Col. John Stapp led the U.S. Air Force Aero Medical Laboratory's research into the effects of higher accelerations. On Stapp's final sled run, the sled reached a speed of 284.4 m/s (632 mi/h) and then stopped with the aid of water brakes in 1.4 s. Stapp was barely conscious and lost his vision for several days but recovered.

A)Determine his acceleration while stopping.
B)Determine the distance he traveled while stopping

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force  T  is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force  in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T =  mg -  mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60  { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

The branch must provide an upward force of 227 Newtons to support a gibbon weighing 8.5 kg, swinging with a speed of 3.2 m/s, and a radius of 0.6 m.

To calculate the upward force that the branch must provide, we need to take into account both the gravitational force acting on the gibbon and the centrifugal force due to its circular motion. The weight of the gibbon (gravitational force) is its mass times the acceleration due to gravity (8.5 kg * 9.8 m/s^2), which equals 83.3 N.

Centrifugal force is given by mass times velocity squared divided by radius of the motion (m*v^2/r), which results in (8.5 kg * (3.2 m/s)^2) / 0.6 m = 144 N.

By adding these two forces together, we find the total force the branch must supply at the lowest point: 83.3 N + 144 N = 227 N.The branch must provide an upward force of 227 Newtons to support the swinging gibbon.

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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.

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Answer:

c. compound light microscope

Explanation:

Microorganisms (such as bacteria, fungi and protozoa) can be observed through a compound light microscope, which allows to increase the observed image since 400 times to 1000 times (in more advanced microscopes). This kind of organisms have sizes in the order of micrometers (μm, a million times lesser than a meter), so you cannot observe them with a dissecting microscope, which only increase the image 10-30 times. In fact, dissecting microscopes are employed to observe thin sections of tissues (of plants and animals) and, as the name say it, to "dissect" in very tiny parts.

The other two microscopes, transmission electron and scanning electron microscopes, are used to observe macromolecules, and physical changes in compounds and matter, in the order of nanometers (nm, a thousand million times lesser than a meter).

Final answer:

The scientist should use a compound light microscope to observe and draw living microorganisms from pond water, as it allows for the necessary magnification while preserving the organisms' life for movement study.

Explanation:

To observe and draw the structure of microorganisms in pond water and study their movement, the best type of microscope for a scientist to use would be a compound light microscope. This type of microscope uses visible light that passes through and is bent by the lens system, allowing the user to observe living organisms, which is essential for studying movement. Moreover, light microscopes can magnify cells up to approximately 400 times, which is typically sufficient for viewing microorganisms.

While electron microscopes like the transmission electron microscope and scanning electron microscope provide much higher magnification and resolution, they are not suitable for observing living specimens because the sample preparation process kills the organisms. Dissecting microscopes, on the other hand, provide a three-dimensional view of the specimen but have lower magnification and are more suitable for larger objects such as tissues, not microorganisms in pond water.

Answer:

lower vapor pressure and freezing point and a higher boiling point

Explanation:

All the properties listed above are colligative properties, they depend on the amount of solute present in a solution.

It is known that a solution has a lower vapour pressure than the corresponding pure solvent due to the presence of a solute. Also the freezing point of a solution is decreased while its boiling point is increased when compared with those of the pure solvent.

A salt water solution, compared to pure water, has a lower vapor pressure and freezing point, but a higher boiling point due to the impact of the salt, a non-volatile solute.

Compared to pure water, a salt water solution has lower vapor pressure, lower freezing point, and a higher boiling point. The presence of salt, a non-volatile solute, reduces the vapor pressure of the solution, leading to a decreased freezing point and an increased boiling point. This is due to a phenomenon known as colligative properties, which states that adding a solute to a solvent affects the physical properties of the solvent, regardless of the nature of the solute.

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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:

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:

t=0.47s

Explanation:

the ball has uniformly accelerated movement due to gravity

Vo=initial speed=4.6m/s

g=gravity=-9.8m/s^2

Vf=final speed=0, the player must wait for the ball to stop. so the final speed will be 0

we can use the following ecuation

T=(Vf-Vo)/g

T=(0-4.6)/-9.8m/s^2

T=0.47s

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:

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:

97.03%

Explanation:

The equation for volumetric expansion due to thermal expansion is as follows

V/Vo=(1+γΔT)

V=final volume

Vo=initial volume

γ=coefficient of volume expansion=3.2 × 10–5 K–1

ΔT=

temperature difference

assuming that the earth is a sphere the volume is given by

V=(4/3)pi R^3

if we find the relationship between the initial and final volume we have the following

[tex]\frac{V}{Vo}  =\frac{ \frac{4}{3} \pi r^{3} }{ \frac{4}{3} \pi ro^{3}}=\frac{r^{3} }{ro^{3}}[/tex]

taking into account the previous equation

r/ro=(1+γΔT)^(1/3)

r/r0=(1-3.2x10-5(3000-300))^(1/3)=

r/ro=0.9703=97.03%

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:

17.7 days

Explanation:

Since the ship accelerates from the rest, its initial velocity would be equal to 0.

So,

[tex]v_{i}=0[/tex]

Acceleration of the star-ship = a = 1 g = 9.8 m/s²

We need to find how many days will it take the ship to reach 5% of the speed of light. Speed of light is [tex]3 \times 10^{8}[/tex] m/s.

5% of the speed of light = [tex]0.05 \times 3 \times 10^{8}=1.5\times 10^{7}[/tex] m/s

This means, the final velocity of the star-ship will be:

[tex]v_{f}=1.5\times 10^{7}[/tex]

We have the initial velocity, final velocity and the acceleration. We need to find the time(t). First equation of motion relates these quantities as:

[tex]v_{f}=v_{i}+at[/tex]

Using the values in this equation, we get:

[tex]1.5 \times 10^{7}=0+9.8(t)\\\\ t=\frac{1.5\times10^{7}}{9.8}\\\\ t=1,530,612.245[/tex]

Thus, the star-ship will take 1,530,612.245 seconds to reach to 5% the speed of light. Now we need to convert this time to days.

Since, there are 60 seconds in a minute:

1,530,612.245 seconds = [tex]\frac{1,530,612.245}{60}=25510.20[/tex] minutes

Since, there are 60 minutes in an hour:

25,510.20 minutes = [tex]\frac{25,510.20}{60}=425.17[/tex] hours

Since, there are 24 hours in a day:

425.17 hours = [tex]\frac{425.17}{24}=17.7[/tex] days

Thus, it will take approximately 17.7 days (approximately 17 days and 17 hours) to reach to 5% the speed of light

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.

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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.

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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.

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.

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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:

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

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:

Q=22.6L/s

Explanation:

First you must consider the continuity equation at points 1 and 2, which indicates that both flows are of equal value, in this way you get an equation between the two flow rates.

Then you raise the Bernoulli equation taking into account that the height is the same, which makes the term h1-h2 zero.

Using the equations above to calculate one of the speeds.

Finally you find the flow by multiplying the speed by the area.

I attached procedure

Answer:

A

Explanation:

Forces are vector quantities because they have both magnitude and direction.

A is the right answer i just got it right for A P E X

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]

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

Yes. According to the coach's mathematical criteria, Tommy had a great season.

Sadly, Tommy doesn't even know it. His tone-deaf coach decided to describe success in terms that are absurd for 7-yr-olds, as well as for most of their parents.

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:

Final speed of the racer, v = 16.55m/s

Explanation:

It is given that,

Initial velocity of the racer, u = 12 m/s

Acceleration, [tex]a=0.65\ m/s^2[/tex]

time taken, t = 7 s

We need to find the final velocity of the racer. Let it is given by v. It can be calculated using first equation of motion as :

[tex]v=u+at[/tex]

[tex]v=12+0.65\times 7[/tex]

v = 16.55 m/s

So, the final speed of the racer is 16.55 m/s. Hence, this is the required solution.

Answer:

The building is 20.68m high.

Explanation:

In order to solve this problem we must first do a drawing of what the situation looks like. (Look at the attached picture). Then, we determine what was the velocity of the ball when it reached the top of the window (let's call this [tex]V_{0A}[/tex] and when it reached the bottom of the window (let's call this [tex]V_{0B}[/tex]. We can start by finding the velocity the ball has at the top of the window [tex]V_{oA}[/tex] when it's falling.

In order to do so, we can use the following equation:

[tex]\updelta y=V_{0A}t+\frac{1}{2} at^{2}[/tex]

We can now solve this equation for [tex]V_{0A}[/tex] so we get:

[tex]v_{0A}=\frac{\updelta y-\frac{1}{2} at^{2}}{t}[/tex]

or when simplified we get:

[tex]v_{0A}=\frac{2\updelta y-at^{2}}{2t}[/tex]

now we can substitute values in the formula so we get:

[tex]v_{0A}=\frac{2(-1.20m)-(-9.8m/s^{2})(0.14s)^{2}}{2(0.14)}[/tex]

and solve:

[tex]v_{0A}=-7.886m/s[/tex]

Once we got the first velocity, we can proceed and find the second velocity.

We can find that by using the following formula:

[tex]a=\frac{V_{f}^{2}-V_{0}^{2}}{2y}[/tex]

So now we can solve for [tex]V_{f}[/tex]

When solving for [tex]V_{f}[/tex], we get the following formula:

[tex]V_f=\sqrt{2ya+V_{0}^{2}}[/tex]

So we can now substitute some values.

[tex]V_{0B}=\sqrt{2(-1.20m)(-9.8m/s\{2})+(-7.885m/s)^{2}}[/tex]

When solving this equation we get an answer of:

[tex]V_{0B}=-9.257m/s[/tex]

Once we got those velocities, we can use them to find the distance from the roof to the highest part of the window and the distance between the

lowest part of the window and the floor.

Let's start with the top portion of the window:

[tex]y=\frac{V_{f}^{2}-V_{0}^{2}}{2A}[/tex]

in this case, the initial velocity is 0 because we are dropping the ball from the roof. So we can now substitute values so we get:

[tex]y=\frac{(-7.885m/s)^{2}-(0)^{2}}{2(-9.8m/s^{2}}[/tex]

when solving this for y we get:

y=-3.17m

Now, we can find the height between the lowest part of the window and the floor, so we get:

[tex]\updelta y=V_{0B}t+\frac{1}{2} at^{2}[/tex]

In this case the time is 1.11s because it's half of the total time the ball is beneath the window when it's falling. So when substituting values we get:

[tex]\updelta y=(-9.257m/s^{2})(1.11s)+\frac{1}{2} (-9.8m/s^{2})(1.11s)^{2}[/tex]

When solving that expression we get that:

[tex]\updelta y=-16.31m[/tex]

so now we have enough information to solve the problem, notice that the heights appear as negative. This is because the bal is falling at this time. Since we only care about the magnitud, we can make them positive.

So in order to find the height of the building, we must ad the three lengths we just found, so we get:

h=3.17m+1.20m+16.31m=20.68m

So the building has a height of 20.68m

Answer:

x = 0.237

y = 0.0789

Explanation:

Vector with direction 18.4° and magnitude 0.250 has x and y components of:

x = 0.250 cos 18.4°

x = 0.237

y = 0.250 sin 18.4°

y = 0.0789

Answer: x = 0.237

y = -0.0789

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]

Answer:

0.146 m/s

Explanation:

We can see it in the pic.