Relating the circumference to the force applied on the balloon, make a conclusion on the relationship of force to the rate of the change of motion of the balloon at a constant mass.​

Following are the answer:

Potential energy is the energy possessed by an object due to its position or configuration relative to other objects. It is the energy that an object has stored in it as a result of its position or state. An object's potential energy is often associated with its ability to do work, which can be released when the object is allowed to move or change its position.

(a) The potential energy stored in a spring stretched by a distance x from its equilibrium position can be calculated using the formula:

\(U = (1/2) k x^2\)

where k is the spring constant.

Substituting the given values, we get:

\(U = (1/2) (440.0 N/m) (0.0415 m)^2U = 0.0424 J\)

Therefore, the potential energy stored in the spring when it is stretched 4.15 cm from equilibrium is 0.0424 J.

(b) Using the same formula as in part (a), but with x = 0.0296 m, we get:

\(U = (1/2) (440.0 N/m) (0.0296 m)^2U = 0.0207 J\)

Therefore, the potential energy stored in the spring when it is stretched 2.96 cm from equilibrium is 0.0207 J.

(c) When the spring is unstretched, its potential energy is zero, since there is no displacement from equilibrium. So the potential energy stored in the spring is zero.

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The formula for potential difference (also known as voltage) is, V = ΔE/q, where V is the potential difference in volts (V), ΔE is the change in electric potential energy in joules (J), and q is the charge in coulombs (C).

Electric potential difference, also known as voltage, is a measure of the electric potential energy per unit of charge required to move a charge from one point to another in an electric circuit. It is the difference in electric potential between two points in an electric circuit.

The formula for potential difference, V = ΔE/q, reflects this relationship. The numerator, ΔE, represents the change in electric potential energy between the two points, while the denominator, q, represents the charge that moves between the two points.

For example, if a charge of +1 C moves from a point A to a point B in an electric circuit, and the electric potential energy at point B is greater than at point A by 1 J, then the potential difference between points A and B is 1 V. This means that it takes 1 J of energy to move a unit of charge from point A to point B in the circuit.

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The frequency of the sixth overtone would be 576 Hz.

To determine the length of a tube closed at one end, we can use the formula:

L = (v/4f) * (2n - 1)

where L is the length of the tube, v is the speed of sound in air, f is the fundamental frequency, and n is the harmonic number.

Given that the fundamental frequency is 96 Hz, we can substitute the values into the formula:

L = (v/4f) * (2n - 1)

L = (v/4*96) * (2*1 - 1)

L = v/384

To find the length at 0ºC, we need to know the speed of sound at that temperature. The speed of sound in air depends on temperature and other factors. At 0ºC, the speed of sound in dry air is approximately 331.4 m/s.

L = 331.4/384

L ≈ 0.862 m

Therefore, the tube closed at one end should have a length of approximately 0.862 meters when the air temperature is 0ºC.

To find the frequency of the sixth overtone, we multiply the fundamental frequency by six: Frequency of sixth overtone = 96 Hz * 6 = 576 Hz.

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Quantities that can be used to describe motion are B: speed and acceleration.

Motion is a change in position of an object over time. Motion is described in terms of displacement, distance, velocity, acceleration, time and speed.

Mass remain same , no matter where the object is taken . Whether the object is in motion or rest , its mass will remain constant , hence mass can never be used to describe the motion of an object and weight can not describe motion as weight is defined as the force of gravity with which a body is attracted to Earth , hence will not describe motion .

speed is the the rate at which someone or something moves or operates or is able to move or operate hence speed is used to describe the motion of an object

Acceleration is defined as the change in the velocity of an object with respect to time hence , it also have velocity in its formula which describes motion

hence correct option is B: speed and acceleration

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A line of best fit is a straight line drawn through the most points on a scatter plot, with an equal number of points above and below the line. It is used to investigate the nature of the relationship between two variables. The correct option is d.

A straight line with the best fit is one that minimizes the distance between it and some data.

In a scatter plot of varying data points, the line of best fit is used to express a relationship. It is a result of regression analysis and can be used to forecast indicators and price movements.

It is important to note that your line does not need to pass through any of the points on the plot; it only needs to bisect the area that contains the data points.

Thus, none of the options are correct, the correct one is d.

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The heat loss from the object in 13 minutes is \(Q = -1.503684 * 10^{-3} Joules\)

Heat loss can be calculated using this formula:

\(\frac{Q}{t} = oeA(T^{4} _{2} - T^{4} _{1})\\\)

It can be found that:

Q = heat loss in joules

t = time in seconds, 13 minutes = 780seconds

σ = (5.67 x 10⁻8J/s . m² . K⁴)

e = emissivity, 1

A = surface area of 1m²

T₂ = temperature of a room,  296 K

T₁ = temperature of an object, 330 K

Find \(Q\), where \(Q\) = \(\frac{Q}{t} = oeA(T^{4} _{2} - T^{4} _{1})\\\) = \(Q = oeA(T^{4} _{2} - T^{4} _{1}) t\\\)

\(Q = (5.67 x 10^{-8} J/s . m^{2} . K^{4} ) (1) (1) (296 K^{4} - 330 K^{4} ) (780)\\\)

\(Q = -0.001503684J\)

\(Q = -1.503684 * 10^{-3} Joules\)

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Answer:

x = 13552.6 m

Explanation:

This is a projectile throwing exercise

Let's start by looking for the time it takes for the pump to reach the ground y = 0

         y = y₀ + v_{oy} t - ½ g t²

as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t²

         t = \(\sqrt{2y_o/g}\)

         t = \(\sqrt{2 \ 10000/9.8}\)

         t = 45.175 s

with this time we can find the distance it travels horizontally

         x = v₀ₓ t

         x = 300 45,175

         x = 13552.6 m

the bomb must be dropped at this distance before hitting the target

Answer: 7

Explanation: 7 is the superior number

Answer:

1,269 N

Explanation:

I guessed and got the question right on CK-12's work practice.

Answer:

the distance x increases as the student increase the mass of the cannon.

Explanation:

a) From the question ; we get to understand that for each launch, the students use a different mass which is launch at  speed v relative to the ground.

This changes in the mass used brought about a change in the momentum at the same speed v ; perhaps an increase in momentum. However; since the conservation of the momentum is considered at each launching.

The momentum of the marble = momentum of the cannon

But since the momentum of the cannon increase ; therefore the same equivalent changes takes place in its kinetic energy . Therefore , the kinetic energy will increase and the distance will also increase in the bid to quench the amount of energy generated. Thus, the distance x increases as the student increase the mass of the cannon.

b)We all know that  conservation of the momentum will definitely  takes place after launching of the cannon.

Let assume that \(\rho\) is the momentum of the cannon with mass \(M_C\)

The kinetic energy of the canon will be:

\(\frac{\rho ^2 }{2 M_C}\)

Also the frictional force acting on the cannon is :

\(f = \mu mg\)

If the cannon move at an  additional distance x; the frictional force acting at this area quench the amount of the energy generated and consume the kinetic energy of the cannon;

So;

\(fx = K.E\)

\(fx = \frac{\rho ^2 }{2 M_C}\)

\(\mu mg x = \frac{\rho ^2 }{2 M_C}\)

\(x = \frac{\rho ^2 }{2 \mu mg M_C}\)

\(x = \frac{m_m^2 V^2 }{2 \mu mg M_C}\)

Thus; it is consistent with the answer in (a) as increase in the mass of the marble will bring about an increase in distance x

The increase in the mass of the marble increase the distance of the cannon.

Momentum:

Momentum of an object is equal to the product of the mass and the velocity of the object.

The kinetic energy of the cannon

\(K_C = \dfrac {\rho^2}{2 M_C}\)

Where,

\(\bold{\rho}\) - momentum of the cannon

Mc -  mass

The frictional force on cannon

\(\bold{F_f = \mu mg}\)

Cannon move a distance x,

Hence,

\(\bold {F_f \times x = K_C}\\\\\bold {\mu m g \times x = \dfrac {\rho^2}{2 M_C }}\\\\\bold {x = \dfrac {\rho^2}{2\mu m g M_C }} }\\\\\bold {x = \dfrac {m^2 V^2}{2\mu m g M_C } }\)

Therefore, The increase in the mass of the marble increase the distance of the cannon.

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Given data:

* The time taken by the box to hit the ground is 6.45 s.

* The initial velocity of the box is 15 m/s.

Solution:

By the kinematics equation, the height of the cliff is,

where u_y is the vertical initial velocity of the box,

The vertical initial velocity of the box is zero,

Substituting the known values,

Thus, the magnitude of the height of cliff is 204 meter.

Hence, the cliff was - 204 meter tall (measured in upward direction), and 1st option is the correct answer.

Answer:

1.142

Explanation:

Answer:

For people to reduce the inclined nature of a mountain, they make the roads longer by putting curved in the way. Riding a bicycle up a hill can be a very challenging task but riding it on a zig-zagged road makes it easier to go up and down. Making it curved ensures that it's flatter even though it becomes longer.

Answer:

Part A: \(1.97ms^{-1}\) (2 s.f.)
Part B: 0.33m (2 s.f.)

Explanation:

Part A:

Frequency = \(\frac{1}{period} = \frac{1}{3}\) Hz

Wavespeed = frequency x wavelength
Wavelength = distance between two crests = 5.90

Freq = 1/3 Hz

Therefore: Wavespeed = 1/3 x 5.90

                  Wavespeed = 1.967 ms^-1

Part B:
Amplitude = \(\frac{peak-to-peak- amplitude }{2}\)

Peak to peak amplitude = 0.65m

Amplitude = 0.65/2 = 0.325m = 0.33m 2sf

Answer:

If I say the correct answer is B.280

When two objects held close together and then released, they move toward each other. The phenomenon can be explained by the Coulomb's Law. Coulomb's Law is given by: $F=k\frac{q_{1}q_{2}}{r^2}$ where F is the electrostatic force, q1 and q2 are the magnitudes of the charges, r is the distance between the centers of the charged objects, and k is Coulomb's constant.

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Answer:

a)   F = 3.3 10⁴ N,  b)  F = 2.2 10³ N,  c) force when rigid is 15 times greater than when bending the knees

Explanation:

For this exercise we can use the relationship between work and the variation of kinetic energy

          K₀ = ½ m v²

the final kinetic energy is zero because the person is stationary

          W = (∑ F) x

          W = W x - F w

the weight is in the same direction of the displacement therefore the work is positive and the force applied, by the floor, is in the opposite direction to the displacement, consequently the work is negative

we substitute

          ( W-  F ) x = 0-K₀

           F = W + K₀ /x  

          F = mg + \(\frac{1}{2} \frac{mv^2 }{2x}\)  

          F = m ( g+  \(\frac{v^2 }{2x }\)  )

Let's use kinematics to find the velocity of the person when reaching the floor

           v² = v₀² - 2g (y + y₀)

the initial velocity is true and when reaching the ground y = 0

          v² = -2 g (0-yo)

we calculate

          v = \(\sqrt{2 ] 9.8 \ 0.800}\)

          v = - 3.96  m/s

the direction of this velocity is vertical down

let's calculate

a) x = 1.50 cm = 0.0150 m

           F = 62.0 (3.96² / 2 0.0150 + 9.8)

           F = 3.3 10⁴ N

b) x = 30 cm = 0.30 m

           F = 62.0 (3.96² / 2 0.30 + 9.8)

           F = 2.2 10³ N

c) to compare the force let's look for the relationship between the two

           \(\frac{F_{rigid} }{ F_{flexible} }\) = 3.3 10⁴ / 2.2 10³

           \frac{F_{rigid}  }{ F_{flexible} } = 15

therefore see that the force when rigid is 15 times greater than when bending the knees

Answer:

A. Increases

Explanation:

As altitude decreases, the amount of gas molecules in the air increases - the air becomes less dense. As altitude increases, the amount of gas molecules in the air decreases—the air becomes less dense.

Answer:

It decreases so it is B

Explanation:

As altitude rises, air pressure drops.

The velocity function is v(t)=1/2 t^2+4t + 3

The acceleration function is given by a(t) = t + 4 where v(0) = 3. The acceleration function a(t) is a known function of time. The velocity function is integral of the acceleration function plus a constant of integration. The time derivative of the velocity function is acceleration, hence d/dt  v(t)=a(t)

Taking indefinite integral of both sides,

∫d/dt  v(t)dt=∫a(t)dt+C1

Where C1 is the constant of integration. Since

∫d/dt  v(t)dt=v(t)

Hence,

v(t)=∫a(t)dt + C1

Inserting the acceleration function, a(t) = t + 4

v(t)=∫(t +4)dt + C1

v(t)=1/2 t^2+4t + C1

In order to find C1, let v = 3 when t = 0 as given

3=1/2 (0)^2+4(0) + C1

3=0 + C1

C1=3

Hence,

v(t)=1/2 t^2+4t + 3

Note: The question is incomplete. The complete question probably is: The acceleration function (in m/s-) and the initial velocity are given for a particle moving along a line a(t) = t + 4, v(0) = 3, 0 ≤ t ≤ 11 (a) Find the velocity at time t.

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Explanation:

Let the initial speed of 2nd stone is u2 and final speed of it be v2.

Using the equation

v= u + at for the 2nd stone

We get

14.7=u2 + 10t (a=g, acceleration due to gravity)

14.7=0 +10t

Or, t= 1.47s

This is the time taken by 2nd stone to reach 14.7m/s

Now,

Total time for 1st stone=1.04 + 1.47= 2.51s

Using the equation

\(s = ut + \frac{1}{2} at {}^{2} \)

for distance travelled by 1st stone

s1=u2t + 1/2at²

u2=0 (initially it was also at rest)

\(s1 = \frac{1}{2} \times 10 \times 2.51 {}^{2} \)

therefore S = 31. 5005m

Distance travelled by 2nd stone

\(s2 = u2 + \frac{1}{2} at {}^{2} \)

\(s2 = \frac{1}{2} \times 10 \times 1.47 {}^{2} \)

therefore S2 = 10.8m

Therefore,

Distance between the stones is= s1 - s2 = 31.5005–10.8 = 20.7005

Ans:- 20.7m

Based on the observations during the experiment, the car's velocity can be described as follows. .

The car had a   constant speed of approximately 60 km/h throughout the experiment,indicating a consistent rate of motion.

In terms of   direction, the car initially traveledin a straight line towards the east.

However, after a certain point,   it made a sharp turn towards the north, changing its direction but maintaining thesame speed.

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Answer:

The car takes approximately 6.3 seconds to slow from 22m/s to 3m/s with a constant acceleration of -2.1m/s². This is calculated using the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

Explanation:

Answer:

The car takes approximately 6.3 seconds to slow from 22m/s to 3m/s with a constant acceleration of -2.1m/s². This is calculated using the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time

Explanation:

Answer:

its constant i think

Explanation:

or its stable dunno which term will they be using

When a truck is moving north word having mass 2000kg at a constant speed of 50m/s the momentum of the truck is 100000 kg.m/s.

Momentum is defined as product of mass and velocity of the body. It is denoted by letter p and it is expressed in kg.m/s.

Mathematically p = mv. it discuss the moment of the body. body having zero mass or velocity has zero momentum.

The dimensions of the momentum is [M¹ L¹ T⁻¹].

In this problem velocity and mass of the truck is not given, Consider the mass of the truck is 2000kg and speed is 50m/s.

Given,

m = 2000kg

v = 50m/s

The momentum p = mv = 2000×50 =100000 kg.m/s

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Answer:

a spark

Explanation:

a spark would be the most smartest

If the universe were to suddenly begin shrinking rather than continuing to expand, it would have a significant effect on the cosmic microwave background radiation (CMB).

The CMB is the afterglow of the Big Bang and is observed as a nearly uniform background radiation in all directions. It is thought to have been emitted when the universe was about 380,000 years old and had cooled enough for neutral atoms to form.

If the universe were to suddenly begin shrinking, the photons in the CMB would lose energy as they travel through the contracting space. This would cause the CMB radiation to shift to shorter wavelengths, which is known as blue-shifting.

Therefore, the correct answer is B. It would blue-shift.

Answer:

he sphere that uses less time is sphere A

Explanation:

Let's start with ball A, for this let's use the kinematics relations

        v² = v₀² - 2g (y-y₀)

indicate that the sphere is released therefore its initial velocity is zero and when it reaches the floor its height is zero y = 0

         v² = 0 - 2 g (0- y₀)

         v = \(\sqrt{2g y_o}\)

         v = \(\sqrt{2 \ 9.8\ H}\)

         v = 4.427 √H

Now let's work the sphere B, in this case it rolls down a ramp, let's use the conservation of energy

starting point. At the highest point, before you start to move

         Em₀ = U = m g y

final point. At the bottom of the ramp

         Em_f = K = ½ m v² + ½ I w²

notice that we include the kinetic energy of translation and rotation

energy is conserved

          Em₀ = Em_f

          mg H = ½ m v² + ½ I w²

angular and linear velocity are related

          v = w r

          w = v / r

the momentorot of inertia indicates that it is worth

          I = \(\frac{2}{5}\) m r²

we substitute

           m g H = ½ m v² + ½ (\(\frac{2}{5}\)  m r²) (\(\frac{v}{r}\) )²

           gH = \(\frac{1}{2}\)  v² + \(\frac{1}{5}\)  v² = \(\frac{7}{10}\)  v²

           v = \(\sqrt{\frac{10}{7} \ g H}\)

           v = \(\sqrt{ \frac{10}{7} \ 9.8 \ H}\)

           v=3.742 √H

Taking the final speeds of the sphere, let's analyze the distance traveled, sphere A falls into the air, so the distance traveled is H.  The ball B rolls in a plane, so the distance (L) traveled can be found with trigonometry

           sin θ = H / L

           L = H /sin θ

we can see that L> H

In summary, ball A arrives with more speed and travels a shorter distance, therefore it must use a shorter time

Consequently the sphere that uses less time is sphere A

A) The rover’s coordinates and its distance from the lander at t = 2.0 s are; (1, 4) and 4.1 m

B) The rover’s displacement and average velocity vector during the interval are; s = (-1, 4) and v = (-0.5, 2) m/s

C) The magnitude and direction of the instantaneous velocity are; 2.24 m/s and 117°

The rover's x and y coordinates are given as;

x = 2.0m − (0.25 m/s²)t²

y = (1.0m/s)t + (0.25 m/s³)t³

A) At t = 2 s, the rovers coordinates are;

x = 2.0m − (0.25 m/s²)2²

x = 1 m

y = (1.0m/s)2 + (0.25 m/s³)2³

y = 4 m

Distance from the lander is;

s = √[(1 - 2)² + 4²]

s = 4.1 m

B) Let us first find the distance coordinates for the interval t = 0.0 s to t = 2.0s. Thus;

s = r - r₀

s = (1 - 2), (4 - 0)

s = (-1, 4)

Thus, average velocity vector is;

v = ¹/₂s

v =  ¹/₂(-1, 4)

v = (-0.5, 2) m/s

C) A general expression for the instantaneous velocity components is;

v_x = -0.5t

v_y = 1 - 0.75t²

Thus, v(2) is;

v_x = -0.5(2) = -1

v_y = 1 - 0.75(2)²

v_y = -2

Instantaneous velocity vector is; v = (-1, -2)

Magnitude of instantaneous Velocity = √(-1² + -2²) = 2.24 m/s

Direction = 180° - tan⁻¹(-2/-1) ≈ 117°

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