A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open at the top. Friction is absent. The spring constant of the spring is 3900 N/m. The piston has a negligible mass and a radius of 0.020 m.

a) When air beneath the piston is completely pumped out, how much does the atmospheric pressure cause the spring to compress?
b) How much work does the atmospheric pressure do in compressing the spring?

Answers

Answer 1
a) We know that:
Pressure = Force / Area

And for a spring,
F = kx, where k is the spring constant and x is the change in length.

P = kx/A; P = 101325 Pa
101325 = 3900x / (0.02²π)
x = 0.033 m or 3.3 cm

b) Work = 1/2 kx²
= 1/2 * 3900 * (0.033)²
= 2.12 Joules

Related Questions

Draw a picture of the bohr atom and label the three particles that comprise it

Answers

smh like that


cheers


)))

Below is a picture of the bohr atom and the three particles that comprise it.

When an electron moves from one energy level to another, it either absorbs or emits a photon of electromagnetic radiation.

The Bohr model of the atom is a description of the structure of atoms, especially that of hydrogen, proposed by the physicist Niels Bohr in 1913. The model was a radical departure from earlier, classical descriptions of the atom, and it was the first that incorporated quantum theory.

Therefore,  Bohr model of the atom consists of a small, dense nucleus surrounded by orbiting electrons. The nucleus is positively charged, and the electrons are negatively charged. The electrons are attracted to the nucleus by the electromagnetic force, but they are also moving around the nucleus, so they are not pulled in.

what is the critical temperature of compound x?

Answers

the critical temperature of component X is 350
Final answer:

The critical temperature of a compound is the point at which its gas phase cannot be liquefied by pressure and for binary mixtures, it represents the temperature where two immiscible liquids become fully miscible.

Explanation:

The critical temperature of a compound is the temperature above which the gas phase of the substance cannot be made to liquefy, no matter how much pressure is applied. For a binary mixture, this can also refer to the upper critical temperature where two immiscible liquids become completely miscible in all proportions. An example provided is that of methyl acetate and carbon disufide, which has a critical temperature of approximately 230 K at one atmosphere. A similar behavior occurs with hexane/nitrobenzene mixtures, which have a critical temperature of 293 K.

These temperatures represent the critical point at which there is no distinction between the liquid and the gas phase of a substance. This is shown by the fact that at the critical temperature and the associated critical pressure, the gas does not condense. Understanding and utilizing the critical temperature of compounds is important in various applications, including the development and use of high-temperature superconductors.

A firefighter mounts the nozzle of his fire hose a distance 36.9 m away from the edge of a burning building so that it sprays from ground level at a 45° angle above the horizontal. After quenching a hotspot at a height of 8.85 m, the firefighter adjusts the nozzle diameter so that the water hits the building at a height of 17.9 m. By what factor was the nozzle diameter changed? Assume that the diameter of the hose stays the same, and treat the water as an ideal fluid.

Answers

Answer:

diameter is decreased by factor 0.91

Explanation:

As we know by the equation of trajectory

[tex]y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}[/tex]

here as per first given situation we know that

x = 36.9 m

y = 8.85 m

[tex]\theta = 45^0[/tex]

now from above equation we have

[tex]8.85 = 36.9 tan45 - \frac{(9.8)(36.9)^2}{2(v^2)cos^245}[/tex]

[tex]8.85 = 36.9 - \frac{13343.8}{v^2}[/tex]

[tex]\frac{13343.8}{v^2} = 28.05[/tex]

[tex]v = 21.8 m/s[/tex]

now similarly after nozzle is adjusted we have

y = 17.9 m

x = 36.9 m

[tex]\theta = 45^0[/tex]

now again from equation we have

[tex]17.9 = 36.9 tan45 - \frac{(9.8)(36.9)^2}{2(v'^2)cos^245}[/tex]

[tex]17.9 = 36.9 - \frac{13343.8}{v'^2}[/tex]

[tex]\frac{13343.8}{v'^2} = 19[/tex]

[tex]v' = 26.5 m/s[/tex]

Now by equation of continuity we can find the change in diameter

as we know that

[tex]A_1v_1 = A_2v_2[/tex]

now we have

[tex]\pi d_1^2 v_1 = \pi d_2^2 v_2[/tex]

[tex]d_1^2 (21.8) = d_2^2(26.5)[/tex]

[tex]\frac{d_1}{d_2} = \sqrt{\frac{26.5}{21.8}}[/tex]

[tex]\frac{d_1}{d_2} = 1.10[/tex]

so we have

[tex]\frac{d_2}{d_1} = 0.91[/tex]

so diameter is decreased by factor of 0.91

The nozzle diameter changed by a factor of 0.907

[tex]\texttt{ }[/tex]

Further explanation

Acceleration is rate of change of velocity.

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

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

[tex]\texttt{ }[/tex]

Given:

horizontal distance = x = 36.9 m

angle of projection = θ = 45°

initial height = y₁ = 8.85 m

final height = y₂ = 17.9 m

Asked:

ratio of nozzle diameter = d₂ : d₁ = ?

Solution:

The motion of the water is a parabolic motion.

Firstly, we will calculate the time taken for the water to reach the hotspot:

[tex]x = (u \cos \theta) t[/tex]

[tex]t = x \div ( u \cos \theta )[/tex]

[tex]t = x \div ( u \cos 45^o )[/tex]

[tex]\boxed {t = \frac{\sqrt{2}x}{u}}[/tex]

[tex]\texttt{ }[/tex]

Next , we could calculate the initial speed (u) of the water as it leaves the nozzle:

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

[tex]y = (u \sin 45^o)( \frac{\sqrt{2}x}{u} ) - \frac{1}{2}g ( \frac{\sqrt{2}x}{u} )^2[/tex]

[tex]y = x - \frac{gx^2}{u^2}[/tex]

[tex]\frac{gx^2}{u^2} = x - y[/tex]

[tex]u^2 = \frac{gx^2}{x - y }[/tex]

[tex]u = \sqrt{ \frac{gx^2}{x - y } }[/tex]

[tex]\boxed {u = x \sqrt{ \frac{g}{x - y} }}[/tex]

[tex]\texttt{ }[/tex]

Finally , we could find the ratio of the diameter by using Continuity Equation as follows:

[tex]u_1 A_1 = u_2 A_2[/tex]

[tex]u_1 \frac{1}{4} \pi (d_1)^2 = u_2 \frac{1}{4} \pi (d_2)^2[/tex]

[tex](d_2)^2 : (d_1)^2 = u_1 : u_2[/tex]

[tex](d_2)^2 : (d_1)^2 = x \sqrt{ \frac{g}{x - y_1} } : x \sqrt{ \frac{g}{x - y_2} }[/tex]

[tex](d_2)^2 : (d_1)^2 = \sqrt { x - y_2 } : \sqrt { x - y_1}[/tex]

[tex]\frac {d_2}{d_1} = \sqrt[4] { \frac {x - y_2} {x - y_1} }[/tex]

[tex]\frac {d_2}{d_1} = \sqrt[4] { \frac {36.9 - 17.9} {36.9 - 8.85} }[/tex]

[tex]\frac {d_2}{d_1} \approx 0.907[/tex]

[tex]d_2 \approx 0.907 \times d_1[/tex]

[tex]\texttt{ }[/tex]

Learn moreVelocity of Runner : https://brainly.com/question/3813437Kinetic Energy : https://brainly.com/question/692781Acceleration : https://brainly.com/question/2283922The Speed of Car : https://brainly.com/question/568302

[tex]\texttt{ }[/tex]

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

You buy a plastic dart gun, and being a clever physics student
you decide to do a quick calculation to find its maximum
horizontal range.
You shoot the gun straight up, and it takes 4.0 seconds for the
dart to land back at the barrel (ignoring air resistance).
(a) What angle must your fire your gun to reach this max distance?
(b) What is the maximum horizontal range of your dart gun?
(c) What are the components of your initial velocity?

Answers

C-consider a projectile launched with an initial velocity of 50 m/s at an angel of 60 degrees such a projectile begins its motion with a horizontal velocity of 25 m/s and a vertical velocity of 43 m/s. These are known as the horizontal and vertical components of the initial velocity.above the horizontal When we shoot the gun straight up, we can find the speed of dart when it leaves the gun: 0 = 2 Where – is a time of a dart’s flight. The horizontal range of an object shouted at angle: = 0 2 sin(2) And the maximum range at 45°: = 0 2 = ( 2 ) 2 = 2 4 = 9.81 2 ⋅ (5.6) 2 4 ≈ 77 Answer:: ≈ 77

Answer:

Part a)

[tex]\theta = 45 degree[/tex]

Part b)

[tex]R = 39.2[/tex]

Part c)

[tex]v_x = 13.86 m/s[/tex]

[tex]v_y = 13.86 m/s[/tex]

Explanation:

Since it took 4.0 s to come back at the same position so we can say

[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]

[tex]0 = v_y(4.0) - \frac{1}{2}(9.8)(4.0)^2[/tex]

[tex]v_y = 19.6 m/s[/tex]

Part a)

Now we know that horizontal range of projectile is given as

[tex]R = \frac{v^2 sin2\theta}{g}[/tex]

Now this range would be maximum if the angle of the projectile is giving maximum value of sine

so we have

[tex]sin(2\theta) = 1[/tex]

[tex]\theta = 45 degree[/tex]

Part b)

For maximum range we have

[tex]R = \frac{v^2}{g}[/tex]

[tex]R = \frac{19.6^2}{9.8}[/tex]

[tex]R = 39.2[/tex]

Part c)

Since we projected at an angle of 45 degree

so the components are given as

[tex]v_x = 19.6 sin45 = 13.86 m/s[/tex]

[tex]v_y = 19.6 cos45 = 13.86 m/s[/tex]

A LASIK vision-correction system uses a laser that emits 10-ns-long pulses of light, each with 3.0 mJ of energy. The laser beam is focused to a 0.80-mm-diameter circle on the corneaWhat is the electric field amplitude of the light wave at the cornea?.

Answers

Final answer:

To calculate the electric field amplitude of light from a Lasik laser, start from the pulse energy and duration to find the power and intensity, then use known constants and the equation I = cε₀E²/2 to derive the amplitude.

Explanation:

The electric field amplitude of the light wave at the cornea from a Lasik laser can be calculated using the energy of light and its other characteristics. Given the pulse energy E = 3.0 mJ and pulse duration t = 10 ns, we first calculate the instantaneous power Pi during one pulse using the relationship: Pi = E / t.

Then, given the diameter of the laser beam d = 0.80 mm, we can calculate the area of the laser beam A = π(d/2)². So, we find the intensity of the pulse by dividing the power by the area: I = Pi / A.

As the intensity of an electromagnetic wave (light wave in this case) is proportional to the square of the electric field amplitude (E), we can deduce the equation I = cε₀E²/2, where c is the speed of light and ε₀ is the permittivity of free space. We solve for E to get the amplitude of the electric field: E = sqrt(2I / cε₀).

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The electric field amplitude of the light wave at the cornea is approximately 3.35 × [tex]10^8[/tex] V/m.

The electric field amplitude of the light wave at the cornea can be calculated using the energy of the light pulse and its other characteristics. Given the pulse energy E = 3.0 mJ and pulse duration t = 10 ns, we first calculate the instantaneous power Pi during one pulse using the relationship:

Pi = E / t = 3.0 mJ / 10 ns = 300 MW

Then, given the diameter of the laser beam d = 0.80 mm, we can calculate the area of the laser beam A = π(d/2)²:

A = π(0.80 mm / 2)² ≈ 0.503 mm² = 5.03 × [tex]10^{-6[/tex] m²

So, we find the intensity of the pulse by dividing the power by the area:

I = Pi / A = 300 MW / 5.03 × [tex]10^{-6[/tex] m² ≈ 5.98 × [tex]10^8[/tex] W/m²

Finally, the electric field amplitude E of the light wave is related to the intensity I by the following equation:

E² = (2I) / (ε₀c)

where ε₀ is the permittivity of free space and c is the speed of light. Substituting the values of I, ε₀, and c, we get:

E² = (2 × 5.98 × [tex]10^8[/tex] W/m²) / ((8.85 × [tex]10^{-12[/tex] F/m) × (3 × [tex]10^8[/tex] m/s)) ≈ 1.12 × [tex]10^{17[/tex]V²/m²

Taking the square root of both sides, we find the electric field amplitude E:

E ≈ 3.35 ×[tex]10^8[/tex] V/m

Therefore, the electric field amplitude of the light wave at the cornea is approximately 3.35 × [tex]10^8[/tex] V/m.

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When blue light of wavelength 470 falls on a single slit, the first dark bands on either side of center are separated by 45.0. determine the width of the slit

Answers

Final answer:

The width of the slit can be determined using the formula θ = λ / (2d), where θ is the angle, λ is the wavelength, and d is the width of the slit. By rearranging the formula and substituting the given values, we find that the width of the slit is approximately 244 nm.

Explanation:

The width of the slit can be determined using the formula for the angle of the first minimum in a single-slit diffraction pattern, which is given by θ = λ / (2d), where θ is the angle, λ is the wavelength, and d is the width of the slit. In this case, the angle is 45.0 degrees and the wavelength is 470 nm. Converting the angle to radians and rearranging the formula, we can solve for the width of the slit:

d = λ / (2 * tan(θ))

Substituting the given values, we have:

d = (470 nm) / (2 * tan(45.0 degrees))

Calculating this, we find that the width of the slit is approximately 244 nm.

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Select the answer choices that describe where plasmas can be found.
(Select all that apply.)

in stars
in fireworks
in lightning
in lightbulbs

Answers

i know stars n lighting is one

Answer:

A + C + D

Explanation:

A) in stars

C) in lightning

D) in light-bulbs

(P.S. I just took a test on this and got it correct)

a pulley of diameter 15.0 cm is driven by a motor that revolves at 10 rpm. the pulley drives a 2nd pulley with diameter 10.0 cm. find the rpm of the 2nd pulley

Answers

6.67prm

diam1/diam2 = rpm1/rpm2
Final answer:

The rpm of the second pulley is 15.

Explanation:

To find the rpm of the second pulley, we can use the formula:

Angular velocity of first pulley * diameter of first pulley = angular velocity of second pulley * diameter of second pulley

Plugging in the given values:

10 rpm * 15.0 cm = rpm of second pulley * 10.0 cm

Dividing both sides by 10.0 cm:

Rpm of second pulley = (10 rpm * 15.0 cm) / 10.0 cm = 15 rpm

A worker pushes a 1.50 x 10^3 N crate with a horizontal force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220.

a) How much work is done by the worker on the crate?
b) How much work is fone by the floor on the crate?
c) What is the net work done on the crate?

Answers

(a) The work done by the worker on the crate is 8,280 J.

(b) The work done by the floor on the crate is 7,920 J.

(c) The net work done on the crate is 360 J.

Work done by the worker

The total work done by the worker is calculated as follows;

[tex]W = Fd\\\\W = 345 \times 24\\\\W = 8,280 \ J[/tex]

Work done by the floor on the crate

The work done by the floor on the crate is calculated as follows;

[tex]W = F_f d\\\\W = \mu F_n d\\\\W = 0.22 \times 1.5 \times 10^3 \times 24\\\\W = 7,920 \ J[/tex]

Net work done on the crate

The net work done on the crate is calculated as follows;

[tex]W = 8,280 - 7,920\\\\W = 360 \ J[/tex]

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Final answer:

The work done by the worker is 8280 Joules. The work done by the floor (accounting for friction) is -7920 Joules. This makes the net work done on the crate 360 Joules.

Explanation:

The question is asking about the concept of work and net work in physics, specifically in the context of friction and forces. To start, we need to understand that work is defined as the force applied on an object times the distance that object moves in the direction of the force (Work = Force x Distance).

a) The work is done by the worker on the crate can be calculated using the formula Work = Force x Distance. Hence, Work = 345 N x 24.0 m = 8280 Joules.

b) The work done by the floor on the crate is the frictional force times the distance. The frictional force can be calculated using the formula Force = Coefficient of friction x Normal force. In this case, the normal force is the weight of the crate (1.50 x 10^3 N). Hence, Frictional force = 0.220 x (1.50 x 10^3 N). Work done by the floor is Frictional force x Distance, so the work done by the floor = Frictional force x 24.0 m = -7920 Joules (It's negative because the force of friction acts in the opposite direction to motion).

c) The net work done on the crate is the sum of the work done by the worker and the work done by the floor. Hence, Net work = 8280 Joules - 7920 Joules = 360 Joules.

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What is an expression for the difference in energy between state c and state d?

Answers

What are the states? Please list them.

The refracting telescope at the Yerkes Observatory has a 1.00-m diameter objective lens of focal length 20.0 m. Assume it is used with an eyepiece of focal length 1.90 cm. (a) Determine the magnification of Mars as seen through this telescope.

Answers

angular magnifiction=[tex]m= \frac{ f_{0} }{ f_{e} } [/tex]
[tex] f_{0}= [/tex]=focal length of object lens=20m
[tex] f_{e}= [/tex]focal length of eye lens=1.90 m
so magnification=m=20/1.9=?

Magnification = (focal length of the objective) / (focal length of the eyepiece)

                     =    (20.0 m)  /  (1.9 cm)

                     =      (2,000 cm)  /  (1.9 cm)

                     =        1,053  (rounded)

A researcher obtains a Pearson correlation of r = 0.43 for a sample of n = 20 participants. For a two-tailed test, which of the following accurately describes the significance of the correlation for alphas of .01 and .05?

Answers

Final answer:

The Pearson correlation coefficient, r, measures the strength and direction of the linear relationship between two variables. In this case, a correlation coefficient of r = 0.43 was obtained for a sample of 20 participants. The correlation is not statistically significant at both alpha levels of 0.01 and 0.05.

Explanation:

The Pearson correlation coefficient, r, measures the strength and direction of the linear relationship between two variables. In this case, the researcher obtained a correlation coefficient of r = 0.43 for a sample of n = 20 participants. To determine the significance of the correlation, a hypothesis test can be conducted at different alpha levels. At an alpha level of 0.01, the critical value for a two-tailed test with 18 degrees of freedom is 2.898.

The calculated t-value using the formula r * sqrt((n-2)/(1-r^2)) is compared to the critical value. If the t-value is less than the critical value, the correlation is not statistically significant. In this example, the calculated t-value is less than 2.898, indicating that the correlation is not statistically significant at the 0.01 level.

At an alpha level of 0.05, the critical value for a two-tailed test with 18 degrees of freedom is 2.101. Since the calculated t-value is less than 2.101, the correlation is not statistically significant at the 0.05 level as well.

Final answer:

To determine if the Pearson correlation of r = 0.43 is significant for n = 20, one must calculate the t-statistic and compare it to the critical t values from a t distribution table for df = 18 for each alpha level (α). These critical values determine whether the correlation is significant at the .05 or .01 level.

Explanation:

To determine the significance of a Pearson correlation coefficient r for a given alpha level (α), one can use the t distribution as a reference. In the given scenario, the researcher has obtained a Pearson correlation of r = 0.43 with a sample size of n = 20 participants. Before evaluating significance, one must calculate the t-statistic using the formula given:

t = r√n-2/√(1-r²)

The degrees of freedom (df) for this calculation could be determined by df = n - 2, which here would be 20 - 2 = 18. Upon comparing the calculated t value to the critical t value from a t distribution table for df = 18, one can establish the significance of the correlation coefficient.

Looking at a standard t distribution table or using statistical software, one would find the critical t values for a two-tailed test. For α = 0.05, the critical t value for df = 18 would typically be around 2.101. For α = 0.01, this value would be higher, representing a stricter criterion for significance. If the calculated t value is greater than the critical value, the null hypothesis, which proposes no significant correlation, is rejected.

Given the details of the Pearson correlation in the question, without the exact t value calculated, it is not possible to definitively say whether the r = 0.43 is significant at the α = 0.05 or 0.01 level. However, provided that the calculated t value is larger than the critical t value at either α level, the null hypothesis would be rejected, indicating a significant correlation.

how fast would a meter stick be moving if the observer measures a length of 60 cm.

Answers

required formula is
[tex]L= L_{0} \sqrt{1- \frac{ v^{2} }{ c^{2} } } [/tex]
where L= length of a stick when it is moving with velocity v
L0=length of that stick when it is at rest
c= velocity of light

What value resistor will discharge a 2.80 μf capacitor to 10.0% of its initial charge in 2.50 ms ?

Answers

Here's the formula for an exponential RC discharge.

(If you were not a Brainly robot, you would have seen this formula before,
in any class that would assign you this problem to solve.)

         Charge = (initial charge) · e ^ -(time / RC) .

In this situation:

                                       0.1  =  (1) · e ^ -(.0025 / 2.8R x 10⁻⁶)

Ln of each side:            ln(0.1) = -.0025 / 2.8R x 10⁻⁶

Multiply each side
by R                            R · ln(0.1)  =  -.0025 / 2.8 x 10⁻⁶

Divide each side
by  ln(0.1) = -2.303:      R  =  0.0025 / (2.303 · 2.8 x 10⁻⁶)

                                 I get     387.8 Ω       

A circular plate of 500-mm diameter is maintained at T1 = 600 K and is positioned coaxial to a conical shape. The back side of the cone is well insulated. The plate and the cone, whose surfaces are black, are located in a large, evacuated enclosure whose walls are at 300 K.
(a) What is the temperature of the conical surface, T2?
(b) What is the electrical power that would be required to maintain the circular plate at 600K?

Answers

Final answer:

The student’s questions involve the principles of thermal radiation and heat conduction in Physics, relating to the determination of the equilibrium temperature in a radiative system and the electrical power needed to maintain a temperature. Calculations require additional data on materials and geometry to provide accurate answers.

Explanation:

The student's questions pertain to the topics of thermal radiation and heat conduction in Physics, which are within the realm of thermodynamics. Specifically, they are asking about (a) the equilibrium temperature of a conical surface and (b) the power required to maintain a certain temperature on a circular plate. These calculations involve understanding Stefan-Boltzmann law, conduction, and power calculations.

Unfortunately, without further information or input parameters for materials and geometries, providing a definitive answer for a system like the one described (circular plate and conical surface in an enclosure) isn't possible. To determine the electrical power required, one would need to know properties like thermal conductivity, surface area, and possibly the geometry of the objects in question to find out the heat transfer rates.






Exercise 28.41

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 24.0cm and carries a clockwise current of 10.0A , as viewed from above, and the outer wire has a diameter of 38.0cm .



Part A

What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?

The currents direction must be clockwise

or

The currents direction must be counterclockwise

Part B

What must be the magnitude of the current in




Exercise 28.41

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 24.0cm and carries a clockwise current of 10.0A , as viewed from above, and the outer wire has a diameter of 38.0cm .



Part A

What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?

The currents direction must be clockwise

or

The currents direction must be counterclockwise

Part B

What must be the magnitude of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?


















I=


A









Answers

(a) The currents direction must be counterclockwise.

(b) The magnitude of the current in the outer wire is 15.83 A.

Direction of the current

The direction of the current will flow in such a way that the magnetic field due to the wires combination will cancel out. Thus, the current will flow in opposite or counterclockwise direction.

Magnitude of the current

The magnitude of the current is calculated using the following formulas;

[tex]\frac{I_1}{D_1} = \frac{I_2}{D_2} \\\\I_2 = \frac{I_1 D_2}{D_1} \\\\I_2 = \frac{10 \times 38}{24} \\\\I_2 = 15.83 \ A[/tex]

Thus, the magnitude of the current in the outer wire is 15.83 A.

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A 2.10-kg block starts from rest at the top of a 30.0° incline and slides a distance of 1.90 m down the incline in 1.40 s.
(a) Find the magnitude of the acceleration of the block.
(b) Find the coefficient of kinetic friction between block and plane
(c) Find the friction force acting on the block.



magnitude

N



direction

---Select--- up the incline down the incline normal to the incline and upward normal to the incline and downward

d) Find the speed of the block after it has slid 1.90 m.

Answers

Final answer:

To find the acceleration, the kinematic equation is used. The kinetic friction coefficient is derived from force equations, and the friction force can be calculated with this coefficient and the normal force. Finally, the block's final speed after sliding down the incline is determined using another kinematic equation.

Explanation:

Finding the Magnitude of Acceleration, Coefficient of Kinetic Friction, Friction Force, and Final Speed

To solve for the acceleration of a 2.10-kg block sliding down a 30.0° incline, we can use the kinematic equation s = ut + (1/2)at², where s is the distance slid which is 1.90 m, u is the initial velocity which is 0 m/s, t is the time which is 1.40 s, and a is the acceleration. Solving for acceleration, a = 2s/t², we find:

a = (2 * 1.90 m) / (1.40 s)²

a = 3.80 m / 1.96 s²

a = 1.9388 m/s²

The coefficient of kinetic friction μ can be determined by using the equation f_k = μ * N, where f_k is the kinetic friction force and N is the normal force. The gravitational component parallel to the incline is mg sin(θ) and perpendicular to the incline is mg cos(θ). Knowing the total force equation, mg sin(θ) - μ*mg cos(θ) = ma, we can isolate μ and solve for it.

The friction force then can be calculated using f_k = μ * mg cos(θ), and the direction of this force is up the incline since it opposes the motion of the block sliding down.

Finally, to find the final speed v of the block after sliding 1.90 m, we use the kinematic equation v² = u² + 2as. Since the initial speed u is 0, this simplifies to v = √(2as), and plugging in the known values for a and s gives us the final speed.

How much energy is stored in a room 3.0 m by 4.0 m by 2.4 m due
to the earth's magnetic field with a strength of 5.0 × 10-5 T? (μ0 = 4π
× 10-7 T · m/A)

Answers

Answer:

The energy stored in a room is 0.0286 Joules.

Explanation:

Given that,

The dimensions of the room are 3 m by 4 m by 2.4 m

The strength of the Earth's magnetic field, [tex]B=5\times 10^{-5}\ T[/tex]

We need to find the energy stored in a room due to the Earth'a magnetic field. It is given by :

[tex]E=\dfrac{B^2}{2\mu_o}V[/tex]

Here,

V is the volume of the room

[tex]E=\dfrac{(5\times10^{-5})^{2}}{2\times4\pi\times10^{-7}}\times(3\times4\times2.4)\\\\E=0.0286\ J[/tex]

[tex]E=0.0286\ J[/tex]

So, the energy stored in a room is 0.0286 Joules. Hence, this is the required solution.


You pull upward on a stuffed suitcase with a force of 105N , and it accelerates upward at 0.710m/s2, what is the wieght and mass of the suitcase?

Suppose the suitcase is on a rocket. The rocket launches with an acceleration of 29.0m/s2, What is the apparent weight of an 86-kg astronaut aboard this rocket?

Answers

I will assume the value of gravitational acceleration is 10m/s^-2 towards the centre of Earth.

[tex]F_{Resultant} =F_{Pull}-F_{Weight}[/tex]
[tex]Mass_{R}* Acceleration_{R}=105-(Mass_{Weight}* Acceleration_{Weight})[/tex][tex]Mass_{R}*0.71 =105-(Mass_{W}*10)[/tex]
Assuming mass remains constant
[tex]0.71M =105-10M[/tex]
[tex]10.71M =105[/tex]
[tex]M =9.8kg[/tex]
[tex]W=Mg[/tex]
[tex]W=9.8*10[/tex]
[tex]W=98N[/tex]

[tex]W_{Apparent}=Mass*(Acceleration_{Weight}+Acceleration_{Rocket})[/tex]
[tex]W_{Apparent}=9.8*(10+29)[/tex]
[tex]W_{Apparent}=9.8*(39)[/tex]
[tex]W_{Apparent}=382.2N[/tex]

1. The weight of the suitcase is approximately 1448.32 N.

2. The apparent weight of the 86-kg astronaut aboard the rocket is approximately 2504 N.

To find the weight and mass of the suitcase, we'll use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

Given:

Force (F) = 105 N

Acceleration (a) = 0.710 m/[tex]s^2[/tex]

We can use the equation F = ma to solve for mass (m):

105 N = m * 0.710 m/s^2

Rearranging the equation, we have:

m = F / a

m = 105 N / 0.710 m/[tex]s^2[/tex]

m ≈ 147.89 kg

So, the mass of the suitcase is approximately 147.89 kg.

To find the weight of the suitcase, we can use the equation W = mg, where W is the weight and g is the acceleration due to gravity (approximately 9.8 m/[tex]s^2[/tex]).

W = 147.89 kg * 9.8 m/[tex]s^2[/tex]

W ≈ 1448.32 N

Therefore, the weight of the suitcase is approximately 1448.32 N.

Now let's calculate the apparent weight of the astronaut aboard the rocket.

Given:

Mass (m) = 86 kg

Acceleration (a) = 29.0 m/[tex]s^2[/tex]

Using the same formula W = mg, we can find the apparent weight (W) of the astronaut:

W = m * a

W = 86 kg * 29.0 m/[tex]s^2[/tex]

W ≈ 2504 N

Therefore, the apparent weight of the 86-kg astronaut aboard the rocket is approximately 2504 N.

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What is the acceleration of a 32-kg object if a 6.75-N force is applied to it?

Answers

∑F = ma
a = ∑F/m
a = 6.75 N / 32 kg
a = 0.21 m/s²

A 15,000-kg train car moving due east at 23 m/s collides with and couples to a 10,000-kg train car that is initially at rest. Find the common velocity of the two-car train after the collision. (East is defined as positive. Indicate the direction with the sign of your answer.)

Answers

13.8 m/s due east

as below (though momentums)

The common velocity of the two-car train after the collision, given that the 15000 kg train car was initially moving due east at 23 m/s, is +13.8 m/s

How to calculate the common velocity of the two-car train after collision?

The question given above talks about conservation of linear momentum of colliding objects.

Thus, we can calculate the common velocity of the two-car train after the collision by using the formula of conservation of linear momentum as shown below:

Mass of first train car (m₁) = 15000 KgInitial velocity of first train car (u₁) = 23 m/sMass of second train car (m₂) = 10000 KgInitial velocity of of second train car (u₂) = 0 m/sCommon velocity of two-car train after collision (v) = ?

Momentum before = Momentum after

[tex]m_1u_1 + m_2u_2 = v(m_1 + m_2)[/tex]

(15000 × 23) + (10000 × 0) = v(15000 + 10000)

345000 + 0 = 25000v

345000 = 25000v

Divide both sides by 25000

v = 345000 / 25000

= +13.8 m/s

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If the speed of light in a medium is 2 x 10^8 m/s, the medium's index of refraction is?

Answers

Final answer:

The index of refraction of the medium is 1.5.

Explanation:

The speed of light in a medium is given by the equation c = v/n, where c is the speed of light in a vacuum, v is the speed of light in the medium, and n is the index of refraction of the medium. Rearranging the equation, we can find the index of refraction as n = c/v. Substituting the given values, the index of refraction is n = (3.00 x 10^8 m/s) / (2.00 x 10^8 m/s) = 1.5.

a 55.0 g bullet is fired vertically with an initial velocity of 123m/s

1) what is the velocity at it's highest point of travel?
2) what maximum height does it reach?
3) at what velocity does it hit the ground?

Answers

1) at highest point of travel it stops. and there v=0 m/s
3) when it returns to the original point it's speed would be again 123m/s
(conservation law and no calculations needed at all)

2) this one is harder

step 1 - finding the time taken from ground to highest point 
v = u + at
v is the speed at highest point (it's 0, as it stops) 
u - speed at origin (123m/s given)
a - acceleration (in our case it's gravitation and it quals 9.8m/s, some people use just 10 m/s, by the way, it's negative)
0 = 123 + (-9.8)* t
t = 123/9.8 = say 12.3 second
it's time taken to travel up

step 2 - finding displacement or heigh
original formula is 
v^2 = u^2 + 2ar
v - speed at point (0 m/s at the top point)
u - initial speed (123m/s)
a - acceleration (it's gravitation = 9.8 or just 10) and it's again negative
r - displacement or height or distance travelled 
0^2 = 123^2 + 2*(-10)*r
123*123=20r
r= 123*123/20
r= 756 meters 

there is also mass given. it's just to confuse you. no matter that the mass is - time taken, max height or range, acceleration and velocity would be the same for any mass. surprised?





can the carbon-14 isotope be used to date a stone tablet

Answers

yes, look at the half life of the carbon-14, you can work it out as there is an exponential decay 

Carbon-14 dating is used to date organic materials, not inorganic objects like stone tablets, which require other methods like potassium-argon dating.

Carbon-14 dating, or radiocarbon dating, is a method used to date materials that were once living and contains carbon. The process measures the decay rate of carbon-14, a radioisotope, to determine the age of organic artifacts up to approximately 60,000 years old. However, this technique is not suitable for dating stone tablets or any other inorganic objects such as pottery or projectile points because they do not contain carbon from living organisms. To date geological materials like stone, other methods such as potassium-argon dating are used. This method relies on the radioactive decay of potassium-40 to argon-40 to estimate the age of igneous and volcanic rocks.

A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.0700 m, an angular speed of 88.0 rad/s, and a moment of inertia of 0.850 kg · m2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of 4.40 s.(a) Find the magnitude of the angular deceleration of the cylinder. rad/s 2 (b) Find the magnitude of the force of friction applied by the brake shoe. N

Answers

Final answer:

The magnitude of the angular deceleration of the cylinder is -10.0 rad/s^2, and the magnitude of the force of friction applied by the brake shoe is -8.50 N.

Explanation:

(a) To find the magnitude of the angular deceleration of the cylinder, we can use the formula:

angular deceleration = (final angular velocity - initial angular velocity) / time

Given that the initial angular speed is 88.0 rad/s, the final angular speed is half of that, and the time is 4.40 s, we can calculate:

angular deceleration = (44.0 rad/s - 88.0 rad/s) / 4.40 s = -10.0 rad/s^2

(b) To find the magnitude of the force of friction applied by the brake shoe, we can use the formula:

force of friction = moment of inertia * angular deceleration

Given that the moment of inertia is 0.850 kg·m², and the angular deceleration is -10.0 rad/s², we can calculate:

force of friction = 0.850 kg·m² * -10.0 rad/s² = -8.50 N

From the small number of impact craters on ____, planetary geologists suspect that virtually all its original surface has been covered over by younger volcanic activity

Answers

Final answer:

Planetary geologists suspect that the original surface of a certain celestial body has been covered by younger volcanic activity, based on the small number of impact craters. Evidence of past impact events is gradually erased by the constant renewal of Earth's crust through plate tectonics.

Explanation:

Based on the small number of impact craters, planetary geologists suspect that virtually all the original surface of a certain celestial body has been covered by younger volcanic activity.

This can be observed in the case of Earth, where the constant renewal of its crust through plate tectonics gradually erases evidence of past impact events.

Geologists have identified eroded remnants of impact craters on Earth, providing evidence for the influence of these impacts on the planet's evolution over time.

Light propagating in the glass n1= 1.65 wall of an aquarium tank strikes the interior edge of the wall with incidence angle 19.0 degrees . What is the angle of refraction in the water?

Answers

Final answer:

To find the angle of refraction in water when light propagates in a glass wall, we can use Snell's law.

Explanation:

To find the angle of refraction in water, we can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the indices of refraction of the two media:



sin(θ1) / sin(θ2) = n2 / n1



Given that the incidence angle in glass is 19.0 degrees and the index of refraction of glass is 1.50, we can substitute these values into the equation:



sin(19.0) / sin(θ2) = 1.333 / 1.50



Solving for θ2, the angle of refraction in water, we find:



θ2 ≈ 13.14 degrees

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Final answer:

The angle of refraction in the water is approximately 24.21 degrees.

Explanation:

When light passes from one medium to another, it changes direction. This change in direction is called refraction. Snell's Law describes the relationship between the angles of incidence and refraction. The equation is: n1*sin(θ1) = n2*sin(θ2), where n1 and n2 are the indices of refraction of the two media, and θ1 and θ2 are the angles of incidence and refraction, respectively.

In this case, light is propagating in glass with an index of refraction of 1.65 and is striking the interface between the glass and water at an angle of 19.0 degrees. The index of refraction of water is 1.333. To find the angle of refraction in the water, we can use Snell's Law:

1.65*sin(19.0) = 1.333*sin(θ2).

Solving for θ2, we find that the angle of refraction in the water is approximately 24.21 degrees.

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Burned Becky stayed out in the sun too long and is now suffering from a terrible sunburn. What caused Becky's sunburn

Answers

The Sun's UV rays is what caused the sun burn.

Answer: radiation from the sun

Explanation: the sun produces ranges if radiation ranging from gamma Ray to Infrared but ultraviolet Ray is the most abundant

She must have been had the sun burn from the head propagated by the ultalraviolet ray from the sunshine

convert 50 years into seconds express your answer in scientific notation

Answers


               (50 years) x (365 da/yr) x (86,400 sec/da)

         =      (50 x 365 x 86,400)  seconds

         =        1.5768 x 10⁹ seconds

50 years converted into seconds would be equivalent to 1.58×10⁹ seconds.

What is a unit of measurement?

A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation. Any additional quantity of that type can be stated as a multiple of the measurement unit.

As given in the problem we have to convert 50 years into seconds,

1 year = 365 days

50 years = 50×365 days

50 years = 50×365×24 hours

50 years =50×365 ×24×60 minutes

50 years =50×365 ×60×60 seconds

               =1576800000 seconds

               =1.58×10⁹ seconds

Thus,50 years converted into seconds would be 1.58×10⁹ seconds

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differnces between uniform and non uniform electric fields

Answers

The electric field strength of a uniform electric field is constant throughout the field. A perfectly uniform electric field has no variations in the entire field and is unattainable in the real world. However, two parallel plates can generate a field that resembles a perfectly uniform field with slight variations near the edge of the plates. Electric fields are represented by drawing field lines that represent the direction of the field, as well as the strength of the field. More field lines represents a higher field strength. In a non-uniform electric field, the field lines tend to be curved and are more concentrated near the charges. In a uniform electric field, since the field strength does not vary, the field lines are parallel to each other and equally spaced. Uniform fields are created by setting up a potential difference between two conducting plates placed at a certain distance from one another. The field is considered to be uniform at the center of the plates, but varies close to the edge of the plates. The strength of the field depends on the potential difference applied to the plates and the distance by which they are separated. A higher potential difference or voltage results in a stronger electric field. The greater the distance between the plates, the weaker the field becomes. The electric field is therefore calculated as a ratio of the voltage between the plates to the distance they are separated by.
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