An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V battery is then disconnected and replaced with a 5.0 V battery, with the positive and negative terminals connected in the same manner as before. Part A How much additional charge flows to the positive plate? Express your answer in microcoulombs.

The answer is 45 degrees.  I am not doing a field experiment for you that involves a cannon and a day's work, for 5 points.

For 45° angle, the range of the pumpkin is maximum with constant initial velocity.

To find the answer, we need to know about the mathematical expression of range in terms of initial velocity and projectile angle.

                         = usin2Ф/g

Thus, we can conclude that the range will be maximum when the angle is 45° with constant initial velocity.

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Acceleration is given by:

[tex]a=\frac{v-u}{t}[/tex]

where

v is the final velocity

u is the initial velocity

t is the time interval

Let's apply the formula to the different parts of the problem:

A) [tex]-20.5 m/s^2[/tex]

Let's convert the quantities into SI units first:

[tex]u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s[/tex]

[tex]v=1600 km/h  \cdot \frac{1000 m/km}{3600 s/h}  =444.4 m/s[/tex]

t = 4.0 min = 240 s

So the acceleration is

[tex]a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2[/tex]

B) [tex]-3.8 m/s^2[/tex]

As before, let's convert the quantities into SI units first:

[tex]u = 444.4 m/s[/tex]

[tex]v=321 km/h  \cdot \frac{1000 m/km}{3600 s/h}  =89.2 m/s[/tex]

t = 94 s

So the acceleration is

[tex]a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2[/tex]

C) [tex]-53.0 m/s^2[/tex]

For this part we have to use a different formula:

[tex]v^2 - u^2 = 2ad[/tex]

where we have

v = 0 is the final velocity

u = 89.2 m/s is the initial velocity

a is the acceleration

d = 75 m is the distance covered

Solving for a, we find

[tex]a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2[/tex]

Final answer:

The rocket's acceleration was -20.49 m/s² during Stage A, -3.78 m/s² during Stage B, and -53.28 m/s² during Stage C, illustrating the varied methods employed to slow the vehicle during its descent onto Mars.

Explanation:

To calculate the rocket's acceleration during each stage of its descent onto Mars, we'll use the formula: a = Δv / Δt, where a is acceleration, Δv is the change in velocity, and Δt is the time taken for that change. It's essential to convert all units to meters and seconds for consistency.

Stage A:

Initial velocity = 19300 km/h = 5361.11 m/s

Final velocity = 1600 km/h = 444.44 m/s

Time = 4.0 min = 240 s

Acceleration = (444.44 m/s - 5361.11 m/s) / 240 s = -20.49 m/s²

Stage B:

Initial velocity = 1600 km/h = 444.44 m/s

Final velocity = 321 km/h = 89.17 m/s

Time = 94 s

Acceleration = (89.17 m/s - 444.44 m/s) / 94 s = -3.78 m/s²

Stage C:

This requires a different formula, a = v² / (2•d), where v is the final velocity (just before stage C, so 89.17 m/s), and d is the distance over which deceleration happens (75 m).

Acceleration = (89.17 m/s)² / (2• 75 m) = -53.28 m/s²

Answer:

(c) The current in each resistor is the same.

Explanation:

When two resistors are connected in series, we have the following:

- The resistors are connected such that the current passing through the two resistors is the same

- The voltage of the battery is equal to the sum of the voltage drops across each resistor

- the equivalent resistance of the circuit is equal to the sum of the individual resistances:

R = R1 + R2

So, let's analyze each statement:

(a) The resistor with the smaller resistance carries more current than the other resistor. --> FALSE. The current through the two resistors is the same.

(b) The resistor with the larger resistance carries less current than the other resistor. --> FALSE. The current through the two resistors is the same.

(c) The current in each resistor is the same. --> TRUE.

(d) The potential difference across each resistor is the same. --> FALSE: the potential difference across each resistor is given by

V=RI

where I (the current) is the same for both resistors, while R (the resistance) is not, so V is also different for the two resistors.

(e) The potential difference is greatest across the resistor closest to the positive terminal --> FALSE. According to

V=RI

the potential difference depends only on the value of the resistance, so it doesn't matter which resistor is connected to the positive terminal.

In a series circuit, the resistors have the same current flowing through them but can have different potential differences. The resistor with smaller resistance carries more current, while the one with larger resistance carries less current. The potential difference across each resistor is not the same.

In a series circuit, the resistors are connected one after the other, so the current passing through each resistor is the same. However, the potential difference (voltage) across each resistor can be different depending on their resistances.

Therefore, the correct statements are:

(a) The resistor with the smaller resistance carries more current than the other resistor. Since the current is the same in both resistors, the one with smaller resistance will have a larger potential difference across it. This means it carries more current.

(b) The resistor with the larger resistance carries less current than the other resistor. Again, since the current is the same in both resistors, the one with larger resistance will have a smaller potential difference across it. This means it carries less current.

(d) The potential difference across each resistor is the same. This statement is incorrect. As mentioned before, the potential difference across each resistor can be different depending on their resistances.

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Answer:

5 m

Explanation:

Work is equal to force times distance.

W = Fd

50 Nm = (10 N) d

d = 5 m

You move the box 5 m.

Answer:

Decreasing the cross-sectional area of the wire will increase the resistance of the wire.

Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.

Increasing the length of the wire will increase the resistance of the wire.

Explanation:

The resistance of a piece of wire is given by:

[tex]R=\rho \frac{L}{A}[/tex]

where

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

We notice that:

- The resistance of the wire is directly proportional to the resistivity of the material and the to the length of the wire

- The resistance of the wire is inversely proportional to the cross-sectional area of the wire

This means the following are correct:

Decreasing the cross-sectional area of the wire will increase the resistance of the wire.

Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.

Increasing the length of the wire will increase the resistance of the wire.

Answer:

21 psi

Explanation:

The weight of the car is:

W = mg

W = 1000 kg * 9.8 m/s²

W = 9800 N

Divided by 4 tires, each tire supports:

F = W/4

F = 9800 N / 4

F = 2450 N

Pressure is force divided by area, so:

P = F / A

P = (2450 N) / (0.13 m × 0.13 m)

P ≈ 145,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 21 psi

The pressure in the car's tires can be calculated using physics principles. By calculating the weight of the car and dividing it by the area of contact each tire has with the ground, the pressure is found to be approximately 21.46 psi.

To determine the tire pressure, we'll use the following physics concept: Pressure = Force / Area. The force in this case would be the weight of the car distributed among the four tires and the area would be the contact area of the tire with the ground.

First, we calculate the weight of the car by using the formula Weight = mass x gravity. Given that the mass of the car is 1000 kg and acceleration due to gravity is 10 m/s^2 (approx), the weight would be 10000 N (Newtons). Since each tire supports a quarter of the car's weight, the force each tire experiences is 2500 N.

Then, we calculate the area of contact the tire has with the ground. The flattened section would approximate a rectangle with length and width equal to 13 cm each, so Area = length x width = 0.13 m x 0.13 m = 0.0169 m^2.

Finally, we calculate the pressure by dividing the force by the area. Pressure = Force / Area = 2500 N / 0.0169 m^2 ≈ 148,000 Pa (Pascal).

Please note that 1 Pascal (Pa) = 0.000145 psi, so the pressure in psi would be approximately 21.46 psi, which is a bit lower than recommended by the tire manufacturer.

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The answer is:

The first option,  the walker traveled 360m more than the actual distance between the start and the end points.

Why?

Since each block is 180 m long, we need to calculate the vertical and the horizontal distance, in order to calculate how farther did the travel walk between the start and the end points (displacement).

So, calculating we have:

Traveler:

[tex]Distance=NorthCoveredDistance+EastCoveredDistance[/tex]

[tex]Distance=4*180m+3*180m=720m+540m=1260m[/tex]

Actual distance between the start and the end point (displacement):

[tex]ActualDistance=\sqrt{NorthDistance+EastDistance}\\\\ActualDistance=\sqrt{NorthDistance^{2} +EastDistance^{2}}\\\\ActualDistance=\sqrt{(720m)^{2} +(540m)^{2}}\\\\ActualDistance=\sqrt{518400m^{2} +291600m^{2}}\\\\ActualDistance=\sqrt{810000m^{2}}=900m[/tex]

Now, to calculate how much farter did the traveler walk, we need to use the following equation:

[tex]DistanceDifference=WalkerCoveredDistance-ActualDistance\\\\DistanceDifference=1260m-900m=360m[/tex]

Therefore, we have that distance differnce between the distance covered by the walker and the actual distance is 360m.

Hence, we have that the walker traveled 360m more than the actual distance between the start point and the end point.

Have a nice day!

Answer:

Work

Explanation:

Anita exerted the same amount of force but for twice the distance.  So she did twice as much work.

The federal government and industry are the main sources of funding for research, providing funding to Principal Investigators (PIs) through their organizations. The U.S. economy has increasingly relied on industry-funded research, but the government still plays a significant role in funding research.

The federal government is one of the main sources of funding for research, along with industry. The government provides funding to Principal Investigators (PIs) through the organizations they work for. Over time, the U.S. economy has relied more heavily on industry-funded research and development (R&D). However, the government still plays a significant role in funding research, especially in areas where private firms are not as active.

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Most funding for research comes from the federal government or industry and is provided to Principal Investigators (PIs) through their organizations.

Explanation:

Most of the funding for research comes from the federal government or industry and is provided to Principal Investigators (PIs) through the organizations for which they work. The federal government, through agencies such as the National Institutes of Health (NIH) and the National Science Foundation (NSF), provides grants for research in various fields. Industry-funded research, on the other hand, is supported by companies and private entities who invest in research and development projects.

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Answer:

[tex]6.60\cdot 10^5 Nm^2/C[/tex]

Explanation:

The electric flux through the rectangle is given by

[tex]\Phi = E A cos \theta[/tex]

where

E is the electric field strength

A is the area of the rectange

[tex]\theta[/tex] is the angle between the direction of the electric field and of the vector normal to the plane of the rectangle

In this problem we have

E = 125 000 N/C

The area of the rectangle is

[tex]A=2.50 m \cdot 5.00 m=12.5 m^2[/tex]

and the angle is

[tex]\theta=65.0^{\circ}[/tex]

so, the electric flux is

[tex]\Phi = (125,000 N/C)(12.5 m^2)(cos 65^{\circ})=6.60\cdot 10^5 Nm^2/C[/tex]

Doctors practice antiseptic medicine to D) prevent introducing infection-causing microbes to the patient. Managing bacterial infections requires careful use of antibiotics to avoid killing beneficial bacteria and prevent antibiotic resistance.

The reason why doctors practice antiseptic medicine is to avoid introducing microbes that could infect the patient. This is an essential part of infection control in medical settings. While antibiotics are valuable in treating bacterial infections by killing the bacteria causing the illness, they also impact the good bacteria that protect the body from infection. When deciding on using antimicrobial drugs, doctors must consider the type of bacterial infection and the potential resistance of bacteria to available antibiotics.

Antibiotic resistance is a significant concern in medicine, as it can lead to infections that are harder to treat, requiring longer and more expensive hospital stays. It emerges from the bacteria's adaptation through natural selection, effectively making certain drugs less effective over time.

Hmmm. Kilowatts should be converted to watts. Simply just move the decimal place to the right three times.

10,000 W / 15.8 A = V

632.9, or 633.

The voltage of the electricity will be 632.9 V. Electric power is found as the multiplication of the voltage and current. Option B is correct.

Electric power is the product of the voltage and current. Its unit is the watt. It is the rate of the electric work done.

The given data in the problem is;

V is the voltage = ? Volt (V)

Electric current (I)= 15.8 amps (A)

P is the power =10.0 kilowatts =10⁴ watt

The formula for the power is given as;

[tex]\rm P= V I \\\\\ 10^4= V \times 15.8 \\\\ V=632.9 \ V[/tex]

The voltage of the electricity will be 63.29 V.

Hence, option B is correct.

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(a) 0.5 s

In a simple harmonic motion, the period is equal to the reciprocal of the frequency:

[tex]T=\frac{1}{f}[/tex]

where f is the frequency

For this simple harmonic oscillator, the frequency is

f = 2.0 Hz

So the period is

[tex]T=\frac{1}{2.0 Hz}=0.5 s[/tex]

(b) 12.56 rad/s

The angular frequency is given by

[tex]\omega = 2 \pi f[/tex]

where

f is the frequency

In this problem,

f = 2.0 Hz

So the angular frequency is

[tex]\omega = 2 \pi (2.0 Hz)=12.56 rad/s[/tex]

(d) 0.419 rad

The displacement of the system can be written as

[tex]x(t) = A cos (\omega t+\phi)[/tex] (1)

where A is the amplitude and [tex]\phi[/tex] is the phase constant.

The velocity is the derivative of the displacement:

[tex]v(t) = x'(t) = -A\omega sin(\omega t+\phi)[/tex] (2)

Here we know that

at t=0, x(5)=5.0 cm and v(t)=-28 cm/s. So we can rewrite the ratio (2)/(1) as

[tex]\frac{v(t)}{x(t)}=\frac{-28 cm/s}{5.0 cm}=\frac{-\omega A sin(\phi)}{A cos(\phi)}=-\omega tan \phi[/tex]

And re-arranging the equation we can find the phase constant:

[tex]\phi = tan^{-1} (\frac{v}{\omega x})=tan^{-1} (\frac{-28 cm/s}{(5.0 cm)(12.56 rad/s)})=0.419 rad[/tex]

(c) 5.47 cm

The displacement of the system can be written as

[tex]x(t) = A cos (\omega t+\phi)[/tex] (1)

at t=0, x=5.0 cm, so using the values we found for [tex]\omega, \phi[/tex] we can now solve the equation to find A, the amplitude:

[tex]A=\frac{x}{cos(\omega t+\phi)}=\frac{5.0 cm}{cos(0.419 rad)}=5.47 cm[/tex]

(e) 68.7 cm/s

The maximum speed in a simple harmonic system is given by

[tex]v=\omega A[/tex]

where in this case we have

[tex]\omega=12.56 rad/s[/tex]

[tex]A=5.47 cm[/tex]

Substituting the numbers into the formula, we find

[tex]v=(12.56 rad/s)(5.47 cm)=68.7 cm/s[/tex]

(f) 862.9 cm/s^2

The maximum acceleration in a simple harmonic system is given by

[tex]a=\omega^2 A[/tex]

where in this case we have

[tex]\omega=12.56 rad/s[/tex]

[tex]A=5.47 cm[/tex]

Substituting the numbers into the formula, we find

[tex]a=(12.56 rad/s)^2(5.47 cm)=862.9 cm/s^2[/tex]

(g) 0.04 J

The total energy of the system is equal to the kinetic energy when the speed of the system is maximum: this occurs at x=0 (equilibrium position), where the elastic potential energy is zero, and all the energy is just kinetic energy:

[tex]E=K=\frac{1}{2}mv_{max}^2[/tex]

where we have

m = 170 g = 0.170 kg is the mass

[tex]v_{max}=68.7 cm/s = 0.687 m/s[/tex] is the maximum speed

Substituting into the equation,

[tex]E=K=\frac{1}{2}(0.170 kg)(0.687 m/s)^2=0.04 J[/tex]

(h) 3.65 cm

The position of the system is given by

[tex]x(t) = A cos (\omega t+\phi)[/tex]

where we have

[tex]\omega=12.56 rad/s[/tex] is the angular frequency

[tex]A=5.47 cm[/tex] is the amplitude

[tex]\phi = 0.419 rad[/tex] is the phase constant

Substituting t=0.4 s, we find the position at this time:

[tex]x = (5.47 cm) cos ((12.56 rad/s)(0.4 s)+0.419 rad)=3.65 cm[/tex]

The SHM the oscillating and periodic motion of an object. a) T = 0.5 s. b) ω = 12.56 rad/seg. c) A = 5.476 cm. d) Φ = 0.4186 rad. e) Vmax = 68.778 cm/s. f) a = 863 m/s². g) Ek = 0.04012 J. h) X = 4.08 cm

It is a rectilinear movement performed by a oscillating and periodic movil.

It is periodic because it repeats in a certain intervale of time.

The time -in seconds- it takes to the movil to make a complete oscillation is called Period, T.

The frequency, f, refers to the number of complete oscillations (cycles) completed per second. f = 1/T (Hz).

When considering a simple harmonic motion, we need to know that there will be always a restoring force (F), that tends to take the movil to the original position when it moves a distance named amplitud.

There are some significant components to consider in this motion,

Available data:

(a) the period, T

T = 1/f

T = 1/2

T = 0.5 s

(b) the angular frequency, ω

ω = 2πf

ω = (2π)2Hz

ω = 12.56 rad/seg

(c) the amplitude, A

It can be cleared from the following displacement formula,

X = A sen (ω t + Φ)

First, we need to get Φ.

Φ = arctan (v/ωx) = tan⁻¹(v/ωx)

Φ = tan⁻¹(-28/12.56x5)

Φ = tan⁻¹(-28/62.8)

Φ = tan⁻¹(0.445)

Φ = 0.4186 rad

So now we can calculate A

X = A sen (ω t + Φ)

5 = A sen (12.56 x 0 + 0.4186)

5 = A sen (0.4186)

A = 5/cos(0.4186)

A = 5/0.913

A = 5.476 cm

(d) the phase constant, Φ

Φ = arctan (v/ωx) = tan⁻¹(v/ωx)

We did this in the previous step.

Φ = 0.4186 rad

(e) the maximum speed, Vmax

Vmax = A ω

Vmax = 5.476 x 12.56

Vmax = 68.778 cm/s

(f) the maximum acceleration

a = ω ² A

a = 12.56² x 5.476

a = 157.75 x 5.476

a = 863 m/s²

(g) the total energy

Total energy = kinetic energy + Potential energy = Ek + Ep

At x=0 ⇒ v=max, Ep = 0 ⇒ Et = Ek + Ep = Ek

Ek = 1/2 mv²

Ek = 1/2 (0.170 kg x 0.687²m/s)

Ek = 0.04012 J.

(h) the position at t = 0.4 s

X = A sen (ω t + Φ)

X = 5.476 cm sen (12.56 rad/seg x 0.4 + 0.4186 rad)

X = 5.476 sen (5.024 + 0.4186)

X = 5.476 sen (5.4426)

X = 4.08 cm

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Answer:

(c) The central bright fringe becomes wider, because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger.

Explanation:

The formula that gives the angle of the first minimum of the diffraction pattern from a single-slit is

[tex]sin \theta = \frac{\lambda}{w}[/tex]

where

[tex]\lambda[/tex] is the wavelength of the light

w is the width of the slit

We see that the angle is inversely proportional to the width of the slit: therefore, if the width of the slit is reduced (so, w is decreased), the angle that locates the first minimum [tex]\theta[/tex] increases, and so the central bright fringe becomes wider.

Reducing the width of a single slit through which light passes leads to a wider central bright fringe. This occurs because the narrowed slit causes greater divergence of light rays, leading to a larger angle for locating the first dark fringe.

The phenomenon being discussed here pertains to physics, specifically light diffraction and interference. When light passes through a single slit, it exhibits a diffraction pattern with a central bright fringe flanked by smaller, dimmer fringes on either side. This pattern is influenced by the width of the slit and the wavelength of the light source.

For the scenario provided in the question, light passes through a single slit that is then narrowed. The correct answer is (c): The central bright fringe becomes wider because the angle that locates the first dark fringe on either side of the central bright fringe becomes larger.

This is explained by the fact that as the slit narrows, the rays of light diverge more upon exiting the slit, due to the wave nature of light. The larger spread of these rays leads to a larger angle for locating the first dark fringe, which subsequently results in a wider central bright fringe.

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(a) 25.2 m/s

Let's take the initial vertical position of the rock as "zero" (reference height).

According to the law of conservation of energy, the speed of the rock as it reaches again the position "zero" after being thrown upwards is equal to the initial speed of the rock, 21 m/s (in fact, if there is no air resistance, no energy can be lost during the motion; and since the kinetic energy depends only on the speed of the rock:

[tex]K=\frac{1}{2}mv^2[/tex]

and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)

This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using

[tex]v^2 = u^2 + 2gd[/tex]

where

u = 21 m/s is the initial speed of the rock as it enters the hole

g = 9.8 m/s^2 is the acceleration due to gravity

d = 10 m is the depth of the hole

Substituting,

[tex]v=\sqrt{u^2 +2gd}=\sqrt{(21 m/s)^2+2(9.8 m/s^2)(10 m)}=25.2 m/s[/tex]

(b) 4.72 s

The vertical position of the rock at time t is given by

[tex]y(t) = v_y t - \frac{1}{2}gt^2[/tex]

where

[tex]v_y = 21 m/s[/tex] is the initial vertical velocity

Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:

[tex]-10 = 21 t - \frac{1}{2}(9.8)t^2\\10+21 t -4.9t^2 = 0[/tex]

which has two solutions:

t = -0.43 s --> negative, so we discard it

t = 4.72 s --> this is our solution

The rock's velocity as it hits the bottom of the hole is approximately 32.19 m/s. The rock is in the air for approximately 2.14 seconds.

a) To find the rock's velocity as it hits the bottom of the hole, we can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. In this case, the rock's initial velocity is 21 m/s, the acceleration is the acceleration due to gravity (-9.8 m/s^2), and the displacement is the depth of the hole (-10 m). Plugging in these values, we get:

v^2 = (21 m/s)^2 + 2(-9.8 m/s^2)(-10 m)

Simplifying, we find that v^2 = 841 + 196 = 1037. Taking the square root of both sides, we get v = √1037 ≈ 32.19 m/s. So, the rock's velocity as it hits the bottom of the hole is approximately 32.19 m/s.

b) To find the time the rock is in the air, we can use another equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 21 m/s, the acceleration is -9.8 m/s^2, and we want to find the time. Plugging in these values, we get:

0 = 21 m/s - 9.8 m/s^2 * t

Simplifying, we find that t ≈ 2.14 seconds. So, the rock is in the air for approximately 2.14 seconds.

Answer:

[tex]v=14.5(m/s)[/tex].

Explanation:

This is a projectile motion problem, so for solving it, we will require some equations of uniform motion and free-fall.

First, we need to recognize the data that we have so we can start solving the problem.

For Free -Fall:

[tex]g= -9.8(m/s^{2})[/tex]

[tex]\Delta y=2.4(m)[/tex]

For uniform motion:

[tex]\Delta x=17(m)[/tex]

The time that takes to the ball to travel 17m horizontally and to hit the crossbar at 2.4m of height is the same, so time is a common variable for free-fall and uniform motion.

Using the equations:

[tex]\Delta y=v_{y0}t+0.5gt^{2}[/tex] and

[tex]\Delta x=v_{x}t[/tex]

and noticing that

[tex]v_{y0}=v*sin(38)[/tex] and

[tex]v_{x}=v*cos(38)[/tex],

we obtain

[tex]\Delta y=v*sin(38)t+0.5gt^{2}[/tex] and

[tex]\Delta x=v*cos(38)t[/tex]

wich is a system of two equations and to variables that can be easily solve.

Making

[tex]v=\frac{\Delta x}{t*cos(38)}[/tex]

we get

[tex]\Delta y=(\frac{\Delta x}{t*cos(38)})sin(38)t+0.5gt^{2}[/tex],

[tex]\Delta y=\Delta x*tan(38)+0.5gt^{2}[/tex],

[tex]\Delta y-\Delta x*tan(38)=0.5gt^{2}[/tex],

[tex]t=\sqrt{\frac{\Delta y-\Delta x*tan(38)}{0.5g}}[/tex],

[tex][tex]t=\sqrt{\frac{2.4-17*tan(38)}{0.5*(-9.8)}}[/tex]

so

[tex]t=1.50s[/tex],

now the speed can be easily compute from one of our equations. Using

[tex]v=\frac{\Delta x}{t*cos(38)}[/tex],

[tex]v=\frac{17}{1.5*cos(38)}[/tex],

[tex]v=14.5(m/s)[/tex].

To find the initial speed of the ball, we can analyze the projectile motion of the ball. We can break down the initial velocity of the ball into horizontal and vertical components, and then use kinematic equations to solve for the initial speed. We can calculate the time it takes for the ball to reach the goal crossbar and then use that time and the horizontal distance to find the initial speed.

To find the initial speed of the ball, we need to analyze the projectile motion of the ball. Given that the ball passes over the goal, 2.4 m above the ground, we can use the kinematic equations to solve for the initial speed. In this case, we know the horizontal distance (17 m) and the vertical distance (2.4 m above the ground). We can break down the initial velocity of the ball into horizontal and vertical components.

The horizontal component of the initial velocity (Vx) can be found using the equation:

Vx = V * cos(theta)

where V is the initial speed and theta is the angle of the kick. The vertical component of the initial velocity (Vy) can be found using the equation:

Vy = V * sin(theta)

Using these values, we can calculate the time it takes for the ball to reach the goal crossbar, which is the same as the time it takes for the ball to hit the ground. The time can be found using the equation:

t = 2 * Vy / g

where g is the acceleration due to gravity. Finally, we can use the time and the horizontal distance to find the initial speed using the equation:

Vx = distance / t

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(a) [tex]1.78\cdot 10^{-6}N[/tex]

Here we want to find the gravitational force exerted on the particle at a distance of 3.7 m from the center of the sphere. Since the radius of the sphere is 1.9 m, we are outside the sphere, so we can use Newton's law of gravitation:

[tex]F=G\frac{mM}{r^2}[/tex]

where

G is the gravitational constant

m = 8.3 kg is the mass of the particle

[tex]M=4.4\cdot 10^4 kg[/tex] is the mass of the sphere

r = 3.7 m is the distance

Substituting into the formula, we find

[tex]F=(6.67\cdot 10^{-11})\frac{(8.3 kg)(4.4\cdot 10^4 kg)}{(3.7 m)^2}=1.78\cdot 10^{-6} N[/tex]

(b) [tex]1.46\cdot 10^{-6}N[/tex]

Here we want to find the gravitational force exerted on the particle at a distance of 0.41 m from the center of the sphere. Since the radius of the sphere is 1.9 m, this time we are inside the sphere, so the formula for the gravitational force is different:

[tex]F=G\frac{mMr}{R^3}[/tex]

where

G is the gravitational constant

m = 8.3 kg is the mass of the particle

[tex]M=4.4\cdot 10^4 kg[/tex] is the mass of the sphere

r = 0.41 m is the distance from the centre of the sphere

R = 1.9 m is the radius of the sphere

Substituting numbers into the formula, we find

[tex]F=(6.67\cdot 10^{-11})\frac{(8.3 kg)(4.4\cdot 10^4 kg)(0.41 m)}{(1.9 m)^3}=1.46\cdot 10^{-6}N[/tex]

(c) [tex]F=G\frac{mMr}{R^3}[/tex]

The magnitude of the gravitational force on the particle when located inside the sphere can be found starting from Newton's law of gravitation:

[tex]F=G\frac{mM'}{r^2}[/tex] (1)

where the only difference compared to the standard law is that M' is not the total mass of the sphere, but only the amount of mass of the sphere enclosed by the spherical surface of radius r centered in the center of the sphere.

The mass enclosed is

[tex]M'=\rho V' = \rho (\frac{4}{3}\pi r^3)[/tex] (2)

where [tex]\rho[/tex] is the density of the sphere and V' is the enclosed volume. We can rewrite the density of the sphere as ratio between mass of the sphere (M) and volume of the sphere:

[tex]\rho=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}[/tex] (3)

where R is the radius of the sphere.

Substituting (3) into (2):

[tex]M' = (\frac{M}{\frac{4}{3}\pi R^3}) (\frac{4}{3}\pi r^3)=\frac{Mr^3}{R^3}[/tex]

And substituting the last equation into (1), we find

[tex]F=G\frac{m(\frac{Mr^3}{R^3})}{r^2}=G\frac{mMr}{R^3}[/tex]

which depends linearly on r.

1. 0.66 s

The current induced in the loop is

I = 0.20 A

while the resistance of the loop is

[tex]R=0.032 \Omega[/tex]

so the emf induced in the loop can be found by using Ohm's law

[tex]\epsilon = RI=(0.032 \Omega)(0.20 A)=0.0064 V[/tex]

We also know that according to Faraday-Newmann-Lenz, the induced emf is

[tex]\epsilon=-\frac{\Delta \Phi}{\Delta t}[/tex] (1)

where

[tex]\Delta \Phi[/tex] is the variation of magnetic flux through the loop

[tex]\Delta t[/tex] is the time interval

We have:

- Initial magnetic field: B = 1.5 T

- Radius of the coil: r = 30 mm = 0.03 m

- So, area of the coil: [tex]A=\pi r^2 = \pi (0.03 m)^2=2.83\cdot 10^{-3} m^2[/tex]

so the initial flux through the coil is

[tex]\Phi_i = B_i A = (1.5 T)(2.83\cdot 10^{-3}m^2)=4.25\cdot 10^{-3} Wb[/tex]

While the final flux through the coil is zero, since the magnetic field at the end is zero, so the change in magnetic flux is

[tex]\Delta \Phi = \Phi_f - \Phi_i = 0-4.25\cdot 10^{-3} Wb=-4.25\cdot 10^{-3} Wb[/tex]

Now re-arranging eq.(1) we find the time interval needed:

[tex]\Delta t = -\frac{\Delta \Phi}{\epsilon}=-\frac{-4.25\cdot 10^{-3} Wb}{0.0064 V}=0.66 s[/tex]

2. Same direction as the external magnetic field

The direction of the induced magnetic field is given by Lenz's law.

In fact, Lenz law states that the induced current in the loop is such that the magnetic field opposes the variation of magnetic flux through the coil.

Here, the magnetic flux through the coil is decreasing, since the external magnetic field is decreasing: this means that the induced magnetic field must be in the same direction as the external magnetic field (in order to restore the flux and to oppose this decrease of flux).

The amount of time it would take for the magnitude of the uniform magnetic field to drop is 0.66 seconds.

Given the following data:

Radius = 30 mm to meters = 0.02 m.

Resistance = 0.032 Ω.

Magnetic field = 1.5 T.

Current = 0.20 A.

In order to determine the amount of time it would take for the magnitude of the uniform magnetic field to drop from 1.5 Tesla to zero (0), we would apply Faraday-Newmann-Lenz equation.

For the wire's area:

[tex]A = \pi r^2\\\\A= 3.142 \times 0.03^2\\\\A=2.83 \times 10^{-3}\;m^2[/tex]

For the initial magnetic flux:

[tex]\phi _i = B_i A\\\\\phi _i = 1.5 \times 2.83 \times 10^{-3}\\\\\phi _i = 4.25 \times 10^{-3}\;Wb[/tex]

For the change in magnetic flux:

[tex]\Delta \phi = \phi_f - \phi_i\\\\\Delta \phi =0-4.25\times 10^{-3}\\\\\Delta \phi =-4.25\times 10^{-3}\;Wb[/tex]

From Faraday-Newmann-Lenz equation, we have:

[tex]\epsilon =-\frac{\Delta \phi}{\Delta t} \\\\\Delta t=-\frac{\Delta \phi}{\epsilon} \\\\\Delta t=-(\frac{-4.25\times 10^{-3}}{0.20 \times 0.032})\\\\\Delta t=\frac{4.25\times 10^{-3}}{0.0064}\\\\\Delta t=0.66\;seconds[/tex]

The direction of the induced magnetic field that is caused by this current is in the same direction as the external magnetic field in accordance with Lenz's law.

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The Compton Shift [tex]\Delta \lambda[/tex] in wavelength when the photons are scattered is given by the following equation:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)[/tex]     (1)

Where:

[tex]\lambda_{c}=2.43(10)^{-12} m[/tex] is a constant whose value is given by [tex]\frac{h}{m_{e}.c}[/tex], being [tex]h=4.136(10)^{-15}eV.s[/tex] the Planck constant, [tex]m_{e}[/tex] the mass of the electron and [tex]c=3(10)^{8}m/s[/tex] the speed of light in vacuum.

[tex]\theta=41\°[/tex] the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are deflected at [tex]41\°[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(41\°))[/tex]   (2)

[tex]\Delta \lambda=\lambda' - \lambda_{o}=5.950(10)^{-13}m[/tex]   (3)

But we are asked to express this in [tex]nm[/tex], so:

[tex]\Delta \lambda=5.950(10)^{-13}m.\frac{1nm}{(10)^{-9}m}[/tex]  

[tex]\Delta \lambda=0.000595nm[/tex]  (4)

The initial energy [tex]E_{o}=265keV=265(10)^{3}eV[/tex] of the photon is given by:

 [tex]E_{o}=\frac{h.c}{\lambda_{o}}[/tex]    (5)

From this equation (5) we can find the value of [tex]\lambda_{o}[/tex]:

[tex]\lambda_{o}=\frac{h.c}{E_{o}}[/tex]    (6)

[tex]\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{265(10)^{3}eV}[/tex]    

[tex]\lambda_{o}=4.682(10)^{-12}m[/tex]    (7)

Knowing the value of [tex]\Delta \lambda[/tex] and [tex]\lambda_{o}[/tex], let's find [tex]\lambda'[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}[/tex]

Then:

[tex]\lambda'=\Delta \lambda+\lambda_{o}[/tex]  (8)

[tex]\lambda'=5.950(10)^{-13}m+4.682(10)^{-12}m[/tex]  

[tex]\lambda'=5.277(10)^{-12}m[/tex]  (9)

Knowing the wavelength of the scattered photon [tex]\lambda'[/tex]  , we can find its energy [tex]E'[/tex] :

[tex]E'=\frac{h.c}{\lambda'}[/tex]    (10)

[tex]E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{5.277(10)^{-12}m}[/tex]    

[tex]E'=235.121keV[/tex]    (11) This is the energy of the scattered photon

If we want to know the kinetic energy of the recoiling electron [tex]E_{e}[/tex], we have to calculate all the energy lost by the photon in the wavelength shift, which is:

[tex]K_{e}=E_{o}-E'[/tex]  (12)

[tex]K_{e}=265keV-235.121keV[/tex]  

Finally we obtain the kinetic energy of the recoiling electron:

[tex]E_{e}=29.878keV[/tex]  

Answer:

The first one:

the energy of the scattered x-ray

The answer for last on:

Kinetic energy of the recoiling electron

Answer:

Wire A has twice the resistance of wire B.

Wire A and wire B have the same resistivity.

Explanation:

- Resistivity is a property of a material, that tells how much is the material able to oppose to the flow of current through it. The value of the resistivity of a wire depends on the material only: this means that two wires made of the same material have same resistivity. Since both wire A and B here are made of copper, they have the same resistivity.

- Resistance of a wire instead is given by

[tex]R=\rho \frac{L}{A}[/tex]

where

[tex]\rho[/tex] is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

Here, the two wires have same resistivity and same cross-sectional area, while wire A is twice as long as wire B (so, L for A is twice the value of L for B): therefore, the resistance of wire A will be twice that of wire B.

Explanation:

The heat (thermal energy) absorbed by the iron skillet can be found using the following equation:

[tex]Q=m.C.\Delta T[/tex]   (1)

Where:

[tex]Q[/tex] is the heat

[tex]m=45 g[/tex] is the mass of the element (iron in this case)

[tex]C[/tex] is the specific heat capacity of the material. In the case of iron is [tex]C=0.444\frac{J}{g\°C}[/tex]

[tex]\Delta T=T_{f}-T_{i}=21\°C - 6\°C= 15\°C[/tex] is the variation in temperature

Knowing this, lets rewrite (1) with these values:

[tex]Q=(45 g)(0.444\frac{J}{g\°C})(15\°C)[/tex]  (2)

Finally:

[tex]Q=299.7 J[/tex]  

Answer:

299.7 J

Explanation:

Answer:

434.0 V/m

Explanation:

The power output of the laser is:

[tex]P=5 mW = 0.005 W[/tex]

while the radius of the beam is

[tex]r=\frac{5 mm}{2}=2.5 mm = 0.0025 m[/tex]

so the cross-sectional area is

[tex]A=\pi r^2 = \pi (0.0025 m)^2=2.0\cdot 10^{-5} m^2[/tex]

So the intensity of the laser beam is

[tex]I=\frac{P}{A}=\frac{0.005 W}{2.0\cdot 10^{-5} m^2}=250 W/m^2[/tex]

The intensity of a laser beam is related to the magnitude of the electric field by

[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]

where

c is the speed of light

[tex]\epsilon_0[/tex] is the vacuum permittivity

Solving the formula for E, we find

[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(250 W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=434.0 V/m[/tex]

Answer:

Latin

Explanation:

In order for the scientists to have a common and official name for a particular thing that can be understood by every scientist in the world, a single language has been established for the purpose. The language chosen is the Latin language. The official scientific names are given in this language, so it is a necessity for the scientists to know and understand this language. The terms that are commonly used are regional, and they come in many different languages, which is why this language has been chosen. Occasionally, the ancient Greek language is used as well, though much less than the Latin.

Answer: English

Explanation: I would say the Answer is English but I’m not Sure!

Answer:

[tex]8.4\cdot 10^7 N[/tex]

Explanation:

The electrostatic force between two objects is given by:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the separation between the two objects

In this problem, we have two boxes separated by

[tex]r = 1\cdot 10^6 m[/tex]

The first box contains [tex]6.02\cdot 10^{23}[/tex] protons, so its charge is:

[tex]q_1 = (6.02\cdot 10^{23})(1.6\cdot 10^{-19} C)=9.63\cdot 10^4 C[/tex]

The second box contains [tex]6.02\cdot 10^{23}[/tex] electrons, so its charge is:

[tex]q_2 = (6.02\cdot 10^{23})(-1.6\cdot 10^{-19} C)=-9.63\cdot 10^4 C[/tex]

We are only interested in the magnitude of the force, so we can neglect the negative sign and calculate the electrostatic force as:

[tex]F=(9\cdot 10^9) \frac{(9.63\cdot 10^4 C)(9.63\cdot 10^4 C)}{(1\cdot 10^6 m)^2}=8.4\cdot 10^7 N[/tex]

Answer:

Nuclear fusion reactions are difficult to accomplish because they only occur at extremely high temperatures

Explanation:

Nuclear fusion is a nuclear process that occurs when nuclei of light elements fuse together forming a nucleus of a heavier element. In this process, the total mass of the final nucleus is smaller than the sum of the masses of the initial nuclei, so part of the mass has been converted into energy according to the equation

[tex]E=\Delta mc^2[/tex]

where

[tex]\Delta m[/tex] is the mass change

c is the speed of light

Due to the huge value of c ([tex]c=3.0\cdot 10^8 m/s[/tex]), we see that even for a small amount of mass change, the energy released in this process is huge.

In fact, nuclear fusion is the process that occurs in the core of the stars, that allow stars to continuously produce huge amount of energy. However, nuclear fusion can only occur at very high temperatures (million of Kelvins), because the nuclei need to have enough kinetic energy in order to overcome the electrostatic repulsion that occurs between them (nuclei are made of protons, so they naturally repel each other). This is the reason why it is currently difficult to accomplish nuclear fusion reactions on Earth, because it is not easy to create environments with such extremely high temperatures.

Several research groups around the world have been able to use fusion reactions on a small scale to produce a net output of energy (more energy is produced than was needed to get the fusion reaction to occur) is true about nuclear fusion. Correct option is d.

a. This statement is not entirely true. Nuclear fusion is considered a potentially promising method for the production of electricity because it has the potential to generate vast amounts of clean energy with abundant fuel (e.g., isotopes of hydrogen). However, one of the challenges of nuclear fusion is dealing with the by-products, which can include high-energy neutrons that can activate surrounding materials, making them radioactive. The radioactivity from fusion by-products may not be as long-lived or as toxic as the by-products of nuclear fission, but it still presents technical challenges that need to be addressed for practical fusion power plants.

b. This statement is not accurate. The reaction responsible for the hydrogen bomb (thermonuclear bomb) is a two-stage process. The first stage involves the fission of uranium or plutonium, which produces the high temperatures and pressures needed to initiate the fusion reaction. The second stage is the fusion of hydrogen isotopes (deuterium and tritium) to form helium and release additional energy.

c. This statement is true. Nuclear fusion reactions require extremely high temperatures and pressures to overcome the electrostatic repulsion between positively charged atomic nuclei. The most promising approach for achieving these conditions is by using magnetic confinement or inertial confinement, which are methods that require sophisticated technology and substantial energy input.

d. This statement is true. Although nuclear fusion is not yet fully realized as a viable large-scale energy source, there have been significant advances in fusion research, and several research groups and experimental reactors have achieved net energy gain, where more energy is produced from the fusion reaction than is used to sustain it. However, it's important to note that the current challenges lie in sustaining and scaling these reactions to achieve continuous, controlled, and economically feasible fusion energy production.

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Complete question is:

Which statement is true about nuclear fusion?

a. Nuclear fusion is not considered a good method for the production of electricity because the by-products of nuclear fusion are even more toxic and long-lived than those from nuclear fission.

b. The reaction responsible for the hydrogen bomb is four hydrogen nuclei combining to form a helium nucleus.

c. Nuclear fusion reactions are difficult to accomplish because they only occur at extremely high temperatures.

d. Several research groups around the world have been able to use fusion reactions on a small scale to produce a net output of energy (more energy is produced than was needed to get the fusion reaction to occur).

Answer:

Niels Bohr

Explanation:

Niels Bohr was scientist whose model showed electrons location around a nucleus and all in an orbit.The orbit had varying energies according to their sizes.This model further indicated that when an electron is lost or gained, radiation is absorbed or emitted.The modern atomic model was prepared from quantum mechanics which shows how matter differ in atomic and subatomic levels.In this model, the chances of finding an electron is higher where the cloud is more dense.Bohrs model resembles the electron cloud model is that they both talk about position of electrons in an atom, and the energy associated with size of the orbit occupied by the electron.

1) 13.8 V

We can use the transformer equation:

[tex]\frac{N_p}{N_s}=\frac{V_p}{V_s}[/tex]

where we have

[tex]N_p = 300[/tex] is the number of turns in the primary coil

[tex]N_s=18[/tex] is the number of turns in the secondary coil

[tex]V_p=230.0 V[/tex] is the voltage in the primary coil

[tex]V_s = ?[/tex] is the voltage in the secondary coil

Solving for Vs, we find

[tex]V_s = \frac{N_s}{N_p}V_p=\frac{18}{300}(230.0 V)=13.8 V[/tex]

2) 5.17 A

For an ideal transformer, the power in input is equal to the power in output, so we can write:

[tex]P_{in} = P_{out}\\V_p I_p = V_s I_s[/tex]

where

[tex]V_p=230.0 V[/tex] is the voltage in the primary coil

[tex]V_s = 13.8 V[/tex] is the voltage in the secondary coil

[tex]I_p=0.31 A[/tex] is the current in the primary coil

[tex]I_s = ?[/tex] is the current in the secondary coil

Solving for Is, we find

[tex]I_s = \frac{I_p V_p}{V_s}=\frac{(0.31 A)(230.0 V)}{13.8 V}=5.17 A[/tex]

Final answer:

The change in kinetic energy of the bullet-block system can be calculated using the principle of conservation of kinetic energy. The initial kinetic energy of the bullet is given by 0.5 * mass of the bullet * (initial velocity of the bullet)^2, and the final kinetic energy is given by 0.5 * mass of the bullet * (final velocity of the bullet)^2. The change in kinetic energy is the difference between the final and initial kinetic energies.

Explanation:

The change in kinetic energy of the bullet-block system can be calculated using the principle of conservation of kinetic energy. The initial kinetic energy of the bullet is given by 0.5 * mass of the bullet * (initial velocity of the bullet)^2, and the final kinetic energy is given by 0.5 * mass of the bullet * (final velocity of the bullet)^2. The change in kinetic energy is the difference between the final and initial kinetic energies.

Using the values given in the question:

Mass of the bullet = 22.3 g = 0.0223 kg

Initial velocity of the bullet = 1000 m/s
Final velocity of the bullet = 100 m/s

Initial kinetic energy of the bullet = 0.5 * 0.0223 kg * (1000 m/s)^2 = 111.5 J
Final kinetic energy of the bullet = 0.5 * 0.0223 kg * (100 m/s)^2 = 11.15 J

Change in kinetic energy of the bullet-block system = Final kinetic energy - Initial kinetic energy = 11.15 J - 111.5 J = -100.35 J

Answer:

[tex]5.23\cdot 10^3 kg/m^3[/tex]

Explanation:

The density of the solid is

[tex]d = 5.23\cdot 10^{-6}kg/mm^3[/tex]

we want to convert it into kg/m^3. We must note that:

[tex]1 m^3 = 1 m \cdot 1 m \cdot \1m =1000 mm\cdot 1000 mm \cdot 1000 mm=1\cdot 10^9 mm^3[/tex]

Therefore, the conversion can be done as follows:

[tex]d=5.23\cdot 10^{-6} \frac{kg}{mm^3} \cdot (1\cdot 10^9 \frac{mm^3}{m^3}) =5.23\cdot 10^3 kg/m^3[/tex]

(A) 40 J

Explanation:

The initial potential energy stored in the spring is:

[tex]U=40 J[/tex]

this energy is stored in the spring when the spring is compressed by a certain amount [tex]\Delta x[/tex], such that the elastic potential energy of the spring is

[tex]U=\frac{1}{2}k(\Delta x)^2[/tex]

where k is the spring constant. On the contrary, when it is at rest, the kinetic energy of the stone is zero:

[tex]K=\frac{1}{2}mv^2 = 0[/tex]

because the speed is zero (v=0).

When the spring is released, the spring returns to its equilibrium position, so that

[tex]\Delta x = 0[/tex]

and

[tex]U=0[/tex]

so the elastic potential energy becomes zero: so the total energy must be conserved, this means that all the potential energy has been converted into kinetic energy of the spring, so 40 J.

(B) 40 J

When the stone starts its motion, its kinetic energy is 40 J:

K = 40 J

While its gravitational potential energy is zero:

U = mgh = 0

where m is the mass of the stone, g is the gravitational acceleration, and h=0 is the height when the stone is thrown up.

As the stone goes up, its gravitational potential energy increases, since h in the formula increases; this means that the kinetic energy decreases, since the total energy must be constant.

When the stone reaches its maximum height, its speed becomes zero:

v = 0

This means that

K = 0

And so all the kinetic energy has been converted into gravitational potential energy, therefore

U = 40 J

(C) 40.8 m

At the maximum height of the trajectory of the stone, we have that the gravitational potential energy is

[tex]U=mgh = 40 J[/tex]

where

m = 0.1 kg is the mass of the stone

g = 9.8 m/s^2 is the acceleration due to gravity

h is the maximum height

Solving the formula for h, we find:

[tex]h=\frac{U}{mg}=\frac{40 J}{(0.1 kg)(9.8 m/s^2)}=40.8 m[/tex]

(D) The initial compression of the spring must be increased by a factor [tex]\sqrt{2}[/tex]

Here we want to double the maximum height reached by the stone:

h' = 2h

In order to do that, we must double its gravitational potential energy:

U' = 2U

This means that the initial potential energy stored in the spring must also be doubled, so

U' = 80 J

The elastic potential energy of the spring is

[tex]U' = \frac{1}{2}k(\Delta x)^2[/tex]

We see that the compression of the spring can be rewritten as

[tex]\Delta x = \sqrt{\frac{2U'}{k}}[/tex]

And we see that [tex]\Delta x[/tex] is proportional to the square root of the energy: therefore, if the energy has doubled, the compression must increase by a factor [tex]\sqrt{2}[/tex].