A plane flying from Townsville, Australia, has an air speed of 264 m/s in a direction 5.0° south of west. It is in the jet stream, which is blowing at 37 m/s in a direction 15° south of east. In this problem you are going to be asked about the velocity of the airplane relative to the Earth.

Answer:

The answer to your question is: Cx = -3.0 m

                                                     Cy = -5.2 m

Explanation:

Vector C is in the third quadrangle then Cx and Cy are negatives. The answer were both components are negatives is letter D. But let's do the operations to prove it.

cos Ф = os/hyp   clear os  

os = hyp x cosФ

os = 6 x cos 60

os = 6 x 0.5 = 3 but is negative   os or Cx = -3 m

sen Ф = as / hyp clear as

as = hyp x sen Ф but is negative as or Cy = -5.2 m

as = 6 x sen 60

as = 6 x 0.87

as = 5.2 m

A vector consists of two components: magnitude and direction. In this case, for vector C, the components are Cx and Cy, where Cx represents the component in the x-direction, and Cy in the y-direction. In the context of this question, the components of vector C are Cx = -5.20m and Cy = 3.00m.

In the context of physics, vectors consist of both magnitude and direction, represented in components. The components of a vector are the projections of the vector along the axes. Here, Cx is the component of the vector C in the x-direction, and Cy is the component in the y-direction. Considering the options given, vector C is defined by the components: Cx = -5.20m, Cy = 3.00m if we consider the first option (A). Please note the choice would depend upon the context or given system of coordinates.

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Answer: The student’s understanding of all four terms (speed, velocity, scalar, and vector) is correct.

Explanation:

Let's start by explaining that a vector is one that has a numerical value along with its units (called a module) and a direction, while a scalar is only determined with a number and its corresponding units, without direction.

Then, speed is the distance an object travels in a given time. That is, it only takes into account the distance traveled, dividing it by time to know how fast it moves, therefore it is a scalar.

Instead, velocity refers to the time it takes for an object to move in a certain direction. So, by involving the direction of movement, velocity is a vector.

In short, the speed does not take into account the direction of the object, while the velocity does.

Therefore, as the student understands this four concepts, the correct option is:

The student’s understanding of all four terms (speed, velocity, scalar, and vector) is correct.

Answer:

Maximum elevation, h = 160 meters

Explanation:

Initially, the rocket is at rest, u = 0

Acceleration of the rocket, [tex]a=20\ m/s^2[/tex]

Time, t = 4 s

We need to find the maximum elevation reached by the rocket. It can be calculated using second equation of motion as :

[tex]h=ut+\dfrac{1}{2}\times a\times t^2[/tex]

[tex]h=\dfrac{1}{2}\times 20\times (4)^2[/tex]

h = 160 meters

So, the maximum elevation reached by the rocket is 160 meters. Hence, this is the required solution.

Answer:

 486.5 m  

Explanation:

Initial velocity is zero as the rocket is fired from rest. u = 0.

Displacement of the rocket during this time:

s = ut +0.5 at²

s = 0+0.5 ×20×4²

s = 160 m

The final velocity at the end of 4 s is:

v = u + at

v = 0 + (20)(4)

v = 80 m/s

This will become the initial velocity for the next half of the motion.

At the maximum elevation, velocity is zero. v = 0

Acceleration due to gravity always acts downwards.

[tex]s=\frac{v^2-u^2}{2a}\\s=\frac{0-80^2}{2\times -9.8} = 326.5 m[/tex]

Thus, the maximum elevation that test rocket would reach is:

326.5 m + 160 m = 486.5 m

Answer : 29.7 mL are in a fluid ounce based on this data.

Explanation :

As we are given that 300 mL and 10.1 oz are equivalent. That means,

300 mL = 10.1 oz

or,

10.1 oz = 300 mL

Now we have to determine the volume of fluid in milliliters.

As, 10.1 oz of fluid = 300 mL

So, 1 oz of fluid = [tex]\frac{1oz}{10.1oz}\times 300mL[/tex]

                         = 29.7 mL

Therefore, 29.7 mL are in a fluid ounce based on this data.

Based on the given data, there are approximately 29.7 milliliters in one fluid ounce. Conversion takes place from ounce to millimeters.

To find the number of milliliters in a fluid ounce based on the given data, we can set up a proportion using the information provided:

300 mL corresponds to 10.1 oz.

300 mL / 10.1 oz = x mL / 1 oz

300  × 1  = 10.1  × x

300  = 10.1 × x

Dividing both sides by 10.1:

300  / 10.1 = x

x ≈ 29.7 mL

Therefore, based on the given data, there are approximately 29.7 milliliters in one fluid ounce.

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Answer:

The velocity of water at the bottom, [tex]v_{b} = 28.63 m/s[/tex]

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, [tex]P_{gauge} = 2.90 atm[/tex]

Solution:

Now,

Atmospheric pressue, [tex]P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa[/tex]

At the top, the absolute pressure, [tex]P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa[/tex]

Now, the pressure at the bottom will be equal to the atmopheric pressure, [tex]P_{b} = 1 atm = 1.01\times 10^{5} Pa[/tex]

The velocity at the top, [tex]v_{top} = 0 m/s[/tex], l;et the bottom velocity, be [tex]v_{b}[/tex].

Now, by Bernoulli's eqn:

[tex]P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b} [/tex]

where

[tex]h_{t} -  h_{b} = 12.8 m[/tex]

Density of sea water, [tex]\rho = 1030 kg/m^{3}[/tex]

[tex]\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}[/tex]

[tex]\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}[/tex]

[tex]v_{b} = 28.63 m/s[/tex]

Answer: 6 times further

Explanation:

Initial velocity is the same that she uses on earth vertically and horizontally

Vertically we can say

V = -g t

If g is 1/6 earth gravity, then t is 6 times earth time.

Then, horizontally, with the same initial time, no incidence of gravity and knowing t

X= v t

Distance in the moon is 6 times distance in the earth.

Answer:

The average velocity is [tex]v_{average}=25[/tex].

Explanation:

The average velocity is calculated in the following way:

If [tex]s(t)[/tex] represents the position of an object at time [tex]t[/tex],

and [tex]s(t_{1})=a[/tex] ; [tex]s(t_{2})=b[/tex], the average velocity is defined in that interval as:

[tex]v_{average}= \frac{final.position-initial.position}{elapsed.time}=\frac{b-a}{t_{2}-t_{1}}[/tex]

Taking the data from the question:

[tex]v_{average}=\frac{173-123}{3-1}=\frac{50}{2}=25[/tex].

Answer:

[tex]|B|=27.00425726m[/tex]

[tex]\alpha =210.3781372[/tex]°

Explanation:

Let's use the component method of vector addition:

[tex]A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844[/tex]

Now, we know:

[tex]C_x=A_x+B_x\\\\C_y=A_y+b_y[/tex]

So:

[tex]B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774[/tex]

Now lets calculate the magnitude of the vector B:

[tex]|B|=\sqrt{(B_x)^{2} +(B_y)^{2}  }=27.00425726m[/tex]

Finally its angle is given by:

[tex]\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344[/tex]°

Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.

Vector is quantity. The magnitude of vector B is 4.644 m while the angle from the positive x-direction is 302.88°.

A Vector is a quantity in physics that has both magnitude and direction.

We know that in order to add to vector we need to divide the vector into two parts, a sine(Vertical) and a cosine(Horizontal), therefore,

The vertical addition of the vectors A and B can be written as,

[tex]\vec A_y +\vec B_y = \vec C_y[/tex]

[tex]\vec A(Sin\ \theta_A) +\vec B(Sin\ \theta_B) = \vec C(Sin\ \theta_C)[/tex]

[tex]13.6(Sin\ 40.2^o) +\vec B(Sin\ \theta_B) = 13.8(Sin\ 20.7^o)\\\\\vec B(Sin\ \theta_B ) =-3.9[/tex]

The Horizontal addition of the vectors A and B can be written as,

[tex]\vec A_x +\vec B_x = \vec C_x[/tex]

[tex]\vec A(Cos\ \theta_A) +\vec B(Cos\ \theta_B) = \vec C(Cos\ \theta_C)[/tex]

[tex]13.6(Cos\ 40.2^o) +\vec B(Cos\ \theta_B) = 13.8(Cos\ 20.7^o)\\\\\vec B(Cos\ \theta_B ) =2.5215[/tex]

As the value of Sin is negative and the value of Cos is positive, therefore, Vector B will lie in the fourth quadrant.

The angle of Vector B,

[tex]\dfrac{\vec B\ Sin\ \theta_B}{\vec B\ Cos\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\\dfrac{\ Sin\ \theta_B}{\ Cos\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\Tan\ \theta_B} = \dfrac{-3.9}{2.5215}\\\\\theta_B = Tan^{-1}\ \dfrac{-3.9}{2.5215}\\\\\theta_B = -57.12^o[/tex]

Thus, the angle of vector B is 57.12° clockwise from the -x direction.

In order to make the angle positive, we can deduct the value from 360°,

[tex]\theta_B = -57.12^o\\\\\theta_B = 360^o -57.12^o\\\\\theta_B = 302.88^o[/tex]

The magnitude of Vector B,

We know the value of the Perpendicular component of vector B,

[tex]\vec B(Sin\ \theta_B ) =-3.9\\\\\vec B(Sin\ -57.12^o ) =-3.9\\\\\vec B= 4.644\rm\ m[/tex]

Hence, the magnitude of vector B is 4.644 m while the angle from the positive x-direction is 302.88°.

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Answer:

The correct equation for measuring the average microscopic weight  for 3 isotopes is multiply the rate of abundance by each weight and add them.

Explanation:

To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.

This gives the required  average microscopic weight of the three isotopes.

Final answer:

To determine the magnitude and direction of the second run of beetle 2 in order to end up at the new location of beetle 1, vector addition can be used. The magnitude can be found using the Pythagorean theorem, and the direction can be calculated using trigonometry.

Explanation:

To determine the magnitude and direction of the second run of beetle 2 in order to end up at the new location of beetle 1, we can break down the given runs and use vector addition. Beetle 1 runs 0.50 m due east and then 0.80 m at 30° north of due east. Beetle 2 runs 1.6 m at 40° east of due north. Adding these vectors, we can find the resultant vector, which represents the displacement from the starting point to the new location of beetle 1. This resultant vector has both magnitude and direction.

To find the magnitude of the resultant vector, we can use the Pythagorean theorem. The sum of the squares of the magnitudes of the individual vectors is equal to the square of the magnitude of the resultant vector. Using trigonometry, we can calculate the angle that the resultant vector makes with the east direction. This angle represents the direction of the second run of beetle 2.

Answer:

Y = 40.94m

Explanation:

The initial speed of the sandbag is the same as the balloon and so is its position, so:

[tex]Y = Yo + Vo*t-\frac{g*t^2}{2}[/tex]

Replacing these values:

Yo = 40m     Vo = 5m/s     g = 9.81m/s^2     t = 0.25s

We get the position of the sandbag:

[tex]Y = 40+5*(0.25)-\frac{9.81*(0.25)^2}{2}[/tex]

Y = 40.94m

Answer:

3. 0.5 sec.

Explanation:

A bullet fired horizontally follows a projectile motion, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with  constant acceleration, g = 9.8 m/s^2, towards the ground

The time taken for the bullet to reach the ground can be calculated just by considering the vertical motion:

[tex]y(t) = h + v_{0y} t - \frac{1}{2}gt^2[/tex]

where y is the vertical position at time t, h is the initial height, and [tex]v_{0y}[/tex] is the initial vertical velocity of the bullet.

Since the bullet is fired horizontally, [tex]v_{0y}=0[/tex]. So the equation becomes

[tex]y(t) = h - \frac{1}{2}gt^2[/tex]

And the time that the bullet takes to reach the ground can be found by requiring y=0 and solving for t:

[tex]t=\sqrt{\frac{2h}{g}}[/tex]

As we can see, in this equation there is no dependance on the initial speed of the bullet: therefore, if the bullet is fired still horizontally but with a different speed, it will still take the same time (0.5 s) to reach the ground.

We have that for the Question "A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and neglecting air resistance and the earth’s curvature, it would have hit the ground in1. There is no way to tell from the information given." it can be said that the time will remain the same

T=0.5

Option 2

From the question we are told

A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and neglecting air resistance and the earth’s curvature, it would have hit the ground in1. There is no way to tell from the information given.

1. less than 0.5 sec.

2. 0.5 sec.

3. more than 0.5 sec.

Generally

When speed is increased and the Range also increased with respect to the speed

Therefore'

The time will remain the same

T=0.5

Option 2

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Answer:

we see that force of collision is the same in both the case.

Explanation:

Let the time of impact in both the situation be t and mass of each be m

Applying conservation of momentum in the first case

m₁v₁ + m₂v₂ = (m₁ +m₂ ) v

m x 20 + m x 10 = 2m x v

v = 15 mph.

So the speed of B will be reduced from 20 to 15 mph  and speed of A will be increased from 10 to 15 mph.

Considering impact on  B only

Impulse on B is equal to change in momentum

F X t = m ( 20 - 15 )

F is force of collision .

F = 5m / t

In the second case ,

Applying conservation of momentum in the second case

m₁v₁ + m₂v₂ = (m₁ +m₂ ) v

m x 40 + m x 30 = 2m x v

v = 35 mph.

So the speed of B will be reduced and speed of A will be increased.

Considering impact on  B only

Impulse on B is equal to change in momentum

F X t = m ( 40 - 35 )

F is force of collision .

F = 5m / t

So we see that force of collision is the same in both the case.

Answer:

the decrease in intensity is due to the conservation of energy in the wavefront.

Explanation:

Electromagnetic waves are transverse waves that comply with the principle of conservation of energy, these are formed by the variation of an electric and / or magnetic field and travels spherically from their point of origin.

 By the principle of conservation of energy after the wave is emitted, the energy of it is distributed throughout the space, generally in spherical form. To conserve energy the density should decrease as the radius of the sphere increases, which is the inverse of the radius squared (1 / r²)

  The previous decrease is observed in the decrease of the amplitude of the wave, since the intensity is the square of the electric field.

In summary, the decrease in intensity is due to the conservation of energy in the wavefront.

Final answer:

The amplitude of light decreases as the distance from the source increases due to the inverse square law, where the wave's energy spreads out over a larger area as it propagates.

Explanation:

The amplitude of an electromagnetic wave, such as light, is the maximum field strength of the electric and magnetic fields. In terms of the dependence of amplitude on the distance from the source, as light travels away from the source, the amplitude decreases. This happens because the wave's energy spreads out as the wave propagates, and because the energy of the wave is directly related to its amplitude, a reduction in energy with distance leads to a decrease in amplitude. If we consider a point source of light, the intensity (and therefore the amplitude) of the light diminishes in proportion to the square of the distance from the source, following the inverse square law.

Answer:

A brighter light

Explanation:

Light waves travel through space via light particles called photons. This particles have in essence 2 properties: 1. Amplitude and 2.Frequency. The first one has to do with the intensity of light we see and the second one has to do with the energy (color). If we change only the amplitude, we will see a lighter or darker light and will keep the same color in all amplitude changes. But if we modify the frequency, the intensity will keep the same and the color changes as we move into the light spectrum.

Thus, increasing the amplitude, we will perceive a brigher light.

Answer:

yes

Explanation:

u = 30 m/s

θ = 45°

h = - 100 m (below)

d = 50 m

g = - 9.8 m/s^2

Use second equation of motion in vertical direction

[tex]h=u_{y}t +\frac{1}{2}a_{y}t^{2}[/tex]

[tex]-100=30\times Sin45\times t -0.5\times 9.8t^{2}[/tex]

[tex]-100 = 21.21 \times t -4.9 \times t^{2}[/tex]

[tex]t=\frac{21.21\pm \sqrt{21.21^{2}+4\times4.9 \times 100}}{9.8}[/tex]

By solving we get

t = 7.17 s

The horizontal distance traveled in this time

= u Cos45 x t = 30 x 0.707 x 7.17 = 152.1 m

This distance is more than the width of the river, So the ball crosses the river.

The term that describes the number of times an analog wave is measured each second during an analog-to-digital conversion is called the 'Sampling Rate'. It refers to the number of samples per second taken from a continuous signal to make a discrete signal.

The number of times an analog wave is measured each second during an analog-to-digital conversion is represented by option D. Sampling Rate. This term is used in digital signal processing and refers to the number of samples per second (or per other unit) taken from a continuous signal to make a discrete or digital signal. For audio, this is typically done in hertz (Hz).

For example, the standard sampling rate for audio is 44.1 kHz (kilohertz, or thousands of hertz), meaning the original wanalog signal is sampled over 44,000 times per second. This is to ensure a faithful reproduction of the sound when it's converted back into an analog signal for playback.

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Answer:

35.71 N

Explanation:

The elevator starts from the rest means its initial velocity is zero.

Given that, the height achieved by the elevator in 1.4 s will be, [tex]S=1m[/tex]

Given that the mass of the bundle which is hold by passenger is, [tex]m=3.3 kg[/tex]

Now according to second equation of motion.

[tex]S=ut+\frac{1}{2}at^{2}[/tex]

Here, S is the height, u is the initial velocity, t is the time taken, and a is the acceleration.

Now initial velocity is zero therefore,

[tex]S=\frac{1}{2}at^{2}\\a=\frac{2S}{t^{2} }[/tex]

According to the free body diagram tension and acceleration in upward direction and weight is in downward direction.

So,

[tex]ma=T-mg\\T=m(g+a)[/tex]

Put the value of a from the above

[tex]T=m(g+\frac{2S}{t^{2} })[/tex]

Put all the variables.

[tex]T=3.3(9.8+\frac{2\times 1}{1.4^{2} })\\T=3.3(9.8+1.02)\\t=35.71N[/tex]

This the required tension.

Final answer:

To determine the tension in the cord as the elevator accelerates, the acceleration of the elevator is first calculated using the kinematic equation and is found to be approximately 1.02 m/s². Then, the tension is calculated using the formula T = mg + ma and is determined to be approximately 36.04 N.

Explanation:

To calculate the tension in the cord as the elevator accelerates, we first need to determine the acceleration of the elevator. Using the distance traveled and the time it took, we can apply the kinematic equation s = ut + 0.5at2 to find the acceleration 'a', where s is the distance (1 m), u is the initial velocity (0 m/s), and t is the time (1.4 s). After finding 'a', we can calculate the tension (T) in the cord using the formula T = mg + ma, where m is the mass of the bundle (3.3 kg), g is the acceleration due to gravity (9.81 m/s2), and a is the acceleration of the elevator.

First, we find the acceleration:

s = u t + 0.5at2
1 = 0 + 0.5a(1.4)2
a ≈ 1.02 m/s2

Next, we calculate the tension:

T = mg + ma
T = 3.3 x 9.81 + 3.3 x 1.02
T ≈ 32.67 + 3.366
T ≈ 36.04 N

Answer:

0.517 mV

Explanation:

Length of wire = 1.07 m

For square:

Perimeter = 1.07 m

Let a be the side of square

So, 4a = 1.07

a = 0.2675 m

Area of square, A 1 = side x side = 0.2675 x 0.2675 = 0.07156 m^2

For circle:

Circumference = 1.07 m

Let r be the radius of circle

So, 2 π r = 1.07

2 x 3.14 x r = 1.07

r = 0.1704 m

Area of circle, A 2 = π r^2 = 3.14 x 0.1704 x 0.1704 = 0.09115 m^2

Change in area, dA = A2 - A1 =   0.09115 - 0.07156  = 0.0196 m^2

Time taken in changing the area, dt = 4.36 s

Magnetic field, B = 0.115 T

According to the Farady's law of electromagnetic induction

[tex]e = \frac{d\phi }{dt}=\frac{dBA }{dt}=B\frac{dA }{dt}[/tex]

[tex]e = 0.115\times \frac{0.0196}{4.36}[/tex]

e = 5.17 x 10^-4 V

e = 0.517 mV

Thus, the induced emf is 0.517 mV.

Answer:

+1

Explanation:

An atom in the neutral state has the same number of protons and electrons. Since protons carry the positive charge and electrons carry negative charge of equal magnitude as that of protons, so, in neutral state the overall charge on the atom is zero.

Atomic number of Lithium is 3. Under neutral state it has 3 protons and 3 electrons. So, its overall electric charge is 0.

If an atom of Lithium loses one of its outermost electron, it is left with 2 electrons and 3 protons. Since, number of protons is 1 more than the number of electrons, the electrical charge on Lithium atom would be positive and the magnitude of charge will be equal to the number of electrons lost, which is 1 in this case. The magnitude can also be calculated as difference in the number of protons and electrons.

Therefore, on losing one electron, the electric charge on Lithium atom would be +1.

Answer:

The car travels 36.8 m before coming to stop after the light changes

Explanation:

The car moves at a constant speed of 59.0 km/h for 0.750 s before the driver hits the brake.

The equation for the position of an object moving at constant speed is:

x = x0 + v t

where:

x = position at time t

x0 = initial position

v = speed

t= time

Let´s consider the initial position as the position at which the driver sees the traffic light and decides to brake. That will make x0 = 0. Then, the position after 0.750 s will be:

x = 59.0 km/h * 0.750 s (1 h /3600 s) = 0.0123 km (1000 m / 1 km) = 12.3 m

while braking, the car has a negative acceleration, then, the speed is not constant. The position of the car will be given by the following equation:

x = x0 + v0 t + 1/2 a t² ( where a = acceleration and v0 = initial speed)

and the speed can be expressed as follows:

v = v0 + a t

from this equation, we can calculate how much time it takes the car to stop (v = 0):

0 = v0 + a t

-v0 = a t

-v0 / a = t

v0 is the speed of the car as the driver hits the brake (59.0 km/h) and "a" is the acceleration (5.50 m/s²) that will be negative because the car is losing speed. Then:

-59.0 km/h (1000 m / 1 km) (1 h / 3600 s) / (-5.50 m/s²) = 2.98 s

Now, we can calculate the position at this time to know the minimum distance the car travels before coming to stop:

x = x0 + v0 t + 1/2 a t²

now x0 will be the distance traveled after the driver sees the light but before braking ( 12.3 m).  v0 will be the speed before braking, 59.0 km / h or 16.4 m/s. Then:

x = 12.3 m + 16.4 m/s * 2.98 s +1/2 (-5.50 m/s²) * (2.98 s)²

x = 36.8 m

Final answer:

The minimum distance the car travels before coming to a stop is 1225 m.

Explanation:

To find the minimum distance the car travels before coming to a stop, we need to consider the distance traveled during the driver's reaction time and the distance traveled while the car is decelerating.

First, we calculate the distance traveled during the reaction time using the equation:

distance = speed × time

distance = 59.0 km/h × 0.750 s = 44.3 m

Next, we calculate the distance traveled while decelerating using the equation:

distance = (initial velocity² - final velocity²) / (2 × acceleration)

distance = (59.0 km/h)² / (2 × 5.50 m/s²) = 1181 m

The minimum distance the car travels before coming to a stop is the sum of the distances traveled during the reaction time and while decelerating:

minimum distance = 44.3 m + 1181 m = 1225 m

Answer:

7.44 seconds

Explanation:

v = at + v₀

where v is the final velocity,

v₀ is the initial velocity,

a is the acceleration,

and t is time.

Given v = 450 m/s, v₀ = 0 m/s, and a = 60.5 m/s²:

450 = (60.5)t + 0

t ≈ 7.44

It takes approximately 7.44 seconds.

Answer:

1300 m/s²

Explanation:

Average acceleration is the change in velocity over change in time.

a = Δv / Δt

a = (76 m/s − 0 m/s) / 0.060 s

a = 1266.67 m/s²

Rounded to two significant figures, a ≈ 1300 m/s².

Answer:

- A conservative force permits a two-way conversion between kinetic and potential energies.  TRUE

- The work done by a conservative force depends on the path taken.  FALSE

- A potential energy function can be specified for a nonconservative force.  

FALSE

- A potential energy function can be specified for a conservative force.  TRUE

- The work done by a nonconservative force depends on the path taken.  TRUE

- A nonconservative force permits a two-way conversion between kinetic and potential energies.  TRUE

- A conservative force permits a two-way conversion between kinetic and potential energies. FALSE

- A potential energy function can be specified for a conservative force.  TRUE

- The work done by a nonconservative force depends on the path taken. TRUE

Explanation:

A conservative force permits a two-way conversion between kinetic and potential energies.  TRUE

The action of conservative force on a system can produce energy potential and kinetic. Example of this: the gravitational force. This claim that the work by extenal forces ( conservatives and non conservatives) is equal to the variation of kinetic energy of the system so the work made by  conservative forces can modify both potential and kinetic energy.

- The work done by a conservative force depends on the path taken.  FALSE

This kind of force can be obtained from potential function so the work made by this kind of force depend only to initial and final point of teh path made.

 - A potential energy function can be specified for a nonconservative force.  

FALSE

Considering that the work made by kind of force depend of the taken path they kind of forces can not be determined by a potential fuction.

- A potential energy function can be specified for a conservative force.  TRUE

This is as consequence of the definition of conservative force that it can be determined from a potential function.

- The work done by a nonconservative force depends on the path taken.  TRUE

This kind of force can not be obtained from potential function so the work made by this kind of force depend of the path taken to do this.

- A nonconservative force permits a two-way conversion between kinetic and potential energies.  TRUE

A nonconservative force permits conversion to kinetic energy plus potential energy during it made work over the system.  This statement is supported by taking into account the energy conservation for system, this claim that the work by extenal forces ( conservatives and non conservatives) is equal to the variation of kinetic energy of the system so the work made by non conservative forces can modify both potential and kinetic energy.

- A conservative force permits a two-way conversion between kinetic and potential energies. FALSE

As the conservative force is determined  from a potential function it can only modify the potential energy of the system.

- A potential energy function can be specified for a conservative force.  TRUE

This is as consequence of the definition of conservative force that it can be determined from a potential function.

- The work done by a nonconservative force depends on the path taken. TRUE

This kind of force can not be obtained from a potential function so the work made by this kind of force depend of the path taken.

A conservative force permits a two-way conversion between kinetic and potential energies. The work done by a conservative force depends on the path taken. A potential energy function can be specified for a conservative force.

A conservative force permits a two-way conversion between kinetic and potential energies. A potential energy function can be specified for a conservative force. The work done by a conservative force depends on the path taken.

A nonconservative force does not permit a two-way conversion between kinetic and potential energies. The work done by a nonconservative force depends on the path taken. A potential energy function cannot be specified for a nonconservative force.

Answer:

V. It is much less than 200 ml.

Explanation:

The final volume of the sugar-water mixture is gonna be something very close to the volume of water itself. The reason to this contraintuitive answer is that sugar molecules can dissolve and "find spaces inside the water molecular structure". In other words in a sugar beaker or cup the volume is mostly free air, because its a crystalline net structure.

The big change is going to be in the density of the solution and the mass is going always to be preserved. So there will be

180 g from sugar + 100g from water = 280 g of total volume

Density=mass/volume

Density of water=100/100=1

Density of sugar=180/100=1.8

Density of solution aprox=280/100=2.8

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining [tex] \rho_A: [/tex] Ethyl alcohol density and [tex] \rho_G: [/tex] Glycerin density , we can write:

[tex] \rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3 [/tex]

Simplifying:

[tex] \rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1) [/tex]

On the other hand:

[tex] h_1 + h_2 = \Delta h + h_3 [/tex]

Rearranging:

[tex] h_1 - \Delta h = h_3 - h_2 (2) [/tex]

Replacing  (2) in (1):

[tex] \rho_A\times h_1 = \rho_G \times (h_1 - \Delta h) [/tex]

Rearranging:

[tex] \frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h [/tex]

Data:

[tex] h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3} [/tex]

[tex] \frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}}  = \Delta h [/tex]

[tex] 7.5 cm =  \Delta h [/tex]

To find the difference in height between glycerin and alcohol in a U-tube, apply the principle of communicating vessels using their respective densities. Due to glycerin's higher density, its height in equilibrium will be lower than the alcohol's for equal pressures at the base of each arm of the tube.

The difference in height between the top surface of the glycerin and the top surface of the alcohol can be determined by using the principle of communicating vessels and the densities of the two liquids. Since the two liquids do not mix, they will each exert a pressure at the bottom of their respective sides of the U-tube based on their densities and heights.

The density of glycerin is 1.26 g/cm3, and the density of ethyl alcohol is 0.789 g/cm3. Since the pressures at the bottom of the tubes must be equal and the pressure exerted by a column of liquid is given by the product of its density, gravitational acceleration, and height, we can set up the following equation:

density of glycerin × height of glycerin = density of alcohol × height of alcohol.

Using this relationship, we can calculate the heights. However, since the question states that the ethyl alcohol column's height is already 20 cm, we only need to verify if the glycerin column's height will remain at 20 cm or change. As glycerin is denser than ethyl alcohol, for the pressures to be equal, the height of the glycerin column should be less than the height of the alcohol column. However, additional information like the specific heights after the alcohol is added would be necessary to provide a numerical answer.

Answer: B

Explanation:

Newton's first law it's law of inertia.

An object at rest will remain at rest (or an object in motion in a straight line at a constant velocity will remain that way) unless it is acted by an unbalanced force.

In A for the ball to slow down and stop, an external force (like friction with air or the floor) needs to be taken in consideration.

In B we can see how we need to make a force on the wagon to make it move.

In C we have an other Newton 's law, force equals mass times acceleration (2nd law)

In D we can see how it does not move because the forces on the box are balanced.

Answer: a. A rolling ball eventually slows down and comes to a stop.

Answer:

The rate at which bus 1 is going is 55 mph

The rate at which bus 1 is going is 35 mph

Explanation:

As per the question:

Suppose, the distance traveled by Bus 1 be 'd' at the rate R after a time, t = 3h

Thus  

Suppose, the distance traveled by Bus 1 be 'd'' at the rate, R'20 mph slower than the rate of Bus 1 after the same time.

R' = R - 20

The distance is given as the product of rate and time:

d = Rt         (1)

Now, the total distance given is 270 miles:

d + d' = 270

Now, using eqn (1):

Rt + R't = 270

3(R + R - 20) = 270

6R = 270 + 60

R = 55 mph

R' = R - 20 = 55 - 20 = 35 mph

Answer:

speed of the two vehicle are 55 mph and 35 mph

Explanation:

given,

speed of friends vehicle = x mph

speed of your vehicle = (x - 20) mph

when both travel in opposite direction

distance between the two buses = 270 miles

distance = speed × time

270 = 3(x) + 3(x-20)                    

90 = 2 x -20                        

x = 55 mph                    

now, speed of other vehicle is (55-20) = 35 mph

hence, speed of the two vehicle are 55 mph and 35 mph

Answer:

Mass, m = 0.00284 kg

Explanation:

Given that,

Density of mercury, [tex]d=13.5939\ g/cm^3[/tex]

Height of mercury column, h = 1.1 m

Diameter of mercury, d = 0.492 meters

Radius of mercury column, r = 0.246 m

We need to find the mass of a drum. The density is given by :

[tex]d=\dfrac{m}{V}[/tex]

V is the volume of mercury column

[tex]d=\dfrac{m}{\pi r^2h}[/tex]

[tex]m=d\times \pi r^2h[/tex]

[tex]m=13.5939\times 3.14\times (0.246)^2\times 1.1[/tex]

m = 2.84 grams

or

m = 0.00284 kg

So, the mass of a drum full of mercury is 0.00284 kg. Hence, this is the required solution.