the speed of aeroplane is A is 87 km/h faster than B.

So, the speed of aeroplane A is 87 + v.

Distance traveled by A after 7 hours, d1 = (87 + v) x 7

Distance traveled by B after 7 hours, d2 = v x 7

Total distance traveled = 11865 km

So, d1 + d2 = 11865

(87 + v) x 7 + v x 7 = 11865

609 + 14 v = 11865

14 v = 11256

v = 804 km/h

So, the speed of A = 87 + 804 = 891 km/h

Speed of B = 804 km/h

Answer:

Rate of change of area is [tex]75.398ft^{2}/sec[/tex]

Explanation:

[tex]Area=\pi r^{2}\\\\\frac{d(Area)}{dt}=\frac{\pi r^{2}}{dt}=2\pi r\frac{dr}{dt}[/tex]

Applying values we get [tex]\frac{d(Area)}{dt}=2\pi 4\times 3=75.398ft^{2}/sec[/tex]