A ball is projected upward at time t= 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to:____________A. 9.0 sB. 9.4 sC. 9.7 sD. 8.7 sE. 10 s

To calculate the resultant force on the aircraft as it accelerates from the catapult, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

Since all of the kinetic energy at take-off is from the work done on the aircraft by the catapult, we can equate the work done by the catapult to the kinetic energy of the aircraft at take-off, that is:

Work done by catapult = kinetic energy of aircraft at take-off

The work done by the catapult is given by the force provided by the catapult multiplied by the distance over which it acts, that is:

Work done by catapult = force × distance

The distance over which the force provided by the catapult acts is given as 150 m, so we have:

Work done by catapult = force × 150

The kinetic energy of the aircraft at take-off is given by:

(1/2) × mass × velocity^2

Since the aircraft is initially at rest, its initial velocity is zero, so we have:

kinetic energy of aircraft at take-off = 0.5 × mass × (final velocity)^2

Using the work-energy principle, we can equate the two expressions for work done and kinetic energy, that is:

force × 150 = 0.5 × mass × (final velocity)^2

Solving for force, we get:

force = 0.5 × mass × (final velocity)^2 / 150

Therefore, the resultant force on the aircraft as it accelerates is given by:

force = 0.5 × mass × (final velocity)^2 / 150

Note that we need to know the mass and final velocity of the aircraft in order to calculate the resultant force.

The tension in the cable is equal to the weight of the elevator plus the force required to accelerate it upward. We can use this fact to find the acceleration of the elevator.

Tension in the cable = mg + ma

where m is the mass of the elevator, g is the acceleration due to gravity, and a is the acceleration of the elevator.

Rearranging this equation, we get:

a = (Tension in the cable - mg) / m

Substituting the given values, we get:

a = (35.2 kN - 2415 kg x 9.81 m/s^2) / 2415 kg

a = 0.26 m/s^2

Since the elevator is moving upward, the acceleration is in upward direction, so we enter a positive value. Therefore, the acceleration of the elevator is 0.26 m/s^2.

Newtons 3rd law of motion states there is always equal and opposite reaction.

The answer would be a positive value and the same as the first part.

3.38x10^7

Answer:

\(F=32.24N\)

Explanation:

From the question we are told that:

Height \(h= 2.75 m\)

Length\(l = 5.25 m\)

Mass \(m=45kg\)

Final speed \(v_f=6.81\)

Generally the equation for Potential Energy P.E is mathematically given by

\(P.E=mgh\)

Therefore

Initial potential energy

\(P.E_1=45*9.8*2.75 \\\\P.E_1= 1212.75 J\)

Generally the equation for Kinetic Energy K.E is mathematically given by

\(K.E=0.5mv^2\)

Therefore

Final kinetic energy

\(K.E_2= 1/2*45*6.81*6.81 \\\\K.E_2= 1043.46J\)

Generally the equation for Work_done is mathematically given by

\(W=P.E_1-K.E_2\\\\W=169.3\)

Therefore

\(F=\frac{W}{d}\\\\F=\frac{169.3}{5.25}\)

\(F=32.24N\)

According to the question: the net force on the roof is 8.45 kN.

The vector sum of the forces operating on a particle of object is known as the net force in mechanics. The original forces' impact on the motion of the particle is replaced by the net force, which is a single force. In accordance with Newton's second rule of motion, it causes the particle to accelerate at a rate equal to the sum of all those actual forces.

The net force on the roof is equal to the air pressure multiplied by the surface area of the roof. The air pressure is equal to the density of the air multiplied by the wind speed squared. Therefore, the net force on the roof can be calculated as follows:
Force = (density x wind speed²) x surface area
Force = (1.29 kg/m² x (40 m/s)²) x (250 cm²)
Force = 8.45 kN
Therefore, the net force on the roof is 8.45 kN.

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The work-energy theorem states that the change in the kinetic energy of an object will be equal to the net work done on the object.

Mathematically, it can be expressed as;

ΔKE = W

Where; ΔKE represents the change in kinetic energy of the object,

W represents the net work done on the object.

This theorem states that when work is done on an object, it results in a change in its kinetic energy. If work is done on an object, its kinetic energy increases, and if work is done by an object, its kinetic energy decreases.

This theorem is a fundamental principle in physics that relates the concepts of work and energy, and it is often used to analyze the motion and behavior of objects in various physical systems.

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Let the circumference of the circle be 10L.

A moves at 10L/2 = 5L per hour

B moves at 10L/5 = 2L per hour

Therefore it takes 10L/(5L+2L) = 10/7 hours

Answer:

Transverse wave

Explanation:

I hope this helps you! :)

Answer:

Ions

Explanation:

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.

Answer:

The mass of the object is 0.53 kg.

Explanation:

We can use the internal energy of an ideal gas equation do evaluate the mass.

\(\sf \Delta U=mc \Delta T\)

Where

\(\sf \Delta U\) is the change in internal energy

\(\sf m\) is the mass of the object

\(\sf c\) is the specific heat capacity of the object

\(\sf \Delta T\) is the change in temperature

We can rearrange the equation to isolate the mass (\(\sf m\)).

Divide both sides of the equation by \(\sf c \Delta T\).

\(\sf \dfrac{\Delta U}{\sf c \Delta T} = \sf \dfrac{mc \Delta T}{\sf c \Delta T}\)

\(\sf c \Delta T\) cancels out on the right side leaving us with

\(\sf \dfrac{\Delta U}{\sf c \Delta T} = \sf m\)

Numerical Evaluation

In this example we are given

\(\sf \Delta U=1943\) J

\(\sf c=90\) J/°C/kg

\(\sf \Delta T=41\) °C

Substituting our given values into the equation yields

\(\sf m=\sf \dfrac{1943}{90 \cdot 41}\)

\(\sf m=0.526558265\)

Rounding to 2 decimal places leaves us with

\(\sf m=0.53\)

The region a represents the distance of the electron from the nucleus.

According to the wave mechanical model of the atom, the probability of finding an electron within a given volume element (representing the atom) is the square of the wave function psi.

Since a is the region in space where there is the greatest probability of finding the electron in the atom, it follows that distance of the electron form the atom is always a.

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Answer:

The frequency of light can be calculated using the formula:

`c = λv`

Where `c` is the speed of light in a vacuum, `λ` is the wavelength of light, and `v` is the frequency of light.

The speed of light in a vacuum is `3.00 × 10^8 m/s`.

To convert the wavelength from nanometers to meters, we need to divide by `1 × 10^9`.

Thus, the frequency of light with a wavelength of 655 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(655 × 10^-9 m)`

`v = 4.58 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`.

Similarly, the frequency of light with a wavelength of 515 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(515 × 10^-9 m)`

`v = 5.83 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz`.

Finally, the frequency of light with a wavelength of 475 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(475 × 10^-9 m)`

`v = 6.32 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

So, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz` and the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

Answer:

\(1277.867\ \text{N}\)

Explanation:

m = Mass of person = 80 kg

r = Radius of swing = 12 m

g = Acceleration due to gravity = \(9.81\ \text{m/s}^2\)

The tension in the rope would be

\(T=mg+\dfrac{mv^2}{r}\\\Rightarrow T=80\times 9.81+\dfrac{80\times 8.6^2}{12}\\\Rightarrow T=1277.867\ \text{N}\)

The tension in the rope is \(1277.867\ \text{N}\).

Answer:The formula for kinetic energy is

(1/2) M V^2.

With M in kg and V in m/s, the answer will be in Joules.

K.E = 40joule

Explanation:

Answer:

2Ω

Explanation:

Ohm's law:

V = IR

10 V = (5 A) R

R = 2 Ω

Answer:

40 m

Explanation:

4 m/s   * 10 s   = 40 m  

  ( see how the 's' cancels out and you're left with "m" as your answer?)

Mark's work decreases when he climbs the same flight of stairs again in the same amount of time without the cello.

The correct answer is option D.

The amount of work Mark does depends on the weight of the cello, as well as the distance he climbs and the time it takes. Work is calculated using the formula :

Work = Force × Distance.
In the given scenario, Mark is carrying a full-sized cello while climbing the stairs. The weight of the cello adds to the force he exerts. So, the total force Mark exerts is the weight of the cello plus his own weight (375 N).

When Mark climbs the stairs with the cello, he is doing work against the force of gravity.

The work done is equal to the force exerted multiplied by the distance climbed (375 N + weight of cello) × 4 m.

Now, if Mark were to climb the same flight of stairs again in the same amount of time (3 seconds), but this time without the cello, the amount of work he does would decrease. This is because without the cello, the force exerted would only be Mark's weight (375 N), which is less than the total force exerted with the cello.

Therefore, mark's work decreases.

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To find the distance between two charges, we can use Coulomb's law, which states that the force between two charges is proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

The formula for Coulomb's law is:

F = (k * |q1| * |q2|) / r^2

where:

F is the force between the charges,

k is the electrostatic constant (k = 8.99 × 10^9 N m^2/C^2),

|q1| and |q2| are the magnitudes of the charges, and

r is the distance between the charges.

Given:

|q1| = 2 × 10^-7 C

|q2| = 4.5 × 10^-7 C

F = 0.1 N

k = 8.99 × 10^9 N m^2/C^2

We can rearrange the equation to solve for r:

r^2 = (k * |q1| * |q2|) / F

Plugging in the values:

r^2 = (8.99 × 10^9 N m^2/C^2 * 2 × 10^-7 C * 4.5 × 10^-7 C) / 0.1 N

r^2 = (8.99 × 2 × 4.5) * (10^9 * 10^-7 * 10^-7) / 0.1

r^2 = 80.91 * (10^9 * 10^-7 * 10^-7) / 0.1

r^2 = 80.91 * 10^(-7 + 9 - 1)

r^2 = 80.91 * 10^1

r^2 = 809.1

Taking the square root of both sides:

r = √809.1

r ≈ 28.46

Therefore, the distance between the two charges is approximately 28.46 units.

Answer: c

Explanation:

bc x=1 if x^1

To solve this problem, we need to determine the frictional force acting on the block with different magnitudes of the applied force.

First, we need to find the normal force on the block, which is equal to the weight of the block. The weight of the block is given by:

W = mg = 2.5 kg x 9.8 m/s^2 = 24.5 N

Next, we need to find the force of the applied vertical force, which is given in the problem as "is". We can use trigonometry to find the vertical component of the force:

Fv = is sinθ

where θ is the angle between the force and the horizontal surface. Since the problem does not give us the value of θ, we will assume it to be 0°, which means the force is purely horizontal.

(a) If the magnitude of the applied force is 8.0 N, then the frictional force can be calculated as:

Ff = μsFn = μs(mg - Fv) = 0.40(24.5 - 0) = 9.8 N

(b) If the magnitude of the applied force is 10 N, then the frictional force can be calculated as:

Ff = μsFn = μs(mg - Fv) = 0.40(24.5 - 10) = 5.8 N

(c) If the magnitude of the applied force is 12 N, then the frictional force can be calculated as:

Ff = μkFn = μk(mg - Fv) = 0.25(24.5 - 12) = 3.1 N

Therefore, the magnitude of the frictional force acting on the block is 9.8 N, 5.8 N, and 3.1 N, for applied forces of 8.0 N, 10 N, and 12 N, respectively.

(a) When the horizontal force is 8 N the frictional force is 11.8 N.

(b) when the applied force is 10 N; the frictional force is 13.8 N.

(c) when the applied force is 12 N; the frictional force is 15.8 N.

(a) The magnitude of the frictional force on the block when the horizontal force is 8 N is calculated as;

F - Ff = ma

where;

F - μmg = ma

6 - 0.4 x 2.5 x 9.8 = 2.5 a

2.5 a = -3.8

a = -3.8/2.5

a = -1.52 m/s²

when the applied force is 8 N;

8 N - Ff = -1.52 m/s² x 2.5 kg

Ff = 11.8 N

(b) when the applied force is 10 N;

10 N - Ff = -1.52 m/s² x 2.5 kg

Ff = 13.8 N

(c) when the applied force is 12 N;

12 N - Ff = -1.52 m/s² x 2.5 kg

Ff = 15.8 N

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Answer:

a increases

Explanation:

as distance between two objects increases the gravitational force decreases so when distance decreases the gravitational force increases

Answer:

(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.

Explanation:

Given that,

a ball is spun around in circular motion such that it completes 50 rotations in 25 s.

(1). Let T be the period of its rotation. It can be calculated as follows :

\(T=\dfrac{25}{50}\\\\T=0.5\ s\)

(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,

\(f=\dfrac{50}{25}\\\\f=2\ Hz\)

Hence, this is the required solution.