A person is swimming in a river with a current that has speed vR with respect to the shore. The swimmer first swims downstream (i.e. in the direction of the current) at a constant speed, vS, with respect to the water. The swimmer travels a distance D in a time tOut. The swimmer then changes direction to swim upstream (i.e. against the direction of the current) at a constant speed, vS, with respect to the water and returns to her original starting point (located a distance D from her turn-around point) in a time tIn. What is tOut in terms of vR, vS, and D, as needed?

Answer:

it should be tension for the space

Answer:

tension

Explanation:

I'm not sure if I'm right but if the example is a pulley a pulley applies tension to a belt or rope or chain to keep it tight

Answer:analize a afirmacao a seguir e tudo que envolve o gerenciamento da marca e que ultrapassa as acoes com objetivos economicos e refere se a cultura principios e valores

Explanation:

Answer:

The frequency increases by 4 because it is inversely proportional to the wavelength.

Answer:

793.5 Joules

Explanation:

The work–energy theorem states that the net work done on an object equals the object’s change in kinetic energy.

So we can say W=ΔKE.

The equation for kinetic energy is

\(KE=\frac{1}{2}mv^2\)

Therefore

\(W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2\)

\(W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}\)

We are given

\(m=3\)

\(v_i=0\)

\(v_f=23\)

Lets solve for \(W\).

\(W=\frac{3*23^2}{2}-\frac{3*0^2}{2}\)

\(W=\frac{3*23^2}{2}\)

\(W=\frac{3*529}{2}\)

\(W=\frac{1587}{2}\)

\(W=793.5\)

The final velocity of the 25.8 g marble after the collision is 16.15 cm/s.

The velocity of the 25.8 g marble after the collision is calculated as follows;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

The final velocity of the 25.8 g marble after the collision is calculated as;

( 25.8 x 21 ) + ( 12.4 x 13.8 ) = ( 12.4 x 23.9 ) + ( 25.8v )

712.92 = 296.36 + 25.8v

25.8v = 416.56

v = 416.56 / 25.8

v = 16.15 cm/s

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The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

The observation that electromagnetic radiation with energy above a certain value can eject electrons out of a metal is a piece of evidence that they have particle-like properties.

This observation that electromagnetic radiation behaves like particles is known as the photoelectric effect.

It provides evidence that electromagnetic radiation exhibits particle-like properties. When EMR with sufficient energy (above a certain threshold) interacts with a metal surface, it can cause the ejection of electrons from the metal.

This behavior indicates that EMR behaves as discrete packets of energy called photons, which transfer their energy to the electrons and cause their release. The photoelectric effect supports the particle nature of EMR and is a fundamental concept in the field of quantum mechanics.

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The instantaneous speed of the ant at t=8s is equal to the slope of the segment between t=7s and t=9s, which is 1.5m/s.

To find the instantaneous speed of the ant at time t=8s, we need to calculate the derivative of the displacement function with respect to time. Since the displacement of the ant is given by a piecewise function, we need to differentiate each segment of the function separately and then piece them together. From 0s to 4s, the displacement of the ant decreases linearly from 6m to 2m. The slope of this segment is -1m/s. From 4s to 7s, the displacement of the ant is constant at 2m. The slope of this segment is zero. From 7s to 9s, the displacement of the ant increases linearly from 2m to 5m. The slope of this segment is 1.5m/s. Finally, from 9s to 10s, the displacement of the ant is constant at 5m. The slope of this segment is zero.

Therefore, the instantaneous speed of the ant at t=8s is equal to the slope of the segment between t=7s and t=9s, which is 1.5m/s. This means that the ant is moving upwards at a constant speed of 1.5m/s at time t=8s. It's important to note that the instantaneous speed of the ant tells us how fast it's moving at a specific moment in time. It's not the same as the average speed over a given time interval, which would be calculated by dividing the total displacement by the total time elapsed.

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The constant acceleration of the train is 50/9 m/s².

The two balls will collide at a height of approximately 10.204 meters above the ground.

Using the kinematic equations of motion, we have:

distance = initial velocity * time + 1/2 * acceleration * time^2

For the first phase of acceleration, the initial velocity is zero, the time is 5 minutes = 300 seconds, and the distance traveled is unknown. So we have:

d1 = 0 + 1/2 * a * (300)^2

For the second phase of constant speed, the initial velocity is v, the time is 5 minutes = 300 seconds, and the distance traveled is also unknown. So we have:

d2 = v * 300

For the third phase of deceleration, the initial velocity is v, the time is also 5 minutes = 300 seconds, and the distance traveled is again unknown. So we have:

d3 = v * 300 + 1/2 * (-a) * (300)^2

The total distance traveled is the sum of these three distances:

distance = d1 + d2 + d3 = 1/2 * a * (300)^2 + v * 600 - 1/2 * a * (300)^2 = v * 600

Since the total distance traveled is given as 10 km = 10000 m, we have:

v * 600 = 10000

Solving for v, we get:

v = 10000/600 = 50/3 m/s

Now we can use the second equation above to find a:

d2 = v * 300 = (50/3) * 300 = 5000 m

Therefore, the constant acceleration of the train is:

a = 2 * (5000 - 1/2 * a * (300)^2) / (300)^2 = 50/9 m/s^2

The constant acceleration of the train is 50/9 m/s^2.

Problem 2: The height of the first ball dropped is given as 10m. Let's assume the height of the collision point is h meters above the ground.

Using the kinematic equation for free fall, we have:

h = 10 + 1/2 * g * t^2

where g is the acceleration due to gravity, which is approximately 9.81 m/s^2, and t is the time it takes for the second ball to reach the collision point after being thrown upwards.

The initial upward velocity of the second ball is 10 m/s, and we know that at the collision point, its velocity will be zero, since it will have reached its maximum height and will be momentarily at rest before falling back down.

Using the kinematic equation for motion with constant acceleration, we have:

0 = 10 + (-g) * t

Solving for t, we get:

t = 10/g = 10/9.81 seconds

Substituting this value of t into the first equation, we get:

h = 10 + 1/2 * 9.81 * (10/9.81)^2

Simplifying, we get:

h = 10.204 m

The two balls will collide at a height of approximately 10.204 meters above the ground.

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In a DC generator, the generated electromotive force (emf) is directly proportional to the rotational speed of the generator's armature and the strength of the magnetic field within the generator.

This relationship is described by the equation for the generated emf in a DC generator:

Emf = Φ * N * A * Z / 60

Where:

Emf is the generated electromotive force (in volts),

Φ is the magnetic flux density (in Weber/meter^2\(meter^2\) or Tesla),

N is the number of turns in the armature winding,

A is the effective area of the armature coil (in square meters),

Z is the total number of armature conductors, and

60 is a constant representing the conversion from seconds to minutes.

From this equation, we can see that the generated emf is directly proportional to the magnetic flux density (Φ) and the product of the number of turns (N), effective area (A), and the total number of armature conductors (Z). This means that increasing any of these factors will result in a higher generated emf.

The magnetic flux density (Φ) can be increased by using stronger permanent magnets or increasing the strength of the field windings in the generator.

The number of turns (N) and the effective area (A) are design parameters and can be optimized for a specific generator. Increasing the number of turns or the effective area will result in a higher generated emf.

Similarly, the total number of armature conductors (Z) can be increased to enhance the generated emf.

By controlling and optimizing these factors, the generated emf in a DC generator can be increased, resulting in higher electrical output. However, it is important to note that there are practical limits to these factors based on the design and construction of the generator.

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Answer:

1.0 m/s^2

Explanation: happy to help :)

Answer: \(1\ m/s^2\)

Explanation:

Given

Masses of the block are \(m_1=1\ kg\)  and

\(m_2=2\ kg\)

Force applied by \(1\ kg\) block on \(2\ kg\) block is \(2\ N\)

From the free body diagram of \(2\ kg\) block, the net force on

\(\therefore m_2a=2\\\\\Rightarrow 2\times a=2\\\\\Rightarrow a=\dfrac{2}{2}\\\\\Rightarrow a=1\ m/s^2\)

Thus, the acceleration of two blocks is \(1\ m/s^2\)

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Answer:

4. 10.0 m/s²

Explanation:

I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:

\(1. \ a=\frac{v-v_0}{t};\)

\(2. \ L=\frac{at^2}{2}.\)

II) using these two equations after substitution v₀=0; v=30 and L=45:

\(\left \{{{45 =\frac{at^2}{2}} \atop {a=\frac{30-0}{t} }} \right.\)

\(\left \{ {{at^2=90} \atop {at=30}} \right. \ <=> \ \left \{ {{a=10} \atop {t=3}} \right.\) \(=> \ a=10\frac{m}{s^2}\)

Nothing accelerates a Projectile Horizontally during Motion...

Consequently

Horizontal Acceleration is Always Zero.

The Horizontal Velocity is always Constant too.

Only the vertical Velocity changes by -9.8ms-².

If this is so...

∆x = v'∆t + ½(0)(∆t)²

∆x = v't.

I'd go with Option D.

The distance the ball was thrown is 52.22m by A quarterback throws a pass at an angle of 35° above the horizontal with an initial speed of 25 m/s. The ball is caught by the receiver 2.55 seconds later.

Every motion under constant acceleration is projectile motion

Angle above horizontal Ф = 35°, initial speed v1 = 25m/s , time 2.55s

Substituting value in the below equation

x=x₁ + (v₁*cosθ)(t)+1/2 *a*t²

a= 0 as acceleration in horizontal direction is zero

x= 25*cos(35)*2.55

x=52.22 m

Projectile motion is a form of motion in which object influenced when it is launched into the gravitational force from the surface of Earth along a curved path.

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Answer:

The value is \(F =13.27 \ N\)

Explanation:

From the question we are told that

     The mass is \(m = 3200 \ g = 3.2 \ kg\)

     The speed is \(v = 7.2 \ m/s\)

     The radius is \(r = 12.5 \ m\)

Generally the centripetal force is mathematically represented as

        \(F = m * \frac{v^2}{r}\)

=>     \(F = 3.2 * \frac{7.2^2}{12.5}\)

=>     \(F =13.27 \ N\)

Answer:

The bag of sand

Explanation:

I think it is the bag of sand because according to the definition of impulse, impulse is the average force acting on a particule when an external force is being acted on it.

Answer:

18Ω

Explanation:

If the resistors are connected in series, the equivalent resistance is the sum of each resistance, so

Equivalent resistance = 4Ω + 6Ω + 8Ω

Equivalent resisteance = 18Ω

Therefore, the answer is 18Ω

The more slanted the sun's rays are, the longer they travel through the atmosphere, becoming more scattered and diffuse. In the winter, Earth tilts away from the Sun.

To collect the asteroid an equal magnitude of the force in opposite direction is needed.

What is force ?

The motion or speed of an object depends upon the magnitude and the direction of the force.

If the direction of the force is against the direction of motion, then the object will slow down or stop. The two forces of the same magnitude in the opposite direction cancel each other and the collided objects stop each other.

Therefore, to collect the asteroid an equal magnitude of the force in opposite direction is needed.

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Hypotenuse be H

Apply Pythagorean theorem

\(\\ \sf\Rrightarrow H^2=P^2+B^2\)

\(\\ \sf\Rrightarrow H^2=5^2+9^2\)

\(\\ \sf\Rrightarrow H^2=25+81\)

\(\\ \sf\Rrightarrow H^2=106\)

\(\\ \sf\Rrightarrow H=\sqrt{106}cm\)

Answer:

D

Explanation:

Answer:

At 20 °C (68 °F), the speed of sound in air is about 343 metres per second (1,235 km/h; 1,125 ft/s; 767 mph; 667 kn), or a kilometre in 2.9 s or a mile in 4.7 s.

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

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According to the question, the different devices as to broom is a first-class lever.

Broom is an implement used for sweeping. It consists of a set of bristles attached to a long handle. Brooms have been used for centuries to keep floors and other surfaces clean. The bristles are usually made from plant fibers or plastic. They are also sometimes made from metal or wire. Brooms can be used to sweep up dirt and dust, collect small particles, and even push larger items. They are perfect for quick cleanups and can be used on almost any type of floor or surface. They are also great for removing cobwebs and other debris from corners and other hard-to-reach places. Brooms are a convenient and easy-to-use tool that can help keep your home or workspace clean.

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Answer: Horizontal force.

The ball did not drop straight down under the force of gravity because the man also acted on the ball with  a horizontal force. So option A is correct

The definition of force is the pushing or pulling of anything. Push and pull are the result of two things interacting with one another. Stretch and crush are two more phrases that can be used to describe force.

The ball did not drop straight down under the force of gravity because the man acted on the ball with a horizontal force.

As the man runs, he moves the ball in a horizontal direction, which causes it to deviate from its straight-down trajectory under gravity and miss the bull's-eye target.

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Answer:

8.2N

Explanation:

I took the test and got it right :)

Answer:

E = 1,720,779.221 or 1.720779221 * 10^ 6V/m

Explanation:

The electric field between the parallel conducting plates is given by

E =V / d

where V is the potential difference and d is the distance between the plates.

E = 26.5 kV/ 1.54 cm

Now we have to convert into proper units

26.5 kv= 26.5 * 1000 v=  26500 volts

1  kv= 1000 volts

1.54 cm = 1.54/ 100 m= 0.0154m

1m = 100cm

Now putting the values

E= 26500/0.0154 = 1,720,779.221 V/m

The Electric field is equal to E= 1,720,799.221 or 1.7220799221 * 10 ^6 Volts per meter.

In scientific notation this can be written as 1.7220799221 *10^6 V/m

Answer:

a) \(V_2=8m/s\)

b) \(P_2=9.54*10^4 Pa\)

Explanation

From the question we are told that:

Initial Area of pipe  \(A_1=8.00 cm^2\)

Initial Fluid flow speed \(r_1 =320 cm/s,\approx 320*10^{-2}\)

Initial Pressure of \(\rho_1 =1.40*10^5 Pa\)

Final area of pipe \(A_2 =3.70 *10^{-2} cm^2\)

Density of acid \(\rho=1660kg/m^3\)

a)

Generally the equation for continuity is mathematically given by

\(A_1V_1=A_2V_2\\\\V_2=\frac{A_1*V_1}{A_2}\)

Since volume is directly proportional to rate of flow

\(V_2=\frac{8*320}{3.20} *10^{-2}\)

\(V_2=8m/s\)

b)

Generally the Bernoulli's equation is mathematically given by

\(p_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2\\\\with\ h_1=h_2\\\\p_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2\)

Therefore

\(P_2=P_1+\frac{1}{2}\rho(v_1^2-V_2^2)\\\\P_2=(1.40*10^5)+\frac{1}{2}(1660)(v_1^2-V_2^2)\)

\(P_2=9.54*10^4 Pa\)

Answer:

Explanation:

There is no acceleration in vertical direction

Fn = mg + Fp sin 35

Frictional force = 99.7 N

Net force in forward direction = Fp cos 35 - 99.7

= 168 cos 35 - 99.7

= 137.61-99.7

= 37.91 N

net acceleration of chair = net force / mass

= 37.91 / 50

= .7582 m / s² .