After an average acceleration of 3.53 m/s2 during 2.01 s , a car reaches a velocity of 15.1 m/s . find the car's initial velocity.

We can calculate the spring constant of the bungee cord and the maximum velocity of the bungee jumper during the oscillations.

To start, we can use the formula for potential energy to find the spring constant:

PE = mgh = (1/2)kx^2

where m is the mass of the bungee jumper (45.0 kg), g is the acceleration due to gravity (9.81 m/s^2), h is the distance the bungee jumper falls (33.0 m), k is the spring constant, and x is the distance the bungee cord stretches.

Solving for k, we get:

k = 2mgx^2 / (h * x^2)

Substituting in the given values, we get:

k = 2 * 45.0 kg * 9.81 m/s^2 * (33.0 m) / (5 * (33.0 m)^2)

k = 40.0 N/m

Now we can use the formula for a simple harmonic motion to find the maximum velocity of the bungee jumper:

vmax = A * ω

where A is the amplitude of the oscillation (half the distance between the lowest and highest points, which is (33.0 m) / 2 = 16.5 m), and ω is the angular frequency (2π/T, where T is the period of the oscillation, which is 28.0 s / 6 = 4.67 s).

Substituting in the given values, we get:

vmax = 16.5 m * 2π / 4.67 s

vmax = 22.4 m/s

Therefore, the bungee jumper reached a maximum velocity of 22.4 m/s during the oscillations.

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Answer:

xbzcfhkuvy

Explanation:

dvdvxvdzdx

The velocity of the box 2 seconds later is 10 m/s.

We can find the velocity of the box 2 seconds later using the given acceleration and initial velocity of the box.

We use the following kinematic equation to solve for the velocity:\[v = u + at\]where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

We are given that the box starts from rest and experiences an acceleration of 5 m/s². Thus, the initial velocity of the box is u = 0 m/s, and the acceleration is a = 5 m/s².

The time taken is t = 2 s.

Substituting these values in the above equation,\[v = u + at\] \[v = 0 + 5 \times 2\] \[v = 10 m/s\]

Therefore, the velocity of the box 2 seconds later is 10 m/s.

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Without going above the maximum permitted radial acceleration, the flywheel can store a maximum of kinetic energy of 7,000 J.

The maximum permitted radial acceleration of a location on the flywheel's rim must be taken into account in order to calculate the maximum amount of kinetic energy that can be stored there.

This is crucial because if the acceleration goes beyond this point, the flywheel may distort or even disintegrate.

We can use the formula to determine the maximum permitted radial acceleration.

a = \(v^2\) / r

where a is the radial acceleration,

v is the velocity of a point on the rim, and

r is the radius of the flywheel.

Rearranging the formula, we get

v = \(\sqrt{(ar)}\)

where the square root function is represented by sqrt.

Now that we know the radial acceleration limit, we can figure out the highest velocity that a spot on the rim can reach.

Assume that 100 rad/s is the maximum permitted radial acceleration.  (this value can vary depending on the specific design and materials of the flywheel).

Using this number and the flywheel's radius, which is assumed to be 0.5 meters, we obtain

v = \(\sqrt{(100 \times 0.5)}\) = 10 m/s.
Next, we can use the formula to determine the maximum kinetic energy that the flywheel can store.

E = 1/2 x I x \(w^2\)

where w is the angular velocity,

I is the flywheel's moment of inertia, and

E is the kinetic energy.

The moment of inertia for a solid disc is 1/2 x m x \(r^2\)

where m is the mass of the flywheel and

r is the radius.

Plugging in the values (m = 70.0 kg, r = 0.5 m), we get I = 8.75 kg \(m^2\).
The maximum angle at which the flywheel can rotate before a spot on the rim's velocity exceeds the predetermined limit must now be determined.

Since the angular velocity times the radius are equivalent to the linear velocity of a point on the rim, we obtain

w x r = 10 m/s.

Solving for w, we get w = 20 rad/s.
Finally, we can calculate the maximum kinetic energy using the formula

E = 1/2 x I x \(w^2\)

Putting the values in (I = 8.75 kg \(m^2\), w = 20 rad/s),

we get E = 7,000 J.

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The minimum amount of work done by the force on the object is 100 joules (J).

The minimum amount of work that a force can do on an object can be determined using the formula:

Work = Force x Distance x cos(θ),

where "Force" is the magnitude of the force applied, "Distance" is the distance over which the force is applied, and "θ" is the angle between the force vector and the direction of displacement.

In this case, the force magnitude is given as 10 N, and the object moves at a distance of 10 m. However, the question asks for the minimum amount of work, which occurs when the force is applied parallel to the direction of motion. In this scenario, θ = 0 degrees, and cos(0) = 1.

Therefore, the minimum amount of work done by the force can be calculated as:

Work = 10 N x 10 m x cos(0) = 100 N·m.

Hence, the minimum amount of work done by the force on the object is 100 joules (J). It is important to note that this value assumes an ideal scenario where there is no loss of energy due to factors like friction or other resistive forces.

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Answer:

The answers as per the given problem is solved in the segment below.

Explanation:

According to the question,

(a)

\(F=ILB\)

By putting the values, we get

   \(=10(2\times 10^4)(5\times 10^{-5})\)

   \(=10 \ N\)

(b)

\(W=Fx\)

or,

    \(=Fvt\)

    \(=10(7.80\times 10^3)(3600)\)

    \(=2.81\times 10^8 \ J\)

(c)

\(\frac{1}{2}mv'^2 =\frac{1}{2}mv^2-\omega\)

or,

      \(v=[v^2-\frac{2 \omega}{m} ]^\frac{1}{2}\)

         \(=[(7.8\times 10^3)^2-\frac{2(2.81\times 10^8)}{1\times 10^5} ]^{\frac{1}{2} }\)

         \(=7.79\times 10^3 \ m/s\)

now,

\(\Delta v=v-v'\)

     \(=0.36 \ m/s\)

(d)

To calculate the magnitude of the drift velocity of the electrons ,

The drift velocity of electrons in a conductor is given by the formula:

v = I / (neA)

where 'v' is the drift velocity of electrons,

'I' is the current flowing through the wire,

'n' is the number of free electrons per unit volume,

'e' is the charge on each electron, and

'A' is the cross-sectional area of the wire.

Therefore, The current-carrying capacity of the 10 gauge copper wire is

 I = 21 A which is a given statement.

For copper, the number of free electrons per unit volume is approximately \(8.5*10\)²⁸ electrons/m³, and the charge on each electron is 1.6 x 10⁻¹⁹ C.

The cross-sectional area of a 10 gauge copper wire is approximately 5.26 mm²= 5.26 x 10⁻⁷ m².

Substituting these values into the formula of drift velocity we get:

v = (21 A) / ((8.5 x 10²⁸ electrons/m³) x (1.6 x 10⁻¹⁹ C/electron) x (5.26 x 10⁻⁷ m²))

= 0.015 m/s

Therefore, the magnitude of the drift velocity of the electrons in the wire is approximately 0.015 m/s.

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1. The change in internal energy of the washer is 17.3 J. The change in internal energy is calculated using the following equation:

ΔU = ncΔT

We can calculate the number of moles of aluminum as follows:

n = m / M

where:

m is the mass of the washer (g)

M is the molar mass of aluminum (g/mol)

n = 0.042 g / 26.98 g/mol = 0.00159 mol

Now we can calculate the change in internal energy:

ΔU = 0.00159 mol * 24.3 J/mol-K * (677 K - 300 K) = 17.3 J

2. The change in the outer radius of the washer is 0.015 mm.

The change in radius is calculated using the following equation:

Δr = αrΔT

where:

Δr is the change in radius (m)

α is the coefficient of linear expansion of aluminum (K-1)

r is the original radius (m)

ΔT is the change in temperature (K)

Δr = 2.22 × 10-5 K-1 * 1.77 cm * (677 K - 300 K) = 0.015 mm

3. The change in the inner radius of the washer is 0.007 mm.

The change in the inner radius is the same as the change in the outer radius, but in the opposite direction. So, the change in the inner radius is -0.007 mm.

The change in internal energy is due to the increase in the kinetic energy of the atoms of aluminum as they are heated. The change in radius is due to the expansion of the aluminum as it is heated.

The expansion is caused by the increase in the distance between the atoms of aluminum as they are heated.

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Answer:

f =-20 cm

Explanation:

Given that,

The magnification of a spherical mirror, m = -1

The image distance, v = 40 cm (for negative magnification)

The magnification of a concave mirror is negative. The mirror showing -1 magnification is a concave mirror.

Let f be the focal length of the mirror. We know that,

\(m=\dfrac{-v}{u}\\\\-1=\dfrac{-v}{u}\\\\v=u\)

Object distance, u = -40 cm

Using mirror's formula i.e.

\(\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}\\\\f=\dfrac{1}{\dfrac{1}{-40}+\dfrac{1}{-40}}\\\\f=-20\ cm\)

So, the focal length of the mirror is 20 cm.

A- 5550 kJ work. B- 4539 kJ work. C- 5550 kJ work. D- 4539 kJ work. E- The work gravity does on the car in part A is 0.

A) The amount of work accomplished while the cable drags the vehicle horizontally can be estimated by dividing the cable's tension (1110 N) by the vehicle's distance travelled (5.00 km). Since the angle between the force and displacement is 0 degrees, the work done is given by the formula:

work = force × displacement × cos(angle).

In this instance, since the angle is 0 degrees (cos(0) = 1), the work done by the cable on the car is 1110 N * 5.00 km = 5550 kJ.

B) The same calculation may be used to calculate the work done when the rope is pulling the car at an angle of 35.0 degrees above the horizontal. The only difference is that the angle is now 35.0 degrees, so the work done is given by:

work = 1110 N × 5.00 km × cos(35.0°) = 4539 kJ.

C) Similar to section A, the work performed by the tow truck's horizontally pulling cable can be estimated. Since there is no angle, the amount of work done is 1110 N × 5.00 km = 5550 kJ.

D) The same method as in part B can be used to calculate the amount of effort that has been done when the cable is pulling the tow truck at an angle of 35.0 degrees above the horizontal. The job is as follows because the angle is 35.0 degrees:

1110 N × 5.00 km × cos(35.0°) = 4539 kJ.

E) In section A, gravity has no effect on the automobile. Gravity pulls downward vertically, but displacement pulls horizontally. As a result, there is no work done by gravity because the angle formed by the force of gravity and the displacement is 90 degrees (cos(90°) = 0).

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The complete question is:

Using a cable with a tension of 1110 N, a tow truck pulls a car 5.00 km along a horizontal roadway.

A- How much work does the cable do on the car if it pulls horizontally?

B- How much work does the cable do on the car if it pulls at 35..0 degree above the horizontal?

C- How much work does the cable do on the tow truck if it pulls horizontally?

D- How much work does the cable do on the tow truck if it pulls at 35.0 degree above the horizontal?

E- How much work does gravity do on the car in part A?

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Answer:

Resultant force of the given forces: approximately \(11.2\; {\rm N}\) at approximately \(53^{\circ}\) west of south.

Explanation:

The resultant force of these forces will be:

Refer to the diagram attached. The resultant \(5\; {\rm N}\) force to the south and the \(10\; {\rm N}\) force to the west are perpendicular to each other, forming the two legs of a right triangle. The hypotenuse of this right triangle will be the net effect of these two forces.

Apply Pythagorean's Theorem on this triangle to find the magnitude of this net effect:

\(\begin{aligned}(\text{length of hypotenuse}) &= \sqrt{10^{2} + 5^{2}} \approx 11.2\end{aligned}\).

Hence, the magnitude of this net effect will be approximately \(11.2\; {\rm N}\).

Let \(\theta\) denote the angle between this resultant force and west. In this right triangle:

\(\begin{aligned} \tan(\theta) &= \frac{(\text{opposite})}{(\text{adjacent})} \\ &= \frac{5\; {\rm N}}{10\; {\rm N}} \\ &= \frac{1}{2}\end{aligned}\).

\(\begin{aligned} \theta &= \arctan\left(\frac{1}{2}\right) \approx 27^{\circ}\end{aligned}\).

Hence, the angle between this resultant force and south will be approximately \((90^{\circ} - 27^{\circ}) = 63^{\circ}\). This resultant force will be at approximately \(63^{\circ}\) west of south.

Answer:

B is the right answer

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The main answer for expressing the function n3/1000 − 100n2 − 100n + 3 in terms of Θ-notation is Θ(n3).

To find the Θ-notation of the given function, we need to determine the highest degree term in the polynomial expression. In this case, the highest degree term is n3/1000.

Since the constant factor (1/1000) does not affect the order of growth, we can simplify the expression to n3.

Therefore, the function is Θ(n3).

Considering the dominant term in the given function, we can express it in Θ-notation as Θ(n^3).

Summary: The function n3/1000 − 100n2 − 100n + 3 can be expressed in terms of Θ-notation as Θ(n3), which represents the order of growth of the function as n cubed.

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Answer:

F = 987600 N

Explanation:

Force Equation

F=ma

Newton's second law states that force is proportional to what is required for an object of constant mass to change its velocity. This is equal to that object's mass multiplied by its acceleration. We use Newtons, kilograms, and meters per second squared as our default units, although any appropriate units for mass (grams, ounces, etc.) or velocity (miles per hour per second, millimeters per second2, etc.) could certainly be used as well - the calculation is the same regardless.

Solve

F = m * a

F = 100 kg * 9876 m/s2

F = 987600 N

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

Answer:

the force pulls each other cause its just physics

Answer:

Kinetic

Explanation:

Its a moving action

When a bicyclist is pedaling up a hill, it is a form of potential energy. The correct option is option potential energy.

An object's position or configuration in relation to other things might provide it potential energy. The potential energy in this situation is gravitational potential energy.

Because they are being lifted against gravity as they ascend the hill, the cyclist gains gravitational potential energy. The potential energy of the cyclist increases as they ascend.

The potential energy will be transformed into kinetic energy once the cyclist begins pedaling or moving downward.

As a result of their increased height above the ground, cyclists are building up potential energy as they bike up a hill. They gain speed as they cycle downhill, transforming that potential energy into kinetic energy.

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Answer:

A. Alpha decay

Explanation:

Answer:

Beta decay (electron).

Explanation:

(a) The horizontal distance travelled by the bomb is 308.39 m.

(b) The velocity of the bomb just before it hits the ground is 116 m/s.

The horizontal distance travelled by the bomb is calculated by determining the time of motion of the bomb from the height.

h = vt + ¹/₂gt²

where;

The velocity, 350 km/h = 97.22 m/s

200 = (97.22 x sin25)t  +  ¹/₂(9.8)t²

200 = 41.1t + 4.9t²

4.9t² + 41.1t - 200 = 0

solve using formula method;

t = 3.5 seconds

The horizontal distance travelled by the bomb is calculated as;

x = ut

where;

x = (97.22 x cos25) x 3.5

x = 308.39 m

The final velocity of the bomb is calculated as follows;

vyf = vyi + gt

vyf = (97.22 x sin25) + (9.8 x 3.5)

vyf = 75.39 m/s

vxf = 97.22 x cos25

vxf = 88.11 m/s

vf = √ (75.39² + 88.1²)

vf = 116 m/s

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Answer:

first

Explanation:

Newton's Newton's first law of motion states that for every action and there's an equal and opposite reaction

Answer:

D. Third

Explanation:

Answer:summer

Explanation:

Answer:

first day of summer

Explanation:

There is a force exerted on your hand by the apple when the apple is in contact with your hand and you are holding it up against the force of gravity.

When you hold an apple in your hand, the apple is pressing against your hand with a force equal and opposite to the force your hand is applying on the apple. This force is due to the interaction between the atoms and molecules of your hand and the apple, which is known as the normal force.

As long as the apple remains in contact with your hand, there is a normal force acting on the apple and your hand, which prevents the apple from falling due to the force of gravity. Therefore, the force exerted on your hand by the apple is present when the apple is in contact with your hand and you are holding it up against the force of gravity.

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On prevent the puppies from tearing down the board, Ginger must apply 32 N of force to the opposite side.

The total force acting on the system is equal to the acceleration of the center of mass times the system's total mass. When applied to an extended object, Newton's second law, F = ma, predicts the motion of a specific reference point for this object.

The vector sum of all forces acting on an object is known as the net force. In other words, the net force is the sum of all the forces, taking into mind that a force is a vector and that two forces that have the same magnitude but facing the opposite direction will cancel each other out.

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The force of friction acting on him is 122 N. The correct answer is option C

Friction is a force that opposes the motion of a static or a moving object.

Given that Barry slides across an icy pond. The coefficient of kinetic friction between his shoes and the ice is 0.15. If his mass is 83 kg

The given parameters are;

The normal reaction N on the body = mg

N = 83 x 9.8

N = 813.4 N

The Frictional force formula is  \(F_{r}\) = μN

\(F_{r}\) = 0.15 x 813.4

\(F_{r}\) = 122.01 N

Therefore, the force of friction acting on him is 122 N approximately.

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Answer:

Yes, in case of uniform velocity

Explanation:

This is the case of uniform velocity. If a body covers equal displacement in equal intervals of time, then the velocity of a body is said to be ‘Uniform Velocity’. It meas that the velocity of a body remains constant during the motion and it does not change.

Since, acceleration is defined as the rate of change of velocity.

Therefore, if there is no change in velocity or in other words the change in velocity is zero, then the acceleration is also zero.

a = ΔV/t = 0/t

a = 0 m/s²

So, the acceleration of the body is 0 m/s², but it has a uniform velocity

Hence, it is possible for an object that, object with zero acceleration have velocity, which is the case case of uniform velocity.

266 m distance will be covered by a car uniformly accelerating from 12m/s to 26m/s in 14 seconds.

Uniform acceleration: If a body is travelling by increasing its velocity in equal intervals of time then that body is said to be in uniform acceleration.

s=ut+1/2 at² where as s= distance , u = initial velocity, t= time

to calculate the uniform acceleration a=\(\frac{v-u}{t}\) here v= final velocity u= initial velocity.

we have v=26ms and u=12m/s

a=\(\frac{26-12}{14}\) = 1 a=m/s².

now substitute a =1 in formula s=ut+1/2 at²

we get s=12(14)+1/2(1)(14)²= 266m

 Hence the distance covered by the  car which is uniformly accelerating from 12m/s to 26m/s in 14 seconds is 266 m.

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Given:

The resistance is,

The capacitance is,

The applied voltage is,

The frequency of the AC is,

To find:

The capacitive reactance

Explanation:

The capacitive reactance is,

Hence, the capacitive reactance is 0.106 ohm.