If the fundamental frequency of this harmonic is 7 Hz, at which frequency is this created? (Just type the number, not the units)​

A projectile motion is any object thrown into space upon which the only acting force is gravity. The primary force acting on a projectile is gravity.The velocity's horizontal component is 24.25 m/s at time t = 2 seconds.The velocity's horizontal component is 24.25 m/s at time t = 3 seconds.

This doesn’t necessarily mean that other forces do not act on it, just that their effect is minimal compared to gravity.The particle is moving vertically (downwards) along the y-axis due to uniform acceleration.A particle's vertical and horizontal projectile motions can both accelerate: The only force acting on a particle when it is launched into the air at some speed is the acceleration brought on by gravity (g). The downward motion of this acceleration is vertical.

347u = 28 m/s for the starting velocity

projecting at a 30° angle

The horizontal component of velocity is constant since there is no acceleration in the horizontal direction.

Vertical component of speed, u cos = 28 x Cos 30 = 24.25 m/s

The velocity's horizontal component is 24.25 m/s at time t = 2 seconds.

The velocity's horizontal component is 24.25 m/s at time t = 3 seconds

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The relative density of kerosene is 0.006, or 6 kg/m³.

1 Response. Employ the Archimedes' Principle to your advantage: the buoyant force of a fluid is equal to the apparent loss of weight in water and is also equal to the weight of the displaced fluid. Weight of the displaced water equals 60 - 40 N, or the apparent loss in weight of the metal block. Water displacement mass is calculated as 20 / 9.8 = 2.04 kg.

Buoyant force = Weight of the displaced water = 20 N - 12 N = 8 N

Volume of water displaced = Buoyant force / Density of water = 8 N / 9.8 m/s² = 0.8163 kg

Density of metal = Weight of metal in air / Volume of metal = 20 N / (density of air x volume of metal)

Density of metal = 20 N / 0.8163 kg = 24.5 kg/m³

Buoyant force = Weight of the displaced kerosene = 20 N - 14 N = 6 N

Volume of kerosene displaced = Buoyant force / Density of kerosene = 6 N / 9.8 m/s² = 0.6122 kg

Density of kerosene = Mass of kerosene / Volume of kerosene displaced = 14 N / 0.6122 kg = 22.86 kg/m³

Relative density of kerosene = Density of kerosene / Density of water = 22.86 kg/m³ / 1000 kg/m³ = 0.006, or 6 kg/m³.

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The acceleration of the car was 6.88 [tex]m/s^2[/tex]. The negative sign indicates that the car was decelerating, or slowing down.

To find the acceleration of the car, we can use the following formula:

acceleration = (final velocity - initial velocity) / time

However, we are not given the time it took for the car to undergo the displacement. To find the time, we can use the following formula:

displacement = (final velocity + initial velocity) / 2 * time

Solving for time, we get:

time = displacement / ((final velocity + initial velocity) / 2)

Plugging in the given values, we get:

time = [tex]-105 / ((-10.9 - 27.7) / 2) = 2.29 s[/tex]

Now that we have the time, we can use the first formula to find the acceleration:

acceleration = [tex](-10.9 - (-27.7)) / 2.29 = 6.88 m/s^2[/tex]

Therefore, the acceleration of the car was [tex]6.88 m/s^2[/tex]. The negative sign indicates that the car was decelerating, or slowing down.

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The log moves a distance of 2.0 m relative to the shore when Ernie reaches Burt.

Since the log is uniform, we can treat it as a system of three point masses: one at each end representing Burt and Ernie, and one at the center representing the center of mass of the log. We can use conservation of momentum to solve this problem.

Initially, the momentum of the system is zero since everything is at rest. When Ernie walks to Burt's end of the log, he exerts a force on the log, which in turn exerts an equal and opposite force on Ernie. This force is internal to the system and does not change the momentum of the system. Therefore, the momentum of the system is still zero after Ernie reaches Burt.

Total momentum = 0 (since the center of mass is at rest)

Burt's momentum = 0 (since he is at rest)

Ernie's momentum = 0 (since he is at rest)

After Ernie walks:

Total momentum = 0 (since the center of mass remains at rest)

Burt's momentum = 0 (since he remains at rest)

Ernie's momentum = (39.0 kg) * v, where v is the velocity of Ernie and the log in the opposite direction

Since the total momentum is conserved, we can equate the momentum before and after Ernie walks:

0 = (39.0 kg) * v

v = 0 m/s

the distance that the log moves when Ernie reaches Burt is simply the distance between the initial and final positions of Ernie, which is half the length of the log:

distance = (1/2) * (4.0 m) = 2.0 m

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Answer:

1000 N

Explanation:

The electric force between two charges is given as

F = kq'q/r²......................... Equation 1

Where F = electric force between the charges, q' = magnitude of the first charge, q = magnitude of the second charge, r = distance between the charges, k = coulombs constant

Given: q' = q = 1 C, r = 3000 m, k = 9×10⁹ Nm²/kg²

Substitute into equation 2

F = 1²×(9×10⁹)/3000²

F = 1000 N.

Hence the magnitude of the electric force between them = 1000 N

Since the positive x-direction was established to be to the right, the sleigh's velocity is in this direction (the positive x-direction).

The whole horizontal momentum is conserved since there is no external force operating on the system (the sleigh plus Jonathan and Jane) in the horizontal direction.

Let's say that the positive x-direction is to the right and the positive y-direction is up. Then, we may divide Jonathan and Jane's velocities into their x- and y-components as follows:

vx1 = 5.00 cos 30.0° = 4.33 m/s and vy1 = 5.00 sin 30.0° = 2.50 m/s are the speeds of Jonathan.

Jane's speed is given by the formulas vx2 = 7.00 cos 36.9° = 5.61 m/s and vy2 = 7.00 sin 36.9° = 4.16 m/s.

The sleigh is initially at rest, hence there is no initial total momentum in the x-direction. The total x-direction momentum after Jonathan and Jane jumps out is:

px = (−800 N)(−4.33 m/s) + (600 N)(5.61 m/s) = 6430 N·s

The final momentum in the x-direction is:

p'x = (1000 N + 800 N + 600 N)vx'

where vx' is the final velocity of the sleigh in the x-direction.

Therefore, we can solve for xv:

vx' = px / (1000 N + 800 N + 600 N) = 6430 N·s / 2400 N = 2.68 m/s

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The measure that helps your body to stretch is flexibility, so the correct answer is D.

Flexibility is the ability of your body to move your joints and muscles through their full range of motion. It is an essential component of physical fitness and can help improve posture, balance, and coordination. Regular stretching exercises can increase flexibility and reduce muscle stiffness, which can lead to improved physical performance, decreased risk of injury, and better overall health.

Muscular strength and muscular endurance are related to the ability of your muscles to generate force or sustain effort over time, respectively. While they are important for overall fitness, they are not directly related to flexibility.

In summary, flexibility is the key measure that helps your body to stretch, and regular stretching exercises can improve your flexibility and overall physical fitness.

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Answer: just say no and never talk to that person again.

Explanation:

because that is what i would do to prevent drugs and/or nicotine to enter my body.

The density of the object is is 4000 kg/m³

Density is the ratio of mass to volume of a body.

To calculate the density of the obeject, we use the formula below

Formula:

Where:

From the question,

Given:

Susbtitute these values into equation 1

Hence, the density is 4000 kg/m³

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Answer:

An energy source – like a battery or mains power. 

An energy receiver – like a lightbulb. 

An energy pathway – like a wire.

Explanation:

Answer:

The statement "The amount increases as the object is lifted higher" is true about the amount of gravitational potential energy an object has. The statement "The amount varies according to the material of the object" is false. The amount of gravitational potential energy only depends on the mass of the object and its elevation from the reference point. The statement "The amount varies according to the mass of the object" is true. The more massive an object is, the more gravitational potential energy it has. The statement "The amount increases the more quickly the object is lifted" is false. The amount of gravitational potential energy only depends on the object's height above the reference point, not the speed at which it is lifted.

Answer: (A)

Explanation: The amount increases as the object is lifted higher.

a) When the metal ball bearing falls from the machine to the concrete floor, its potential energy decreases while its kinetic energy increases.

The potential energy lost by the ball bearing can be calculated using the formula PE = mgh, where m is the mass of the ball bearing (0.005 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the height the ball bearing falls (3.0 m). Thus, the potential energy lost by the ball bearing is:

PE = (0.005 kg)(9.8 m/s²)(3.0 m) = 0.147 J

At the same time, the ball bearing's kinetic energy increases by an amount equal to the potential energy lost. Therefore, the ball bearing's initial kinetic energy is zero, and its final kinetic energy is:

KE = 0.147 J

b) As the ball bearing bounces back up from the concrete floor to its starting point, its kinetic energy decreases while its potential energy increases. The kinetic energy lost by the ball bearing can be calculated as the same value as before:

KE = 0.147 J

The ball bearing's potential energy at its starting point is equal to the potential energy it lost on the way down, which is:

PE = 0.147 J

Therefore, the ball bearing's change in potential energy is 0.147 J (from down to up), and its change in kinetic energy is -0.147 J (from up to down).

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Answer:  a) When the ball bearing falls from the machine to the floor, there is a change in its potential and kinetic energies. The potential energy of an object at a height h above the ground is given by mgh, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the ground. Initially, the ball bearing is at rest on the machine, so its kinetic energy is zero. Therefore, the initial energy of the ball bearing is purely potential energy, given by:

PEi = mgh = (0.005 kg)(9.8 m/s^2)(3.0 m) = 0.147 J

When the ball bearing hits the ground, its potential energy is zero and its kinetic energy is at a maximum. The velocity of the ball bearing just before it hits the ground can be found using the equation:

v^2 = u^2 + 2as

where u is the initial velocity (which is zero), a is the acceleration due to gravity (-9.8 m/s^2), s is the distance fallen (3.0 m), and v is the final velocity just before hitting the ground. Solving for v, we get:

v = sqrt(2as) = sqrt(2*(-9.8 m/s^2)*(3.0 m)) = 7.67 m/s

The kinetic energy of the ball bearing just before it hits the ground is given by:

KEf = (1/2)mv^2 = (1/2)(0.005 kg)(7.67 m/s)^2 = 0.145 J

Therefore, the change in potential energy is:

ΔPE = PEf - PEi = 0 - 0.147 J = -0.147 J

And the change in kinetic energy is:

ΔKE = KEf - KEi = 0.145 J - 0 J = 0.145 J

b) When the ball bearing bounces back up to its starting point, there is another change in its potential and kinetic energies. Just before it reaches its highest point, the ball bearing's velocity is zero, so its kinetic energy is also zero. Therefore, its energy is purely potential energy, given by:

PEf = mgh = (0.005 kg)(9.8 m/s^2)(3.0 m) = 0.147 J

The ball bearing reaches its highest point when all of its initial kinetic energy has been converted to potential energy. At this point, its potential energy is at a maximum and its kinetic energy is at a minimum. The ball bearing then starts to fall back down towards the ground, and its potential energy starts to decrease while its kinetic energy increases. Just before it hits the ground, its kinetic energy is at a maximum and its potential energy is at a minimum, as we saw in part (a). When it bounces back up, the process repeats.

Therefore, the change in potential energy as the ball bearing travels from the floor back up to its starting point is:

ΔPE = PEf - PEi = 0.147 J - 0 J = 0.147 J

And the change in kinetic energy is:

ΔKE = KEf - KEi = 0 J - 0 J = 0 J

Note that the change in kinetic energy is zero because the ball bearing starts and ends at rest.

Explanation: can i get brainliest

The charged spheroid pushes against the line of charge with a force of 0.0181 N.

Its magnitude is determined by the Coulomb's law, F = k q1 q2/r2, when a point charge (a particle with a charge Q) is operating on a test charge q at a distance r.

Force between the ions' magnitude

Apply the electrostatic force equation of Coulomb.

F = (kq1q2)/r²

where;

q1 is charge 1=6.3nC/mx0.2m=1.26 nC

q2 is charge 2=-4 μC

r is the distance between the charges=5 cm= 0.05 m

F = (9x10⁹x1.26x10⁻⁹x4x10⁻⁶)/(0.05)²

F = 0.0181 N.

B) By breaking the force down into its x and y components, you can determine the orientation of the force the charged sphere applies to the line of charge.

θ = arctan(Fy/Fx) = arctan(Fy/0) = arctan(-Fy)

where Fy is the force's y-component. The force's y component can be calculated as follows:

Fy = F * sin(θ) = F * sin(arctan(-Fy))

Solving for Fy, we get:

Fy = -F * sin(arctan(-Fy)) = -F * (-0.05 / r) = 0.05F / √(x² + 0.05²)

where we used the fact that sin(arctan(x)) = x/√(1+x²).

Plugging in the values, we get:

Fy = 0.05 * 4.47 × 10⁻⁴/ √(0² + 0.05²)

≈ 4.00 × 10⁻⁵ N

Therefore, the direction angle is:

θ = arctan(-Fy/Fx) = arctan(4.00 × 10⁻⁵/ 0) = 90°

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The net power radiated by the Earth in free space is approximately

1.2 x 10^ 17 W watts.

To calculate the net power radiated by the Earth in free space, we can use the Stefan-Boltzmann Law:

Power = σ * A * ε * (T^4 - T0^4)

where:

σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)

A is the surface area of the Earth (4πR^2, where R is the radius of the Earth)

ε is the emissivity of the Earth (0.8)

T is the temperature of the Earth surface (270 K)

T0 is the temperature of space (2 K)

Plugging in the values, we get:

Power = 5.67 x 10^-8 * 4πR^2 * 0.8 * (270^4 - 2^4)

The radius of the Earth is approximately 6.37 x 10^6 m, so:

Power = 5.67 x 10^-8 * 4π(6.37 x 10^6)^2 * 0.8 * (270^4 - 2^4)

Power ≈ 1.193 x 10^ 17 W

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The ranking of their accelerations from smallest to largest is: Turtle C, Turtle A, Turtle B.

Based on the given accelerations:

So the ranking of their accelerations from smallest to largest is: Turtle C, Turtle A, Turtle B.

Based on the given velocities:

As for the change in direction:

Turtle A is accelerating in the same direction as its velocity (North), so it's not changing direction, but its speed is increasing.

Turtle B is accelerating in the opposite direction of its velocity (North), so it will eventually change direction and start moving North when the acceleration overcomes the initial Southward velocity.

Turtle C is accelerating in the same direction as its velocity (North), so it's not changing direction, but its speed is increasing.

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Three turtles are walking in the park. The x-axis is positive towards the North direction. The velocities of the turtles are as follows:

Turtle A has a velocity of 2 cm/s to the North and an acceleration of 0.1 cm/s².

Turtle B has a velocity of 1 cm/s to the South and an acceleration of 0.2 cm/s².

Turtle C has a velocity of 3 cm/s to the North and an acceleration of 0.05 cm/s².

Rank the accelerations of the three turtles from smallest to largest based on their given accelerations.

Are the turtles going towards the North or South direction? Are they changing direction based on their accelerations?

Answer:

true it is refraction of visible

The object's distance from the lens is 11.4 cm.

The thin lens equation can be used to determine how a converging lens's object distance (p), image distance (q), and focal length (f) relate to one another:

1/p + 1/q = 1/f

where p denotes the distance to the object, q is the distance to the picture, and f is the focal length.

Let's solve for the object distance using the thin lens equation:

1/p + 1/59.6 = 1/9.56

Combining both sides with p59.69.56 results in:

59.69.56 + p9.56 = p*59.6

Adding and subtracting:

570.176 + 9.56p = 59.6p

50.04p = 570.176

p = 11.4 cm

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Answer:

A person who drives a car for themselves, generally taking themselves back and forth to work or errands, is generally referred to as a driver.

The electric field 0.40 m away from a source charge of 3.00 x 10^-5 C is 1.69 x 10^6 N/C, directed away from the source charge.

The electric field created by a point charge at a distance r from the charge is given by the equation:

E = kQ/r^2

where

E is the electric field in N/C,

k is Coulomb's constant (9.0 x 10^9 N m^2/C^2),

Q is the source charge in Coulombs, and

r is the distance from the source charge in meters.

Substituting the given values into the equation, we get:

E = (9.0 x 10^9 ) x (3.0 x 10^-5) / (0.40)^2

E = 1.69 x 10^6 N/C

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Explanation:

let at a distance x from the c

Answer:52.972ft^3

Explanation:

It is unit conversion based and for a volume of a bin from cubic meter to cubic foot as we know 1 meter =3.281 foot.

where volume=1.5m^3

multiply 1.5*(3.281)^3 ft^3

v=52.972 ft^3

the magnitude of the friction force on the disk is approximately 1.16 N.

We can use the conservation of energy to find the friction force on the disk. The initial kinetic energy of the wheel is equal to the work done by the friction force on the disk:

K_i = W_friction

The initial kinetic energy of the wheel can be found from its moment of inertia and angular velocity:

K_i = (1/2) I [tex]ω^2[/tex]

where I is the moment of inertia, ω is the angular velocity, and the factor of 1/2 comes from the rotational kinetic energy formula.

The final kinetic energy of the wheel is zero, since it comes to a stop. The work done by the friction force can be found from the distance over which it acts:

W_friction = F_friction d

where F_friction is the friction force and d is the distance over which the pads act on the disk. We can find the distance from the angular displacement of the wheel:

θ = ω t

where θ is the angle through which the wheel rotates, t is the time for the wheel to come to a stop, and the factor of 1/2π converts from rotations to radians. The distance over which the pads act is then:

d = r θ = 0.071 m × (5/2π) ≈ 0.562 m

Now we can put everything together:

K_i = W_friction

(1/2) I [tex]ω^2[/tex] = F_friction d

We can solve for the friction force:

F_friction = (1/2) I [tex]ω^2[/tex] / d

Plugging in the given values:

F_friction = (1/2) ×[tex]0.097 kg⋅m^2[/tex] × [tex](5/2.2π rad/s)^2[/tex] / 0.562 m ≈ 1.16 N

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The average power required to accelerate the truck with a weight of 20,000 N is approximately 116,930.68 W.

The average power required to accelerate an object is given by the formula:

Power = Work / Time

where Work is the change in kinetic energy of the object and Time is the time interval over which the change in kinetic energy occurs.

The change in kinetic energy of the truck is given by:

ΔK = 1/2 * m *[tex]v^2[/tex]

where m is the mass of the truck and v is its final velocity.

Since the truck starts from rest, its initial kinetic energy is zero, so the work done on the truck is equal to its change in kinetic energy. Therefore, the average power required to accelerate the truck is:

Power = Work / Time = ΔK / Time

Substituting the given values, we get:

(a) For weight w = 10,000 N (approximately 1,020 kg):

m = w / g = 10000 N / [tex]9.81 m/s^2[/tex] = 1019.3 kg

ΔK =[tex]1/2 * m * v^2[/tex] = 1/2 * 1019.3 kg * [tex](18.5 m/s)^2[/tex]= 333,036.6 J

Power = ΔK / Time = 333036.6 J / 5.7 s ≈ 58,426.84 W

Therefore, the average power required to accelerate the truck with a weight of 10,000 N is approximately 58,426.84 W.

Note: g is the acceleration due to gravity, which is approximately 9.81 [tex]m/s^2.[/tex]

(b) For weight w = 20,000 N (approximately 2,040 kg):

m = w / g = 20000 N / 9.81 m/s^2 = 2041.2 kg

ΔK = 1/2 * m *[tex]v^2[/tex] = 1/2 * 2041.2 kg * [tex](18.5 m/s)^2[/tex]= 666,073.3 J

Power = ΔK / Time = 666073.3 J / 5.7 s ≈ 116,930.68 W.

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A = wL = 63 cm²

m = (2.5 g/cm²)(63 cm²)

m = 157.5 g = 0.1575kg

[tex]I=\frac{1}{3} mL^2\\\\I = \frac{1}{3}(0.1575kg)(0.09m)^2[/tex]

I = 4.2525×10⁻⁴ kg/m²

The magnitude of the velocity of S' with respect to S is therefore approximately  [tex]9 * 10^(11)[/tex]m/s. then the magnitude is approximately[tex]9 * 10^(11)[/tex] m/s.

the velocity of the system S' with respect to S as v. Since the two events are simultaneous in the system S', they must have the same time coordinate in that system. Let's call this time coordinate t' and the position of the two events in the S' system as (x1', t') and (x2', t').

Using the Lorentz transformation equations, we can relate the coordinates (x1, t1) and (x2, t2) in the S system to the coordinates (x1', t') and (x2', t') in the S' system:

x1' = γ(x1 - vt1)

t' = γ(t1 - vx1/[tex]c^2[/tex])

x2' = γ(x2 - vt2)

t' = γ(t2 - [tex]vx2/c^2[/tex])

where γ = 1/√(1 - [tex]v^2/c^2[/tex]) is the Lorentz factor.

Since the two events are separated by 900 m and a time interval of 10^-6 s, we have:

x2 - x1 = 900 m

t2 - t1 = [tex]10^-6[/tex] s

Using the above equations, we can solve for v. First, we can eliminate t' by equating the two expressions for t':

γ(t1 - vx1/[tex]c^2[/tex]) = γ(t2 - vx2/[tex]c^2[/tex])

Simplifying this expression, we get:

v =[tex]c^2[/tex](x2 - x1)/(t2 - t1)(x1 + x2)

Plugging in the given values, we get:

v = c^2(900 m)/([tex]10^-6[/tex] s)(0 m + 900 m) = 9 × [tex]10^11[/tex] m/s

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the initial momentum of the system of block m1 and block m2 is

Pi= m1v1 + m2v2

the final momentum of the combine blocks is

Pf= (m1+m2)V

according to the law of convervation of momentum

Pi = Pf

m1v1 + m2v2 = (m1+m2)V

1.3 × 1 + 5m2 = 1.3 × 2 + 2m2

m2= 1.3/3 kg

Answer:

Explanation:

To calculate the mechanical advantage of a lever, we use the formula:

Mechanical Advantage = Output Force / Input Force

In this case, the output force is 100 Newton and the input force is 20 Newton, so:

Mechanical Advantage = 100 N / 20 N

Mechanical Advantage = 5

Therefore, the mechanical advantage of the lever is 5. This means that for every 1 Newton of input force applied to the lever, the lever will produce 5 Newtons of output force. So, in this case, an input force of 20 Newtons applied to the lever would produce an output force of 100 Newtons.

It takes 0.3 seconds for a force of 1000 N to halt a 10,000 kg goods car moving at 3.00 m/s along a track.

We must first establish the car's starting kinetic energy in order to calculate the time required to stop the vehicle:

Kinetic Energy (KE) is equal to half of mass times speed, or 10,000 kg times 3.00 m/s.

KE = 45,000 J

Then, we may use the designed with the intent, which asserts that an object's change in kinetic energy equals the jobs performed by an external force: Work equals Force x Distance x Change in KE.

The gain in kinetic energy is equal to the starting kinetic energy because the car is coming to a stop: KE Change = -45,000 J

As a result, the external force's work is: Work equals force times distance, or -45,000 J.

When we solve for distance, we obtain: Work / Force = -45,000 J / 1000 N Distance

Location = -45 m

Because the force is against the direction of the car's motion, you'll see that the range is negative.

Finally, we can calculate the travel time using the kine model for uniformly accelerated motion: Distance is equal to 1/2*acceleration*time2. Time is calculated as sqrt(2 * Distance / Acceleration) as well as sqrt(2 * 45 m / (1000 N / 10,000 kg)).

time equals sqrt(0.09 s2/kg).

time = 0.3 s

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Answer:

120 V

Explanation:

Given:

I (total) = 0,5 A

R1 = 110 Ω

R2 = 130 Ω

Find: V (total) - ?

The resistors are connected in series

That means, the current in the circuit is the same for every resistor:

I (total) = I1 = I2

Now, we need to find the voltage in each resistor:

V1 = I × R1

V1 = 0,5 × 110 = 55 V

V2 = I × R2

V2 = 0,5 × 130 = 65 V

Now, since the connection is in series, in order to find the total voltage in the circuit, we have to add the voltages of the resistors:

V (total) = V1 + V2

V (total) = 55 + 65 = 120 V

The statement that best fits the graph is: D) Car B is faster than A because B's motion is accelerating and A is constant speed.

From the graph, we can see that the distance traveled by both cars increases over time. However, the slope of the distance-time graph for car B is steeper than that of car A. This means that car B covers more distance in the same amount of time as car A, indicating that car B is traveling faster than car A.

Furthermore, we can see that the speed-time graph for car B is a straight line with a positive slope, indicating that its speed is increasing over time. On the other hand, the speed-time graph for car A is a horizontal line, indicating that its speed is constant.

Therefore, we can conclude that car B is faster than car A because car B's motion is accelerating, while car A is moving at a constant speed.

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