-Our balance is maintained, at least in part, by the endolymph fluid in the inner ear. Spinning displaces this fluid, causing dizziness. Suppose a dancer (or skater) is spinning at a very fast 2.6 revolutions per second about a vertical axis through the center of his head. Although the distance varies from person to person, the inner ear is approximately 7.0 cm from the axis of spin. part A What is the radial acceleration (in m/s^2 ) of the endolymph fluid? part B What is the radial acceleration (in g's) of the endolymph fluid? 2-A model of a helicopter rotor has four blades, each of length 4.00m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 540rev/min . A-What is the linear speed of the blade tip? B-What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity, g?

Answer:

4.933m/s

Explanation:

the wagon has a weight of 35.1kg*9.81m/s2 = 343.98N

of that weight 343.98N*sin(18.3)=108N are parallel to the hill and oposit tothe tension of the rope.

then, the force that is moving the wagon is 125N-108N=17N

F=m*a then 17N=35.1kg*a

a=0.4843 m/s2

we have two equations

[tex]v=a.t\\x=v.t+\frac{1}{2} . a . t^{2}[/tex]

then

[tex]x=a.t^{2} +\frac{1}{2}. a t^{2} \\x=\frac{3}{2}. a t^{2}[/tex]

[tex]t=\sqrt{\frac{2x}{3a} }[/tex]

t=10.188s

[tex]v=a.t[/tex]

v=4.933 m/s

Answer:

[tex]v_o = 2761 m/s[/tex]

Explanation:

Since there is no external Force or Torque on the system of bullet and rod

So we can say that total angular momentum of the system will remain conserved about its center

So we will have

[tex]mv_o\frac{L}{2}sin\theta = (I_{rod} + I_{bullet})\omega[/tex]

here we know that

[tex]I_{rod} = \frac{mL^2}{12}[/tex]

[tex]I_{rod} = \frac{4.10\times 0.70^2}{12} [/tex]

[tex]I_{rod} = 0.167 kg m^2[/tex]

[tex]I_{bullet} = mr^2[/tex]

[tex]I_{bullet} = (0.003)(0.35^2)[/tex]

[tex]I_{bullet} = 3.675 \times 10^{-4} kg m^2[/tex]

[tex]\omega = 15 rad/s[/tex]

[tex]\theta = 60 [/tex]

now we have

[tex]0.003(v_o)(0.35)sin60 = (0.167 + 3.675 \times 10^{-4})15[/tex]

[tex]v_o = 2761 m/s[/tex]

This physics question is about calculating the speed of a bullet before it strikes a rotating rod, based on the conservation of angular momentum. After setting up the equations for initial and final angular momentum, by equating these two, we can find the speed of the bullet just before it hits the rod.

This is a physics problem regarding the conservation of angular momentum during a collision. We are given a scenario where a bullet impacting and lodging into a rotating rod which is initially at rest. According to the conservation of angular momentum, the initial angular momentum prior to the collision should equal the angular momentum after the collision, if no external torque is acting on the system.

The initial angular momentum (just before the collision) is the product of the bullet's mass, speed, and distance from the axis of rotation (which is half of the rod's length), and the cos(θ), where θ is the angle the bullet's path makes with the rod. In this case, angular momentum is mbrvb cos(θ)r

The final angular momentum (just after the collision) is the moment of inertia of the system (bullet plus rod) times the ensuing angular velocity, which is (mb(r^2) + (1/12)M(r^2))ω

By setting the initial and the final angular momentum equations equal to each other and arranging for vb, we will find the speed of the bullet just before impact.

https://brainly.com/question/1597483

#SPJ11

Answer:

0 km/h vertically, horizontally yet 350 km/h

Explanation:

The bundle was inside the plane, so it start a free fall movement from rest, i.e., 0km/h, but of course the speed increases with gravity, at 9,8m/s^2 over time, but since here it is asked the vertical speed at the beginning, it is 0 km/h.

Answer:

[tex]\theta=41.52^{\circ}[/tex]

Explanation:

Given that,

Velocity of the  airplane, v = 240 m/s

Angle with horizontal, [tex]\theta=30^{\circ}[/tex]

The altitude of the plane is 2.4 km, d = 2400 m

Vertical speed of the airplane, [tex]v_y=v\ sin\theta=240\ sin(30)=120\ m/s[/tex]

Horizontal speed of the airplane, [tex]v_x=v\ cos\theta=240\ sin(30)=207.84\ m/s[/tex]

So, the equation of the projectile for the flare is given by :

[tex]4.9t^2+120t-2400=0[/tex]

On solving the above equation, we get the value of t as:

t = 13.04 seconds

Horizontal distance travelled,

[tex]d=v_x\times t[/tex]

[tex]d=207.84\times 13.04[/tex]

d = 2710.23 m

Let [tex]\theta[/tex] is the angle with which it hits the target. So,

[tex]tan\theta=\dfrac{2400}{2710.23}[/tex]

[tex]\theta=41.52^{\circ}[/tex]

Hence, this is the required solution.

To solve this problem, you would apply principles of physics like projectile motion and trigonometry. We calculate the horizontal and vertical velocities using the given initial velocity and angle. The final total velocity and angle can be found by using these calculations as the horizontal velocity does not change.

The subject requires the application of the concepts of physics, specifically kinematics and trigonometry. Understanding the question in context, we are given that the airplane is flying at a velocity of 240m/s at an angle of 30.0° with the horizontal. A flare is released from the plane when it is at an altitude of 2.4km, and it hits a target on the ground. The problem needs us to find the angle θ.

Considering the fact that the time for projectile motion is completely determined by vertical motion, we set up the problem in the following way: We break down the initial velocity into components using the initial angle. The horizontal velocity (Vx) can be calculated using Vx = V*cos(θ), and the vertical velocity (Vy) can be calculated using Vy = V*sin(θ), where V is the initial velocity and θ is the initial angle.

Since the horizontal motion is constant and the initial position is known, we can use these two vertical and horizontal velocities to find the total velocity and the angle it makes with the horizontal. The trick here is to remember that since the x component (horizontal velocity) doesn't change, we can determine the final total velocity and its angle using these components.

https://brainly.com/question/29545516

#SPJ11

Answer: Ok, the acceleration decreases means if you have an object with acceleration equal 100 meters for second square, and it starts to decrease slowly to 90, 80... etc, there you still have positive acceleration, so your object's velocity keeps increasing, but more slower than before.

remmember, acceleration means the change in velocity, if you have positive acceleration, your velocity will increase.

the correct answer is D, the air resistance will fight against the gravity, but the object will keep accelerating (with less intensity) ence will go faster and faster.

An object's velocity can increase even as its acceleration decreases, such as in the case of an object falling with air resistance where the rate of speed increase slows down due to the opposing force of air resistance.

It is indeed possible for an object's velocity to increase while its acceleration decreases. This can occur, for example, during a scenario where an object continues to speed up, but the rate at which it is speeding up is decreasing over time. A typical example of this is an object in free fall with air resistance. Initially, as the object begins to fall, it accelerates due to gravity. However, as the object's speed increases, air resistance begins to counteract some of the acceleration due to gravity, causing the acceleration to decrease. Despite the reduction in acceleration, the object's velocity can continue to increase up until it reaches terminal velocity, at which point the acceleration will become zero, but the velocity remains constant and positive.

It is also important to note that acceleration is not always in the direction of motion. When an object is moving and its acceleration is in the direction of motion, it speeds up, while if the acceleration is in the opposite direction of motion, it will slow down or decelerate.

Answer:

The natural, fundamental frequency of this string is 24 hertz.          

Explanation:

Length of the string, l = 2 m

Number of segments, n = 5

Frequency, f = 120 Hz

Let f' is the natural fundamental frequency of this string. The frequency for both side ended string is given by :

[tex]f=\dfrac{nv}{2l}[/tex]

[tex]v=\dfrac{2fl}{n}[/tex]

[tex]v=\dfrac{2\times 120\times 2}{5}[/tex]        

v = 96 m/s

For fundamental frequency, n = 1

[tex]f'=\dfrac{v}{2l}[/tex]

[tex]f'=\dfrac{96}{2\times 2}[/tex]

f' = 24 Hertz

So, the natural, fundamental frequency of this string is 24 hertz. Hence, this is the required solution.

The natural, fundamental frequency of this standing wave is equal to 24 Hertz.

Given the following data:

Length of the string = 2.0 meters.

Number of segments = 5.

Frequency = 120 Hertz.

First of all, we would determine the velocity of the standing wave by using this formula:

[tex]F=\frac{nV}{2L} \\\\V=\frac{2FL}{n} \\\\V=\frac{2 \times 120 \times 2.0}{5}\\\\V=\frac{480}{5}[/tex]

V = 96 m/s.

Now, we can calculate the natural, fundamental frequency by using this formula:

[tex]F'=\frac{V}{2L} \\\\F'=\frac{96}{2 \times 2.0} \\\\F'=\frac{96}{4.0}[/tex]

F' = 24 Hertz.

Read more on frequency here: brainly.com/question/3841958

Answer: 23000 frames/s

Explanation:

The rest of the statement of the question is presented below:

The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this?

We know the maximum initial speed at which the seeds are dispersed is:

[tex]V_{i}=4,7 m/s[/tex]

In addition, we know the maximum distance at which the seeds move between photographic frames is:

[tex]d_{max}=0.20 mm \frac{1m}{1000 m}=0.0002 m[/tex]

And we need to find the minimum frame rate of the camera with these given conditions. This can be found by finding the time [tex]t[/tex] for each frame and then the frame rate:

Finding the time:

[tex]t=\frac{d_{max}}{V_{i}}[/tex]

[tex]t=\frac{0.0002 m}{4.6 m/s}[/tex]

[tex]t=0.00004347 s/frame[/tex] This  is the time for each frame

Now we need to find the frame rate, which is the frequency at which the photos are taken.

In this sense, frequency [tex]f[/tex] is defined as:

[tex]f=\frac{1}{t}[/tex]

[tex]f=\frac{1}{0.00004347 s/frame}[/tex]

Finally:

[tex]f=23000 frames/s[/tex]

Hence, the minimum frame rate is 23000 frames per second.

Answer:

True

Explanation:

Distance is defined as the length of the actual path traveled by the body.

Displacement is defined as the minimum distance between the two points.

the magnitude of displacement is always less than or equal to the distance traveled by the body.

As a deer runs from A to b , so it means the distance traveled by the deer is either equal to the magnitude of displacement or always greater than the magnitude of displacement of the deer.

Displacement can never be greater than the distance.

Thus, the option is true.

Answer:

The required radius of its motion is [tex]0.084m[/tex].

Explanation:

The formula for calculating the required  radius of its motion is given by

[tex]F = (mv^2)/r[/tex]

Where m= mass  

V= moving velocity

F=frictional force

r = radius of its motion

Then the required radius of its motion is given by

[tex]r =(mv^2)/F[/tex]

Given that

mas =0.0818 kg

Frictional force= 0.108 N

Moving with Velocity of  = 0.333 m/s

radius of its motion = [tex]\frac{[0.0818 kg \times (0.333 m/s)^2]}{0.108 N}[/tex]

Hence the required radius of its motion is r = [tex]8.4 cm=0.084m[/tex]

Answer:

0.084

Explanation:

Acellus

Answer:

New force, F' = F

Explanation:

Given that, two small balls, A and B, attract each other gravitationally with a force of magnitude F. It is given by :

[tex]F=G\dfrac{m_Am_B}{r^2}[/tex]

If we now double both masses and the separation of the balls, the new force is given by :

[tex]F'=G\dfrac{2m_A\times 2m_B}{(2r)^2}[/tex]

F' = F

So, the new force remains the same as previous one. Hence, the correct option is (d) "F"

Answer:

After studying the law of gravitational attraction, students constructed a model to illustrate the relationship between gravitational attraction (F) and distance. If the distance between two objects of equal mass is increased by 2, then the gravitational attraction (F) is 1/4F or F/4. How would this model, situation A, change if the mass of the spheres is doubled?

A)  A

B)  B

C)  C

D)  D

If you came here from usa test prep it is:

Actually A

But for the question given right now is D.

Explanation:

Answer:

A. Scatterplot

Explanation:

because it is

The car will pass the truck after 20 seconds.

To find when the car will pass the truck, we need to determine the time it takes for the car to reach a speed of 25 m/s and the distance traveled by the truck in that time.

The car is accelerating at a rate of 1.25 m/s². Using the equation v = u + at, where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration, and t is the time, we can solve for t: 25 m/s = 0 m/s + 1.25 m/s² * t.

Simplifying the equation, we get t = 20 seconds. Therefore, the car will pass the truck after 20 seconds.

https://brainly.com/question/32366903

#SPJ12

Answer:

C) about 1/10 as great.

Explanation:

We use the relation between Impulse, I, and momentum, p:

[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ \\[/tex]

[tex]F=m(v_{f}-v{o})/t=-mv{o}/t\\ \\[/tex]     the final speed is zero

We can see that the average Force is inversely proportional to the time, so if the time is 10 times bigger, the average Force is 1/10 as great

Average force body experience is about 1/10 as great as  when the legs are kept stiff

from Newtons second law, force. f is the product of mass, m and acceleration, a

f = m * a

a = ( v - u ) / t

f = m * ( v- u ) / t

so force is inverrsely related to time, making the time of the impact 10 times as great as stiff-legged landing impiles mathematically as follows:

for stiff-legged

f1 = m * ( v - u ) / t

for bent knees

f2 = m * ( v - u ) / 10t

f2 = f1 / 10

Read more on Impact force here: https://brainly.com/question/1871060

Answer:

customer loyalty

Explanation:

According to my research on five forces model, I can say that based on the information provided within the question the two strength factors that relate to all three of the competitive forces are switching costs and​ customer loyalty. Customer loyalty is when a customers choose and become loyal to a certain business over their competitors because their organizational culture and customer service.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Answer:

[tex]2LiBr+Pb(NO_3) = PbBr_2+ 2LiNO_3[/tex]  is a double displacement reaction

Explanation:

[tex]2LiBr+PbNO_3 = PbBr_2+2LiNO_3[/tex]

                                                       It is of the form [tex]AX+BY = AY+BX[/tex]

                                                       [tex]Fe+2HCl = FeCl_2+H_2[/tex]

Both oxidation and reduction occur in this reaction. Fe gets oxidized and H gets reduced.

                                          [tex]Fe = Fe^(2+)+2e^-[/tex]

                                          [tex]2H^++2e^- = H_2[/tex]

                                          [tex]CaO+H_2 O = Ca(OH)_2[/tex]

It is an exothermic combination reaction. Calcium reacts with water to produce calcium hydroxide and heat is released in this process.

It is a decomposition reaction where nickel chloride decomposes to nickel and chlorine.

Answer:

18. Double Displacement Reaction

19. Single Displacement Reaction

20. Synthesis Reaction

21. Decomposition Reaction

Explanation:

Double Displacement formula:

AB + CD --> AD + CB

Single Displacement formula:

A + BC --> AC + B

Synthesis formula:

A + B --> AB

Decomposition formula:

AB --> A + B

Answer:

a) [tex]V=14.01 m/s[/tex]

b) [tex]V=8.86 \, m/s[/tex]

c)[tex]t = 2.33s[/tex]

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

[tex]E_m = E_k +E_p[/tex]

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this [tex]E_m[/tex] mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

[tex]E_k = \frac{1}{2} m \, V^2[/tex]

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

[tex]E_p= m\, g\, h[/tex]

where g is the acceleration due to gravity ([tex]g=9.81 \, m/s^2[/tex]) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which [tex]h=0[/tex], a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

[tex]E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m[/tex]

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set [tex]h=0[/tex]

[tex]E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2[/tex]

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

[tex]E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\[/tex]

And so we have found the velocity of the ball as it hits the floor.

[tex]V = \sqrt[]{2g\cdot 10m}=14.01\, m/s[/tex]

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

[tex]V = \sqrt[]{2g\cdot h}[/tex]

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

[tex]V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s[/tex]

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

[tex]V_{avg}\cdot t=h\\t=h/V_{avg}[/tex]

what's the average speed when the ball is descending?

[tex]V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s[/tex]

so the time it takes the ball to go down is:

[tex]t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\[/tex]

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

[tex]V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s[/tex]

and the time it takes to go up is:[tex]t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s[/tex]

When we add both times , we get:

[tex]t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s[/tex]

To find the speed of the tennis ball just before it hits the floor on its way down, we can use the equation y = 0 + voyt - 1/2gt^2. The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec and end at v = -0.98 m/s at t = 0.65 sec. The time the ball is in the air from the time it is dropped until it reaches its maximum height on the first rebound is 2.5 s.

(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation 4.22: y = yo + voyt - 1/2gt^2. If we take the initial position yo to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

(b) The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec (there should be a data point at t = 0.05 sec, v = 4.90 m/s). It should then have a slope of -9.8 m/s², crossing through v = 0 at t = 0.55 sec and ending at v = -0.98 m/s at t = 0.65 sec.

(c) This gives t = 2.5 s. Since the ball rises for 2.5 s, the time to fall is 2.5 s.

https://brainly.com/question/33880642

#SPJ3

Answer:

2.76m/s

Explanation:

The position x for a given time t and constant acceleration a is:

(1) [tex]x=\frac{1}{2}at^2[/tex]

The velocity v:

(2) [tex]v=at[/tex]

Solving equation 2 for time t:

(3) [tex]t=\frac{v}{a}[/tex]

Combining equations 1 and 3:

(4) [tex]x=\frac{v^2}{2a}[/tex]

Solving equation 4 for velocity v:

(5) [tex]v=\sqrt{2ax}[/tex]

For a = 0.465m/s² and x = 8.2m:

v = 2.76m/s

Answer:

No the distance traveled in last 0.1 s is not same as that the distance traveled in first 0.1 s

so it will cover more distance in last 0.1 s then the distance in first 0.1 s

Explanation:

As we know that when stone is dropped from the diving board then its velocity at the time of drop is taken to be ZERO

so here we can say that its displacement from the top position in next 0.1 s is given as

[tex]d_1 = v_y t + \frac{1}{2}at^2[/tex]

[tex]d_1 = 0 + \frac{1}{2}(9.81)(0.1)^2[/tex]

[tex]d_1 = 0.05 m[/tex]

Now during last 0.1 s of its motion the stone will attain certain speed

so we will have

[tex]d_2 = v_y(0.1) + \frac{1}{2}(9.81)(0.1)^2[/tex]

[tex]d_2 = 0.1 v_y + 0.05 m[/tex]

so it will cover more distance in last 0.1 s then the distance in first 0.1 s

Final answer:

The distance a rock travels during a 0.1-second interval while falling will vary because of acceleration due to gravity. Early in its fall, it will travel less distance, and just before impact, it will cover more distance due to increased speed.

Explanation:

The distance traveled by a rock in free fall will not be the same in two intervals of time if these intervals occur at different stages of the fall, because the rock is accelerating due to gravity.

Near the top of its flight, it will have just started to accelerate, so it will cover a smaller distance in the first 0.1-second interval. However, just before the rock hits the water, it will have been accelerating for the entire duration of the fall, meaning it will be traveling much faster and will cover a greater distance in the last 0.1-second interval before impact.

Answer:

the potential energy increases and the electric potential increases

Explanation:

we know here that Voltage i.e. (Electric potential) increases from the initial point  to the final point

consider if 2 object of different charge 1 is twice the charge of other moveing in same distance in electric filed

than object of twice the charge require twice the force so twice amount of work

This work change the potential energy by  equal amount of work done

so electric potential energy is depend on the amount of charge on object

so if work being done on a positive test charge by an external force in moving the charge from one location to another

the potential energy increases and the electric potential increases

When a positive test charge is moved from one location to another, the potential energy increases and the electric potential energy increases.

The work done in moving a unit positive test charge from infinity to a certain point in the electric field is known as potential energy.

V = E x d

V = (F/q) x d

where;

Thus, when a positive test charge is moved from one location to another, the potential energy increases and the electric potential energy increases.

Learn more about potential energy here: https://brainly.com/question/1242059

Explanation:

Given that,

Mass of the object, m₁ = 23 kg

Acceleration of this object, a₁ = 57 m/s/s

Mass of another object, m₂ = 23 kg

We need to find the acceleration of another object. It can be calculated using second law of motion as :

[tex]F=ma[/tex]

[tex]a=\dfrac{F}{m}[/tex]

Here, F is same. So,

[tex]\dfrac{a_1}{a_2}=\dfrac{m_2}{m_1}[/tex]

[tex]a_2=\dfrac{a_1m_1}{m_2}[/tex]

[tex]a_2=\dfrac{57\times 23}{23}[/tex]

[tex]a_2=57\ m/s/s[/tex]

So, another object will create same acceleration as 57 m/s/s. Hence, this is the required solution.

Final answer:

The same force that accelerates an object of 23 Kg at 57 m/s² would also accelerate another object of mass 23 Kg at 57 m/s², as per Newton's second law of motion.

Explanation:

If a certain force accelerates an object of mass 23 Kg at 57 m/s², the same force would produce the same acceleration on another object of mass 23 Kg. This is because acceleration is directly proportional to force and inversely proportional to mass, according to Newton's second law of motion, which can be expressed as F = ma. Since both the force and mass are the same in this scenario, the acceleration would also be the same, which is 57 m/s².

Answer:

These are the answers for 1, 2 and 3

Explanation:

Sorry I couldn't help you with 4 and 5

1.

Answer:

y = 11.48 m

x = 13.0 m

Explanation:

Components of initial velocity is given as

[tex]v_x = 7.2 cos25 = 6.52 m/s[/tex]

[tex]v_y = 7.2 sin25 = 3.04 m/s[/tex]

Now after t = 2 s the vertical position is given as

[tex]y = y_0 + v_y t + \frac{1}{2}a_y t^2[/tex]

[tex]y = 25 + 3.04(2) - \frac{1}{2}(9.8)(2^2)[/tex]

[tex]y = 11.48 m[/tex]

Now horizontal position is given as

[tex]x = v_x t[/tex]

[tex]x = 6.52 \times 2[/tex]

[tex]x = 13.04 m[/tex]

2.

Answer:

d = 26.6 m

Explanation:

Initial position on y axis is given as

[tex]y = 1.5 m[/tex]

velocity of ball in y direction

[tex]v_y = 0[/tex]

now we have

[tex]\Delta y = v_y t + \frac{1}{2}gt^2[/tex]

[tex]1.5 = \frac{1}{2}(9.8) t^2[/tex]

[tex]t = 0.55 s[/tex]

now the distance moved by the ball in horizontal direction is given as

[tex]d = v_x t[/tex]

[tex]d = 48.1 \times 0.55[/tex]

[tex]d = 26.6 m[/tex]

3

Answer:

[tex]A + B = 12.42\hat i + 0.35 \hat j[/tex]

Explanation:

Here we can see the two vectors inclined at different angles

so two components of vector A is given as

[tex]A = 11.3 cos21\hat i - 11.3 sin21\hat j[/tex]

[tex]A = 10.55 \hat i - 4.05 \hat j[/tex]

Similarly for other vector B we have

[tex]B = 4.78cos67 \hat i + 4.78 sin67\hat j[/tex]

[tex]B = 1.87\hat i + 4.4 \hat j[/tex]

now we need to find A + B

so we have

[tex]A + B = (10.55\hat i - 4.05\hat j) + (1.87\hat i + 4.4 \hat j)[/tex]

[tex]A + B = 12.42\hat i + 0.35 \hat j[/tex]

4.

Answer:

d = 81.86 m

Explanation:

Displacement of airplane is given as

[tex]d_1[/tex] = 72 km in direction 30 degree East of North

[tex]d_1 = 72sin30\hat i + 72cos30\hat j[/tex]

[tex]d_1 = 36\hat i + 62.35\hat j[/tex]

[tex]d_2[/tex] = 48 km South

[tex]d_2 = -48\hat j[/tex]

[tex]d_3[/tex] = 100 km in direction 30 degree North of West

[tex]d_3 = -100 cos30\hat i + 100 sin30\hat j[/tex]

[tex]d_3 = -86.6\hat i + 50\hat j[/tex]

so net displacement is given as

[tex]d = d_1 + d_2 + d_3[/tex]

[tex]d = 36\hat i + 62.35\hat j - 48\hat j - 86.6\hat i + 50 \hat j[/tex]

[tex]d = -50.6\hat i + 64.35\hat j[/tex]

now magnitude of displacement is given as

[tex]d = \sqrt{50.6^2 + 64.35^2}[/tex]

[tex]d = 81.86 m[/tex]

5.

Answer:

1) final speed at which it will hit the ground again

2) acceleration during the motion of the ball

Explanation:

As we know that the speed at which the ball is thrown is always same to the speed by which it will hit back on the ground

so we know that final speed will be same as initial speed

Also we know that during the motion the acceleration of ball is due to gravity so it will be

[tex]a = - g[/tex]

Answer:

An annular Solar Eclipse

Explanation:

Solar eclipse is an event that occurs naturally on Earth when the moon in its orbit is positioned between the Earth and the Sun.Solar Eclipse can be total ,partial or annular.In the total solar eclipse, the moon completely covers the sun where as in the annular solar eclipse the moon covers the center of the Sun leaving outer edges of the Sun to be visible forming the ring of fire.In partial solar eclipse the Earth moves through the lunar penumbra as the moon moves between Earth and Sun.The moon blocks only some parts of the solar disk.Annular solar eclipse happens during new moon and the moon is at its farthest position from the Earth called Apogee.

A solar eclipse occurs when the moon blocks the sun. It can either be a total eclipse, where the sun is completely covered, or a partial or annular eclipse when the new moon is too far to entirely cover the sun.

A solar eclipse occurs when the moon moves between the earth and the sun, blocking out sunlight and casting a shadow. This can either result in a total eclipse, where the full face of the sun is covered, or a partial eclipse when part of the sun is still visible. A total eclipse only happens when the moon is close enough to the earth to totally cover the sun. When the new moon is too far from the earth to completely cover the sun, the eclipse can be either partial or become what's known as an 'annular' eclipse. In an annular solar eclipse, the moon is located too far from the Earth to completely cover the sun's disk, resulting in a ringlike appearance.

https://brainly.com/question/4702388

#SPJ3

Answer: Technician B

Explanation: A taillights, and several kind of ilumation devices, works by converting a current flow trought a material in Light. This mean that the Current Flow (Amperes) which goes through a lights is proportional to the light it gives (Lumens).

This means, that if the tailights has a dimly light the current that goes through is less than the one it is suppose to be.

A fuse is a protection device for over-current, if the current at any given time goes beyond a limit (designed on the fuse) the fuse will melt and cutting all the current to the circuit. A blown fuse will cut all the current from the circuit and the tailights will be completely off.

However a open ground circuit is a differente kind of failure. In this cases, there is a pact from where the current on the circuit "escapes" from it. This could be by several reason, unprotected wires the most usual. While the current escapes from his intended course, not all of the energy goes away from the load. This explain why the tailight still has enough energy through it to light dimly.

The estimated diameter of the Moon is approximately [tex]3.536 \times 10^4[/tex] meters, (two significant figures as [tex]3.5 \times 10^4[/tex] meters).

Here, we have to set up a proportion to solve for the diameter of the Moon using the given information:

The pencil's diameter = 0.7 cm

Distance from the eye to the pencil = 0.75 m

Earth-Moon distance = [tex]3.8 \times 10^5[/tex] km = [tex]3.8 \times 10^8[/tex] m

Let, the diameter of the Moon as "D."

When the pencil blocks out the Moon, the ratio of the pencil's diameter to the distance from the eye to the pencil is equal to the ratio of the Moon's diameter to the Earth-Moon distance:

Diameter of Moon / Earth-Moon distance = Pencil's diameter / Distance to pencil

D / (  [tex]3.8 \times 10^8[/tex] m) = (0.7 cm) / (0.75 m)

Now, solve for D:

D = (0.7 cm) * ( [tex]3.8 \times 10^8[/tex] m) / (0.75 m)

D ≈ [tex]3.536 \times 10^6[/tex] cm

Convert the diameter of the Moon to meters:

D ≈ [tex]3.536 \times 10^4[/tex] m

So, the estimated diameter of the Moon is approximately [tex]3.536 \times 10^4[/tex] meters, which can be expressed using two significant figures as [tex]3.5 \times 10^4[/tex] meters.

Learn more about diameter :

brainly.com/question/33266593

#SPJ12

Answer:

(a) The x-component of velocity is 31.55 km/h

(b) The y-component of velocity is 44.92 km/hr

Solution:

As per the solution:

The relative position of ship A relative to ship B is 4.2 km north and 2.7 km east.

Velocity of ship A, [tex]\vec{u_{A}}[/tex] = 22 km/h towards South = [tex]- 22\hat{j}[/tex]

Velocity of ship B, [tex]\vec{u_{B}}[/tex] = 39 km/h Towards North east at an angle of [tex]36^{\circ}[/tex] = [tex]\vec{u_{B}} = 39sin36^{\circ} \hat{j}[/tex]

Now, the velocity of ship A relative to ship B:

[tex]\vec{u_{AB}} = \vec{u_{A}} - \vec{u_{B}}[/tex]

[tex]\vec{u_{A}} = - 22\hat{j}[/tex]

[tex]\vec{u_{B}} = 39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}[/tex]

Now,

[tex]\vec{u_{AB}} = - 22\hat{j} +39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}[/tex]

[tex]\vec{u_{AB}} = 31.55\hat{i} - 44.92\hat{j}[/tex]

Answer:

[tex]E_{th} = 19680 J[/tex]

Explanation:

GIVEN DATA:

Total mass ( rider  + his tube) = 80 kg

tension  Force = 360 N

height of slope  =  30 m

length of slope = 120 m

we know that thermal energy is given as

E_{th}  = W- Ug

W= F*d = (360N*120m)= 43200 J        

Ug= m*g*h = (80kg*9.8m/s2*30m) = 23520 J

[tex]E_{th} = 43200J - 23520 J[/tex]

[tex]E_{th} = 19680 J[/tex]

The amount of thermal energy that is created in the slope and the tube during the ascent is 19680 Joules.

Given the following data:

To find the amount of thermal energy that is created in the slope and the tube during the ascent:

By applying the Law of Conservation of Energy:

[tex]Work\;done = K.E + P.E + T.E[/tex]

Where:

Since the rider and his tube are pulled at a constant speed, K.E is equal to zero (0).

Therefore, the formula now becomes:

[tex]Work\;done = 0 + P.E + T.E\\\\T.E = Work\;done-P.E[/tex]

For the work done:

[tex]Work\;done = Tensional\;force \times displacement\\\\Work\;done = 360 \times 120[/tex]

Work done = 43,200 Joules.

For P.E:

[tex]P.E = mgh\\\\P.E = 80\times9.8\times30[/tex]

P.E = 23,520 Joules.

Now, we can find the amount of thermal energy:

[tex]T.E = Work\;done-P.E\\\\T.E =43200-23520[/tex]

T.E = 19680 Joules.

Read more: https://brainly.com/question/22599382

Answer:

[tex]v = 33.66 m/s[/tex]

Explanation:

Let hockey puck is moving at constant speed v

so here we have

[tex]d = vt[/tex]

so time taken by the puck to strike the wall is given as

[tex]t = \frac{58.2}{v}[/tex]

now time taken by sound to come back at the position of shooter is given as

[tex]t_2 = \frac{58.2}{340}[/tex]

[tex]t_2 = 0.17s[/tex]

so we know that total time is 1.9 s

[tex]1.9 = t + t_2[/tex]

[tex]1.9 = t + 0.17[/tex]

[tex]1.9 - 0.17 = t[/tex]

[tex]t = 1.73 s[/tex]

now we have

[tex]1.73 = \frac{58.2}{v}[/tex]

[tex]v = 33.66 m/s[/tex]

The net electric force acting on particle 3 due to particle 1 and particle 2 is 4.55 N.

To find the net electric force acting on particle 3 due to particle 1 and particle 2, we can use Coulomb's Law, which states that the electric force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is: F = k × (|q1 × q3| / r^2)

Where:

Substituting the given values into the formula:

F = (9 × 10^9 Nm^2/C^2) × (|(-8.2 × 10^-9 C) × (8.0 × 10^-9 C)| / (3.3 × 10^-2 m)^2)

Simplifying the equation, we get:

F = 4.55 N

Therefore, the net electric force acting on particle 3 is 4.55 N.

The net electric force acting on particle 3 is approximately [tex]\( 1.62 \times 10^{-3} \, \text{N} \)[/tex] at an angle of [tex]\( 30^\circ \)[/tex] from the horizontal axis.

Given:

[tex]\( q_1 = -8.2 \, \text{nC} \)\\ \( q_2 = -16.4 \, \text{nC} \)\\ \( q_3 = 8.0 \, \text{nC} \)[/tex]

- Side length of the equilateral triangle, [tex]\( a = 3.3 \, \text{cm} = 0.033 \, \text{m} \)[/tex]

Calculate the Magnitude of the Forces

The magnitude of the force between two charges is given by Coulomb's law:

[tex]\[ F = k_e \frac{|q_1 q_2|}{r^2} \][/tex]

where:

- [tex]\( k_e \)[/tex] is Coulomb's constant, [tex]\( k_e = 8.99 \times 10^9 \, \text{N.m}^2/\text{C}^2 \)[/tex]

- r is the distance between the charges, [tex]\( r = 0.033 \, \text{m} \)[/tex]

Force between [tex]\( q_1 \)[/tex] and [tex]\( q_3 \)[/tex]:

[tex]\[ F_{13} = k_e \frac{|q_1 q_3|}{a^2} = 8.99 \times 10^9 \frac{|-8.2 \times 10^{-9} \times 8.0 \times 10^{-9}|}{(0.033)^2} \]\[ F_{13} = 8.99 \times 10^9 \frac{65.6 \times 10^{-18}}{0.001089} \]\[ F_{13} = 8.99 \times 10^9 \times 6.025 \times 10^{-14} \]\[ F_{13} = 5.41 \times 10^{-4} \, \text{N} \][/tex]

Force between [tex]\( q_2 \)[/tex] and [tex]\( q_3 \)[/tex]:

[tex]\[ F_{23} = k_e \frac{|q_2 q_3|}{a^2} = 8.99 \times 10^9 \frac{|-16.4 \times 10^{-9} \times 8.0 \times 10^{-9}|}{(0.033)^2} \]\[ F_{23} = 8.99 \times 10^9 \frac{131.2 \times 10^{-18}}{0.001089} \]\[ F_{23} = 8.99 \times 10^9 \times 1.205 \times 10^{-13} \]\[ F_{23} = 1.08 \times 10^{-3} \, \text{N} \][/tex]

Determine the Direction of the Forces

Since the triangle is equilateral, the angles between the forces are 60 degrees. The forces are directed along the lines connecting the charges.

- [tex]\( F_{13} \)[/tex] is directed from [tex]\( q_1 \)[/tex] to [tex]\( q_3 \)[/tex] (repulsive force, but [tex]\( q_1 \)[/tex] is negative, so the force direction is towards [tex]\( q_1 \))[/tex].

- [tex]\( F_{23} \)[/tex] is directed from [tex]\( q_2 \)[/tex] to [tex]\( q_3 \)[/tex] (repulsive force, but [tex]\( q_2 \)[/tex] is negative, so the force direction is towards [tex]\( q_2 \))[/tex].

Calculate the Net Force

We need to resolve the forces into components and sum them.

Resolving [tex]\( F_{13} \)[/tex]:

- Along x-axis: [tex]\( F_{13x} = F_{13} \cos(30^\circ) = 5.41 \times 10^{-4} \cos(30^\circ) \)[/tex]

- Along y-axis: [tex]\( F_{13y} = F_{13} \sin(30^\circ) = 5.41 \times 10^{-4} \sin(30^\circ) \)[/tex]

Resolving [tex]\( F_{23} \)[/tex]:

- Along x-axis: [tex]\( F_{23x} = F_{23} \cos(30^\circ) = 1.08 \times 10^{-3} \cos(30^\circ) \)[/tex]

- Along y-axis: [tex]\( F_{23y} = F_{23} \sin(30^\circ) = 1.08 \times 10^{-3} \sin(30^\circ) \)[/tex]

Since both forces have the same y-component direction:

[tex]\[ F_{13x} = 5.41 \times 10^{-4} \times \frac{\sqrt{3}}{2} \approx 4.68 \times 10^{-4} \, \text{N} \]\[ F_{13y} = 5.41 \times 10^{-4} \times \frac{1}{2} \approx 2.70 \times 10^{-4} \, \text{N} \]\[ F_{23x} = 1.08 \times 10^{-3} \times \frac{\sqrt{3}}{2} \approx 9.35 \times 10^{-4} \, \text{N} \]\[ F_{23y} = 1.08 \times 10^{-3} \times \frac{1}{2} \approx 5.40 \times 10^{-4} \, \text{N} \][/tex]

The net force components are:

[tex]\[ F_{net_x} = F_{13x} + F_{23x} = 4.68 \times 10^{-4} + 9.35 \times 10^{-4} \approx 1.40 \times 10^{-3} \, \text{N} \]\[ F_{net_y} = F_{13y} + F_{23y} = 2.70 \times 10^{-4} + 5.40 \times 10^{-4} \approx 8.10 \times 10^{-4} \, \text{N} \][/tex]

Calculate the Magnitude of the Net Force

[tex]\[ F_{net} = \sqrt{F_{net_x}^2 + F_{net_y}^2} = \sqrt{(1.40 \times 10^{-3})^2 + (8.10 \times 10^{-4})^2} \]\[ F_{net} = \sqrt{1.96 \times 10^{-6} + 6.56 \times 10^{-7}} \]\[ F_{net} = \sqrt{2.62 \times 10^{-6}} \]\[ F_{net} \approx 1.62 \times 10^{-3} \, \text{N} \][/tex]

Determine the Direction of the Net Force

The direction of the net force can be found using the angle:

[tex]\[ \theta = \tan^{-1} \left( \frac{F_{net_y}}{F_{net_x}} \right) = \tan^{-1} \left( \frac{8.10 \times 10^{-4}}{1.40 \times 10^{-3}} \right) \]\[ \theta = \tan^{-1} (0.579) \]\[ \theta \approx 30^\circ \][/tex]

Answer:

The acceleration from it's legs is [tex]a=138.20\frac{m}{s^{2} }[/tex]

Explanation:

Let's order the information:

Initial height: [tex]y_{i}=0m[/tex]

Final height: [tex]y_{f}=2.26m[/tex]

The bush accelerates from [tex]y_{i}=0m[/tex] to [tex]y_{e}=0.16m[/tex].

We can use the following Kinematic Equation to know the velocity at [tex]y_{e}[/tex]:

[tex]v_{f}^{2} = v_{e}^{2} - 2g(y_{f}-y_{e})=2ay_{e}[/tex]

where g is gravity's acceleration (9.8m/s). Since [tex]v_{f}=0[/tex],

[tex]2g(y_{f}-y_{e}) = v_{e}^{2}[/tex]

⇒ [tex]v_{e}=6.41\frac{m}{s}[/tex]

Working with the same equation but in the first height interval:

[tex]v_{e}^{2} = v_{i}^{2} + 2(a-g)(y_{e}-y_{i})[/tex]

Since [tex]v_{i}=0[/tex] and [tex]y_{i}=0[/tex],

[tex]v_{e}^{2} = 2(a-g)y_{e}[/tex]

⇒[tex]a-g=\frac{v_{e}^{2}}{2y_{e}}[/tex]

⇒[tex]a=\frac{v_{e}^{2}}{2y_{e}}+g[/tex] ⇒ [tex]a=138.20\frac{m}{s^{2} }[/tex]

The work done in stretching a spring in two stages can be estimated by dividing the stretching process into two stages, calculating the average force in each stage, and then calculating the work done in each stage. The total work done is the sum of the work done in each stage.

To calculate the work done in stretching the spring in two stages, we can assume an average force for each stage. The force exerted by the spring at any displacement from its equilibrium length is given by Hooke's Law, F = kx, where 'F' is the force, 'k' is the spring constant, and 'x' is the displacement from equilibrium.

For stretching the spring from 0.3 m to 0.5 m (stage one), the average force exerted would be F_avg1 = k(x2 + x1) / 2 = 22 * ( 0.5 + 0.3 ) / 2 = 8.8 N. The work done can be then calculated as Work1 = F_avg1 * (x2 - x1) = 8.8 * (0.5-0.3) = 1.76 J.

For stretching the spring from 0.5 m to 0.7 m (stage two), the average force would be F_avg2 = k(x2 + x1) / 2 = 22 * ( 0.7 + 0.5 ) / 2 = 13.2 N. The work done in this stage would be Work2 = F_avg2 * (x2 - x1) = 13.2 * (0.7 - 0.5) = 2.64 J.

Therefore, the total work done in stretching the spring in two stages would be Work_total = Work1 + Work2 = 1.76 J + 2.64 J = 4.4 J.

https://brainly.com/question/29392391

#SPJ3

To estimate the work done in stretching the spring, divide the process into two stages: 0.3 m to 0.5 m and 0.5 m to 0.7 m. The total work done is 4.4 J.

Stage 1: Stretching from 0.3 m to 0.5 m

The initial force (F₁) at 0.3 m is:

The final force (F_mid) at 0.5 m is:

The average force (F_avg₁) for this stage is:

The distance stretched (d1) is:

Thus, the work done (W₁) is:

Stage 2: Stretching from 0.5 m to 0.7 m

The initial force (F_mid) at 0.5 m is:

The final force (F₂) at 0.7 m is:

The average force (F_avg₂) for this stage is:

The distance stretched (d2) is:

Thus, the work done (W₂) is:

Total Work Done in stretching the spring from 0.3 m to 0.7 m is: