g A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When the magnetic field is increased from 250 µT to 700 µT in 0.300 s, an emf of magnitude 60.0 mV is induced in the coil. What is the total length of wire in the coil?

Answer:

Explanation:

Given that,

Pie diameter = 9 in

Then, the circumference of the pie is

P = πd = 9π in

Then rim of the pie rotates 233 in,

Then,

1 Revolution of the pie is 9π in,

So, for 233 in, we will have

233 in / 9π in revolution

8.24 revolution

So, the revolution of the pie is 8.24

1 revolution is 2πrad

Then,

8.24 revolution = 8.24 × 2π = 51.78 rad.

And also, 1 revolution is 360°

Then,

8.24 revolution = 8.24 × 360 = 2966.4°

So,

In revolution, θ = 8.24 revolution

In radian = θ = 57.78 rad

In degree θ = 2966.4°

The angular distance should be

In revolution, θ = 8.24 revolution.

In radian = θ = 57.78 rad.

In degree θ = 2966.4°.

Since

Pie diameter = 9 in

So,  the circumference of the pie should be

P = πd = 9π in

And, rim of the pie rotates 233 in,

So,

1 Revolution of the pie is 9π in,

So, for 233 it should be

= 233 in / 9π in revolution

= 8.24 revolution

Now in the case when

1 revolution is 2πrad

So,

8.24 revolution = 8.24 × 2π = 51.78 rad.

And also, 1 revolution is 360°

So,

8.24 revolution = 8.24 × 360 = 2966.4°

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Answer:

Explanation:

y = y₀ cos[tex]\sqrt{\frac{k}{m} }\times t[/tex]

a )

[tex]\frac{dy}{dt}[/tex] = - y₀ x [tex]\sqrt{\frac{k}{m} }[/tex] sin (  [tex]\sqrt{\frac{k}{m} }\times t[/tex]  )  

b ) If m = 4m

[tex]\frac{dy}{dt}[/tex] = - y [tex]\sqrt{\frac{k}{4m} }[/tex] sin ( [tex]\sqrt{\frac{k}{4m} }\times t[/tex]  )

Magnitude of velocity will be decreased .

c )

[tex]\frac{dy}{dt}[/tex] = - y [tex]\sqrt{\frac{4k}{m} }[/tex] sin ( [tex]\sqrt{\frac{4k}{m} }\times t[/tex]  )

magnitude of velocity will be increased .

d )

velocity = - y₀ [tex]\sqrt{\frac{k}{m} }[/tex] sin( [tex]\sqrt{\frac{k}{m} }\times t[/tex]  )

              =  L [tex]\sqrt{\frac{ms^{-2}}{m} }[/tex]  X 0

               = L s⁻¹

= m /s

unit of velocity is consistent .

Final answer:

The velocity of a mass attached to a spring is found by differentiating the position equation with respect to time, revealing how the system's velocity changes with variations in mass and spring stiffness. Increasing the mass decreases the velocity amplitude, while increasing the spring stiffness increases it.

The units of velocity, m/s, are confirmed through dimensional analysis.

Explanation:

To find the velocity of the mass, we need to differentiate the position equation y = yocos (t √ k/m) with respect to time. Using the chain rule for differentiation, the derivative of y with respect to t gives us dy/dt = -y_0(√ k/m)sin(t √ k/m), where dy/dt represents the velocity of the mass.

This equation tells us the velocity at any given moment for a mass m and spring constant k.

b) If the mass is increased by four times, the equation for velocity becomes dy/dt = -y_0(√ k/(4m))sin(t √ k/(4m)). The increase in mass causes the velocity amplitude to decrease, as √(1/4) is in the equation, indicating that velocity decreases in proportion to the square root of the mass increase.

c) Increasing the spring constant k fourfold results in the new velocity equation dy/dt = -y_0(√ (4k)/m)sin(t √ (4k)/m). This shows an increase in the velocity amplitude, as the increase in k results in a velocity proportional to the square root of the increase in k, thus making the system oscillate faster.

d) To confirm the units of velocity are meters per second (m/s), we substitute the units into the derivative of the position equation: [m]*[s√(kg/s2)/kg] simplifies to m/s, thus showing the units of velocity are indeed consistent and correct.

The frequency of the third normal mode of a guitar string with a length of 40.0 cm, mass of 0.5 g when stretched with a tension of 80N is approximately 948.68 Hz.

To calculate the frequency of the third normal mode (
n = 3) for a guitar string, we can use the formula for the frequency of a string fixed at both ends:

f_n = (n/2L) √(T/μ)

where:

f_n is the frequency of the nth mode,

n is the mode number (which is 3 in this case),

L is the length of the string,

μ is the mass per unit length of the string (linear mass density), and

T is the tension in the string.

Given the length of the string L = 40.0 cm = 0.4 m, the mass m = 0.5 g = 0.0005 kg, and the tension T = 80 N, we first need to calculate the linear mass density:

μ = m/L

In this case,

μ = 0.0005 kg / 0.4 m = 0.00125 kg/m

Now, we use the frequency formula to find f_3:

f_3 = (3/2 0.4 m √(80 N/0.00125 kg/m)
= (3/(0.8 m)) √(64000 N/m)
= 3.75 √(64000 N/m)
= 3.75 √(64000 N/m)
= 3.75 * 252.9822 Hz
= 948.68 Hz

The frequency of the third normal mode of the guitar string is approximately 948.68 Hz.

Answer:

A) 21.1 m

B) 17.3 m

C) 3.267x10^7 m/s

Explanation:

This is a case of special relativity.

Let the relative speed of astronauts ship to my ship be v.

According to my observation,

My craft is 21.1 m long, according to my observation, astronauts craft is 17.3 m long.

If we fix the reference frame as my ship, then the rest lenght of our identical crafts is 21.1 m and the relativistic lenght is 17.3 m

l' = 21.1 m

l = 17.3 m

From l = l'(1 - p^2)^0.5

Where p is c/v, and c is the speed of light

17.3 = 21.1 x (1 - p^2)^0.5

0.82 = (1 - p^2)^0.5

Square both sides

0.67 = 1 - p^2

P^2 = 0.33

P = 0.1089

Revall p = v/c

v/c = 0.1089

But c = speed of light = 3x10^8 m/s

Therefore,

v = 3x10^8 x 0.1089 = 3.267x10^7 m/s

Following are the response to the given points:

a) Its ship travels to the distance of [tex]21.1\ m[/tex]

b) The astronaut's craft would be at a range of  [tex]17.3\ m[/tex]

c) Relativity's use of length contraction:

[tex]\to L=L_0(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to \frac{L}{L_0}=(\sqrt{1-\frac{v^2}{c^2}})[/tex]

Here,

 [tex]\to \frac{L}{L_0}=\frac{17.3}{21.1}=0.81[/tex]

Hence

[tex]\to 0.81=(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to 0.6561=1-\frac{v^2}{c^2}\\\\\to \frac{v^2}{c^2} =1- 0.6561\\\\\to \frac{v^2}{c^2} =0.3439\\\\\to \frac{v}{c} =0.58\\\\\to v= 0.58 c[/tex]

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Answer:

troposphere

Explanation:

because troposphere is the layer of atmosphere closest to the earth . air is thicker at lower altitudes requiring more energy to push themselves to the sky . however , the air is thinner caused flight more fuel efficient .

Answer:

Explanation:

In series connection of resistors, same current flows in the circuit.

Power dissipated by a resistor is

P = i²R

And since same current flows in them, it implies that the power is directly proportional to Resistance, so the higher the resistance of the resistor the higher the power dissipated in the series connection. Also, the lower the resistance, the lower the power dissipated.

2. In parallel connection, same voltage is applied across the resistor.

So, power dissipated by each resistor is

P = V² / R

So, since the same voltage is applied across parallel connection, then, power dissipated in each resistor is inversely proportional to the resistance.,

So, the higher the resistance, the lower the power and the lower the resistance, the higher the power.

Answer:

Find attachments for step by step solution.

A power of 68 KW is required to operate the ski lift. A power of 9660.5W is required to accelerate this ski lift.

The power can be defined as the rate of doing work, it is the work done in unit time. The SI unit of power joules per second (J/s) or Watt (W).

Power is a time-based quantity and the rate at which work is done upon an object. The formula for power can be expressed as mentioned below.

Power = Work/time

P = W/t

Given, the chairs are spaced 20 m apart a length of 1 km = 1000m

Then the number of chairs = 1000/20 = 50

Each chair weighs  = 250 kg

Then the weight of M = 50 × 250 = 12500 Kg

Consider, the initial and final heights, h₁ = 0, h₂ = 200 m

The work needed to raise the chairs, W = mgh,

W = 12500 × 9.81 × (200 - 0)

W = 2.54 × 10⁷ J

The rate of  work done at a distance of 1 km = 10 km/h,

t = 1/10 = 0.1 hr = 360 s

The power needed to operate this ski lift is, P = W/t

P = 2.54×10⁷ / 360

P = 68125 W = 68 kW

Given, the initial velocity, u = 0 m/s, final velocity, v = 10 km/h = 2.78 m/s

and, t = 5sec

Acceleration during it is first turned on is:

a = (v - u)/t

a = 2.78/ 5

a = 0.556 m/s²

The power required to accelerate this ski lift is:

P = ½ m [(v² - u²)/t]

P = ½ × 12500 × [2.78²/5]

P = 6250 × 1.55

P = 9660.5 W

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Answer:

Work needed = 1515.15 KJ

Explanation:

The center of mass of a cylinder lying horizontally on its side would lie on the axis of the cylinder at the center of length l.

Depth of center of mass from ground level;Δh = (r + 5) metres

Now, work done to pump the gasoline out of the tank is equal to the gain in potential energy by gasoline on lifting it from center of mass to the ground level.

Thus;

W = ΔU = mgΔh

We know that mass(m) = volume(V) x density(ρ)

So,

W = (ρV)gΔh

Volume(V) = πr²L

Thus;

W = (ρ(πr²L)) * g(r + 5)

We are given;

Density; ρ = 673 kg/m³

Length; L = 5 m

Radius; r = 1.5 m

Acceleration due to gravity;g = 9.8 m/s²

Thus;

W = (673(π•1.5²•5)) * 9.8(1.5 + 5)

W = 1515154.4 J = 1515.15 KJ

Answer:

b

Explanation:

the earths mass is more than the moon .

Answer:

The object's weight would be less on the moon.

The Object's mass would be the same on the moon

Explanation:

Answer:

The wavelength of the radiation absorbed by ozone is 319.83 nm

Explanation:

Given;

frequency of absorbed ultraviolet (UV) radiation, f = 9.38×10¹⁴ Hz

speed of the absorbed ultraviolet (UV) radiation, equals speed of light, v = 3 x 10⁸ m/s

wavelength of the absorbed ultraviolet (UV) radiation, λ = ?

Apply wave equation for speed, frequency and wavelength;

v = fλ

λ = v / f

λ = (3 x 10⁸) / (9.38×10¹⁴)

λ = 3.1983 x 10⁻⁷ m

λ = 319.83 x  10⁻⁹ m

λ = 319.83 nm

Therefore, the wavelength of the radiation absorbed by ozone is 319.83 nm

Answer:

If Patty remains on Earth then at the time Peggy will come back from her trip, Peggy will be much younger than her sister Patty because of time-dilation.

Explanation:

Peggy and Patty are sisters. Peggy is an astronaut and Patty is an astronomer.

Peggy goes for mission to Alpha Centauri, the nearest star system to the Solar System at 80% of the speed of light, and will come back after 1010 years.

If Patty remains on Earth then at the time Peggy will come back from her trip, Peggy will be much younger than her sister Patty because of time dilation.

This is due to the fact that time moves slower in Alpha Centauri because of its massive gravitational force which bends space time. Moreover, It is known that Peggy's spacecraft moves at 80% of the speed of light, it will result in velocity time dilation since time moves slow if you travel at a speed near to the speed of light.

Final answer:

Peggy, the astronaut twin traveling at 80% of the speed of light, will experience less time due to time dilation, and upon her return will be younger than her Earth-bound twin sister, Patty.

Explanation:

The question is about the relativistic effects that occur when one twin travels at significant fraction of the speed of light while the other remains on Earth.

According to the theory of relativity, time dilation will cause the traveling twin, Peggy, to age more slowly compared to her twin sister, Patty, who remains on Earth.

If Peggy travels to Alpha Centauri, which is 4.3 light years away, at 80% of the speed of light, and assuming the round trip takes 10 years for the Earth-bound twin, we can calculate that the moving twin will experience less than 10 years of elapsed time due to the effects of time dilation.

This happens because the faster Peggy travels, the more pronounced the effect of time dilation will be. This is a well-known result predicted by Einstein's special theory of relativity and has been confirmed through experiments involving high-speed particles and precise clocks.

Thus, when Peggy returns, she will be younger than her twin sister Patty, who has experienced the full 10 years on Earth.

Answer:

9.6 ft

Explanation:

Distance is inversely proportional to weight

distance = k / (weight), where

k is a constant

or you could say,

distance * weight = k

In this scenario,

120 * 16 = 200 * distance

On rearranging, making, distance the subject of formula, we have

Distance = 120 * 16 / 200

Distance = 1920 / 200

Distance = 9.6 ft

So the 200 pounds person should sit 9.6 feet away from the centre to balance the see saw

Answer:

4.8 ft

Explanation:

torque = wt × distance

t1 = 120lb x 8 ft =960

t2 = 200lb x X ft

set them equal to each other.

120(8) = 200x

960 = 200x

x = 4.8 ft

Answer:

Dogs were probably domesticated by accident, when wolves began trailing ancient hunter-gatherers to snack on their garbage. Docile wolves may have been slipped extra food scraps, the theory goes, so they survived better, and passed on their genes. Eventually, these friendly wolves evolved into dogs

Selective breeding, also known as artificial selection, is a process used by humans to develop new organisms with desirable characteristics. Breeders select two parents that have beneficial phenotypic traits to reproduce, yielding offspring with those desired traits.

Hope it helps!

Answer:

A)

The emf is zero because the velocity of the rod is parallel to the direction of the magnetic field, so the charges experience no force.

B)

The emf is vBL

= (2.6 m/s)(4.50 T)(1.3 m)

= 15.21 V.

The positive end is end 2.

C)

The emf is zero because the magnetic force on each charge is directed perpendicular to the length of the rod.

Answer:

Explanation:

The magnitude is 0.5 N

The direction is 0°

Answer:

Magnitude: 0.5 N

Direction: 0 degrees

Explanation:

Going to be honest, I do not understand the magnitude of R, but what I can explain is the direction.

So, following the steps, we find that the magnitude of A (coulomb's constant k times (q1 x q3)/dq13^2) = 1.798 Newtons

The magnitude of B is the same because q1 = q2 in this scenario

Using SohCahToa, we can find the angle of vector A = about 36.9 degrees

since q1 is 0.3 meters away from the x-axis (this is our opposite side) and the distance between q1 and q3 is 0.5 (our hypotenuse because the x-axis and y-axis make a right angle) this means that we can use this for our formula,

the sine of angle A = opposite side / hypotenuse side

Sine of angle A = 0.3m / 0.5m

Angle A = (inverse Sine) of 0.3m/0.5m

Angle A = 36.9 degrees

Using that we can get the x and y components of each vector:

The x component of A will be A cosine angle A which is

Ax = 1.798 N times (cosine(36.9 degrees)) = 1.44 N

The y component of A will be NEGATIVE A sine angle A

The reason why it is negative is because for the A vector, it has a negative slope so its y value is continually decreasing, so it makes sense for it's resulting vector to have the same y decrease.

Ay = (-1.798)(sine(36.9)) = -1.079 N

Now we can do the same with B.

Since B = A IN THIS SCENARIO, the values will be the same,

Bx = 1.798 N times (cosine(36.9 degrees) = 1.44 N

BUT with the y value here, it is actually INCREASING

so:

By = 1.798 N times (sine(36.9 degrees)) = 1.079 N

So the next step is to sum these values.

Rx = Ax + Bx = 1.44 N + 1.44 N = 2.88 N

The x component of vector R is 2.88 N.

Ry = Ay + By = -1.079 N + 1.079 N = 0 N.

The y component of vector R is 0.

Basically, it is traveling straight along the x-axis. This makes sense because the forces of each particle were repelling this one at an EQUAL MAGNITUDE (q1 = q2) So, the direction is 0 degrees because even following the steps, (tan-1 times (Ry over Rx)) is 0/2.88. This will be 0.

Answer:

Apparent

The key discovery about Cepheid variable stars that led in the 1920s to the resolution of the question of whether spiral "nebulae" were separate and distant galaxies or part of the Milky Way Galaxy was the direct relationship between the pulsation period and the absolute brightness or luminosity of the Cepheid variables. A measurement of apparent brightness of a variable star could then be used to determine the distance to the "nebula" containing it.

Explanation:

A variable star is a star with changing apparent brightness. The changes can occur over years or in a fraction of seconds. For example the sun whose energy output varies by approximately 0.1 percent of its magnitude, over an 11-year solar cycle. This variable(apparent brightness) can be used to determine how far a variable star is (distance). Therefore, a measurement of apparent brightness of a variable star could then be used to determine the distance to the "nebula" containing it.

Answer:

A) 30 s, 792 m

B) 10.28 s, 4108.2 m = 4.11 km

Explanation:

A) Traveling with an initial speed of 70 km/h, a car accelerates at 6000km/h^2 along a straight road. How long will it take to reach a speed of 120 km/h? Also, through what distance does the car travel during this time?

Using the equations of motion.

v = u + at

v = final velocity = 120 km/h

u = initial velocity = 70 km/h

a = acceleration = 6000 km/h²

t = ?

120 = 70 + 6000t

6000t = 50

t = (50/6000) = 0.0083333333 hours = 30 seconds.

Using the equations of motion further,

v² = u² + 2ax

where x = horizontal distance covered by the car during this time

120² = 70² + 2×6000×x

12000x = 120² - 70² = 9500

x = (9500/12000) = 0.79167 km = 791.67 m = 792 m

B) At t = 0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When t = 3 s, bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time t, after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?

Bullet A is fired upwards with velocity 450 m/s

Bullet B is fired upwards with velocity 600 m/s too

Using the equations of motion, we can obtain a relation for when vertical distance covered by the bullets and time since they were fired.

y = ut + ½at²

For the bullet A

u = initial velocity = 450 m/s

a = acceleration due to gravity = -9.8 m/s²

y = 450t - 4.9t² (eqn 1)

For the bullet B, fired 3 seconds later,

u = initial velocity = 600 m/s

a = acceleration due to gravity = -9.8 m/s²

t = T

y = 600T - 4.9T²

At the point where the two bullets pass each other, the vertical heights covered are equal

y = y

450t - 4.9t² = 600T - 4.9T²

But, note that, since T starts reading, 3 seconds after t started reading,

T = (t - 3) s

450t - 4.9t² = 600T - 4.9T²

450t - 4.9t² = 600(t-3) - 4.9(t-3)²

450t - 4.9t² = 600t - 1800 - 4.9(t² - 6t + 9)

450t - 4.9t² = 600t - 1800 - 4.9t² + 29.4t - 44.1

600t - 1800 - 4.9t² + 29.4t - 44.1 - 450t + 4.9t² = 0

179.4t - 1844.1 = 0

t = (1844.1/179.4) = 10.28 s

Putting this t into the expression for either of the two y's, we obtain the altitude at which this occurs.

y = 450t - 4.9t²

= (450×10.28) - (4.9×10.28×10.28)

= 4,108.2 m = 4.11 km

Hope this Helps!!!!

Answer:

The magnitudes of the second force is   [tex]Z = 129.9 N[/tex]

The magnitudes of the  resultant force is   [tex]R = 256.047 N[/tex]

Explanation:

From the question we are told that  

    The force is  [tex]F = 187 \ lb[/tex]

     The angle made with second force [tex]\theta_o = 73 ^o 36' = 73 + \frac{36}{60} = 73.6^o[/tex]

     The angle between the resultant force and the first force [tex]\theta _1 = 29 ^o 1 ' = 29 + \frac{1}{60} = 29.0167^o[/tex]

For us to solve problem we are going to assume that

     The magnitude of the second force is  Z N

     The magnitude of the resultant force is R N

According to Sine rule

                [tex]\frac{F}{sin (\theta _o - \theta_1 } = \frac{Z}{\theta _1}[/tex]

Substituting values

             [tex]\frac{187}{sin(73.3 - 29.01667)} =\frac{Z}{sin (29.01667)}[/tex]  

             [tex]267.82 =\frac{Z}{0.4851}[/tex]  

              [tex]Z = 129.9 N[/tex]

According to cosine rule

       [tex]R = \sqrt{F ^2 + Z^2 + 2(F) (Z) cos (\theta _o) }[/tex]

Substituting values

     [tex]R = \sqrt{187^2 + 129.9 ^2 + 2 (187 ) (129.9) cos (73.6)}[/tex]

     [tex]R = 256.047 N[/tex]

Answer:0.15 V

Explanation:

Given

Dimension of coil [tex]4cm\times 5cm[/tex]

Area of coil [tex]A=4\times 5=20\ cm^2[/tex]

Magnetic field [tex]B=0.75\ T[/tex]

Time of rotation [tex]t=0.1\ s[/tex]

No of turns [tex]N=10[/tex]

Initial flux associated with the coil

[tex]\phi_i=N(B\cdot A)[/tex]

[tex]\phi_i=N(BA\cos \theta )[/tex]

where [tex]\theta [/tex]=angle between magnetic field and area vector of coil

[tex]\phi_i=N(BA\cos 90 )[/tex]

Finally when coil is perpendicular to the field

[tex]\phi_f=N(B\cdot A)[/tex]

[tex]\phi_i=N(BA\cos 0 )[/tex]

and induced emf is given by

[tex]e=-\frac{d\phi }{dt}[/tex]

[tex]e=-\frac{\phi_1-\phi_2}{t-0}[/tex]

[tex]e=-\frac{(0-10\times 0.75\times 20\times 10^{-4})}{0.1}[/tex]

[tex]e=0.15\ V[/tex]

Answer:

The force is "19 µN".

Explanation:

The lane's j-component is meaningless, as the current is flowing in the -j line.  

Therefore the power is now in the direction of + z (out of the page if x and y are in the page plane) and has the magnitude.

[tex]\ formula: \\\\\ Forec (F) = mA \\\\ \ F \ = 2.0mA \times \int {0.3} \ y \ dy \rightarrow \ from\ 0 \ to \ 0.25 \\\\\ F \ = \ 2.0mA \times 0.15 * 0.25^{2} m\cdot T \\\\ F = 19 \µN[/tex]

Answer:

0.52rad/s^2

Explanation:

To find the angular acceleration you use the following formula:

[tex]\omega^2=\omega_o^2+2\alpha\theta[/tex]   (1)

w: final angular velocity

wo: initial angular velocity

θ: revolutions

α: angular acceleration

you replace the values of the parameters in (1) and calculate α:

[tex]\alpha=\frac{\omega^2-\omega_o^2}{2\theta}[/tex]

you use that θ=22 rev = 22(2π) = 44π

[tex]\alpha=\frac{(12rad/s)^2-(0rad/s)^2}{2(44\pi)}=0.52\frac{rad}{s^2}[/tex]

hence, the angular acceñeration is 0.52rad/s^2

Final answer:

The speed of the 15.0-kg object as it passes the 10.0-kg object is 13.3 m/s, and the speed of the 10.0-kg object as it passes the 15.0-kg object is -13.3 m/s.

Explanation:

To determine the speeds of the two objects as they pass each other, we can use the principle of conservation of mechanical energy. When the objects are released from rest, the potential energy of the system is converted into kinetic energy as the objects fall. The sum of the kinetic energies of the two objects will be equal to the initial potential energy of the system.

Using the formula for potential energy (PE=mgh), we can calculate the initial potential energy of the system. The 15.0-kg object will fall a distance of 3.00m, so its potential energy is (15.0 kg)(9.8 m/s^2)(3.00 m) = 441 J. The 10.0-kg object will rise a distance of the same amount, so its potential energy is -441 J (we take the negative sign because the object is moving in the opposite direction).

Now, we can equate the sum of the kinetic energies of the two objects to the initial potential energy of the system. Let v1 be the speed of the 15.0-kg object and v2 be the speed of the 10.0-kg object. The kinetic energy of the 15.0-kg object is (1/2)(15.0 kg)(v1^2) and the kinetic energy of the 10.0-kg object is (1/2)(10.0 kg)(v2^2). Setting the sum of these two kinetic energies equal to 441 J, we can solve for v1 and v2.

441 J = (1/2)(15.0 kg)(v1^2) + (1/2)(10.0 kg)(v2^2)

Simplifying the equation, we have 441 J = (7.5 kg)(v1^2) + (5.0 kg)(v2^2). Since the objects are joined by a cord and the pulley does not slip, the speeds of the two objects will be equal in magnitude but opposite in direction. So we can write v2 = -v1 and substitute into the equation. We can then solve for v1:

441 J = (7.5 kg)(v1^2) + (5.0 kg)(-v1^2)

Simplifying further, 441 J = (2.5 kg)(v1^2)

Solving for v1,

v1^2 = 176.4 m^2/s^2

v1 = 13.3 m/s

Therefore, the speed of the 15.0-kg object as it passes the 10.0-kg object is 13.3 m/s, and the speed of the 10.0-kg object as it passes the 15.0-kg object is -13.3 m/s.

Answer:2m/s^2

Explanation:

mass=5kg

Force=10N

Acceleration=force ➗ mass

Acceleration=10 ➗ 5

Acceleration=2m/s^2

Answer: 10.2 kg if g = 9.8, 10 if g = 10.

Explanation:

Weight or the "force of gravity" on a person is simply defined by the equation: F = ma. In this case, the acceleration is g, which is 9.8 but can be rounded up to 10. Based on this, we have:

F = mg

100 = m*9.8

m = 10.2(or 10 if we set g to 10).

The child's mass on the earth  if the force of gravity on the child’s body is 100 N will be equal to 10.2 kg.

The fundamental force of attraction operating on all matter is recognized as gravity, also spelled gravity, in mechanics.

It has no impact on identifying the interior properties of common matter because it is the weakest force known to exist in nature.

The formation and growth of planets, galaxies, and the universe are all under the influence of this long-range, cosmic force, which further determines the trajectories of objects throughout the universe and the entire universe.

As per the given information in the question,

Weight, w = 100 N

Use the formula,

W = m × g

100 N = m × 9.8

m = 100/9.8

m = 10.2 kg

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Answer:

0.05m

Explanation:

Density of water = ρ(w) = 1000 kg/ m³ ;  

Density of Mercury = ρ(m) = 13628.95 kg/ m³  

Total pressure at bottom of cylinder=1.1atm

Therefore, pressure due to water and mercury =1.1-1 =atm

0.1atm=10130pa

The pressure at the bottom is given by,

ρ(w) x g[0.4 - d] + ρ(m) x g x d  = 10130

1000 x 9.8[0.4 - d] + 13628.95 x 9.8 d = 10130

3924 - 9810d + 133416d= 10130

123606d= 6206

d= 6202/123606

d= 0.05m

Depth of mercury alone =  d = 0.05m

We have that for the Question " determine the depth of the mercury."

It can be said that

From the question we are told

the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m^3, and the ambient atmospheric pressure to be 1.013e5 Pa)

Therefore,

Absolute pressure at the bottom of the container =

[tex]P = 1.1 atm = 1.1 * (1.013*105) Pa\\\\= 1.1143 * 10^5 Pa[/tex]

Where,

Height of the cylinder = H = [tex]0.4 m[/tex]

Height of the water in the cylinder = [tex]H_1[/tex]

Height of the mercury in the cylinder = [tex]H_2[/tex]

Therefore,[tex]H = H_1 + H_2\\\\H_1 = H - H_2\\\\P = P_{atm} + \rho_1gH_1 + \rho_2gH_2\\\\P = P_{atm} + \rho_1g(H - H_2) + \rho_2gH_2\\\\1.1143*10^5 = 1.013*10^5 + (1000)(9.81)(0.4 - H_2) + (1.36*10^4)(9.81)H_2\\\\1.013*10^4 = 3924 - 9810H_2 + 133416H_2\\\\143226H_2 = 6206\\\\H_2 = 4.333*10^{-2} m[/tex]

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Answer:2250J

Explanation:

mass(m)=20kg

velocity(v)=15m/s

Kinetic energy=(m x v^2)/2

Kinetic energy =(20 x 15^2)/2

Kinetic energy =(20x15x15)/2

Kinetic energy=4500/2

Kinetic energy=2250J

The kinetic energy of the ball will be 2250 joules.

We have a 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second.

We have to determine its kinetic energy is joules.

The kinetic energy of the body is as follows -

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

According to the question -

Mass of ball = 20 kg

Velocity of ball = 15 m/s

Substituting the values in the above formula, we get -

K.E. = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2} \times 20\times 15\times 15[/tex] = 225 x 10 = 2250 joules.

Hence, the kinetic energy of the ball will be 2250 joules.

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Answer: 20 grams

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed by liquid = 5.23 kcal = 5230 cal    (1kcal=1000cal)

C = heat capacity of liquid = [tex]2.09cal/g^0C[/tex]

Initial temperature of the liquid  = [tex]T_i[/tex] = 101 K

Final temperature of the liquid  = [tex]T_f[/tex]  = 225 K

Change in temperature ,[tex]\Delta T=T_f-T_i=(225-101)K=124K[/tex]

Putting in the values, we get:

[tex]5230=m\times 2.09cal/g^0C\times 124K[/tex]

[tex]m=20g[/tex]

Thus the sample of liquid in grams is 20

Answer:

[tex]W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]

Explanation:

To find the work W to put the negative charge in the new orbit you can use the following formula:

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

That is, the total work is equal to the change in the kinetic energy of the negative charge. Then you calculate the speed of the electron, by using the second Newton Law and the expression for the electrostatic energy:

[tex]F=ma_c\\\\-k\frac{(q_1)(q_2)}{r_1^2}=m\frac{v^2}{r_1}\\\\v^2=k\frac{q_1q_2}{mr_1}[/tex]

r1: radius of the first orbit

m: mass of the negative charge

v: velocity of the charge

k: Coulomb's constant

q1: charge of the fixed particle at point P

q2: charge of the negative charge

Hence, the velocity of the charge in a new orbit with radius r2 is:

[tex]v'^2=k\frac{q_1q_2}{mr_2}[/tex]

Finally the work required to put the charge in the new orbit is:

[tex]W=\Delta K =\frac{1}{2}m[v'^2-v^2]\\\\W=\frac{1}{2}m[k\frac{q_1q_2}{mr_2}-k\frac{q_1q_2}{mr_1}]\\\\W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]

Answer:

v(t) = 21.3t

v(t) = 5.3t

[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]

Explanation:

When no sliding friction and no air resistance occurs:

[tex]m\frac{dv}{dt} = mgsin \theta[/tex]

where;

[tex]\frac{dv}{dt} = gsin \theta , 0 < \theta < \frac{ \pi}{2}[/tex]

Taking m = 3 ; the differential equation is:

[tex]3 \frac{dv}{dt}= 128*\frac{1}{2}[/tex]

[tex]3 \frac{dv}{dt}= 64[/tex]

[tex]\frac{dv}{dt}= 21.3[/tex]

By Integration;

[tex]v(t) = 21.3 t + C[/tex]

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta[/tex]

Taking m =3 ; the differential equation is;

[tex]3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}[/tex]

[tex]\frac{dv}{dt}= 5.3[/tex]

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv[/tex]

The differential equation is :

= [tex]3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v[/tex]

= [tex]3 \frac{dv}{dt}=16 -\frac{1}{3}v[/tex]

By integration

[tex]v(t) = 48 + Ce ^{\frac{t}{9}[/tex]

Since; V(0) = 0 ; Then C = -48

[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]