In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by 0.50 m, the mass that moves downward is 80 kg, and the collision on the floor lasts 0.081 s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

Complete Question:
Two uniform spheres,each of mass 0.260kg are fixed at points A and B

A.)Find the magnitude of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point P and acted on only by forces of gravitational attraction of the spheres at A and B.

B.) Find the direction of the initial acceleration of a uniform sphere with mass 0.010 kg.

Answer:

Friction force is 0.1375 N

Solution:

As per the question:

Radius of the metal disc, R = 4.0 cm = 0.04 m

Magnetic field, B = 1.25 T

Current, I = 5.5 A

Force on a current carrying conductor in a magnetic field is given by considering it on a differential element:

[tex]dF = IB\times dR[/tex]

Integrating the above eqn:

[tex]\int dF = IB\int_{0}^{R}dr[/tex]

[tex]\tau = IB\times \frac{R^{2}}{2} = \frac{1}{2}IBR^{2}[/tex]          (1)

Now the torque is given by:

[tex]\tau = F\times R[/tex]                        (2)

From eqn (1) and (2):

[tex]F\times R = \frac{1}{2}IBR^{2}[/tex]

Thus the Frictional force is given by:

[tex]F = \frac{1}{2}\times 5.5\times 1.25\times 0.04 = 0.1375\ N[/tex]

The new level of sound β2 = β1 + 10 log10(f).

Suppose that a sound has initial intensity level β1 measured in decibels.

This sound now increases in intensity level by a factor f. What is the new level of sound β2?

Formula to calculate the new sound intensity β2 = β1 + 10 log10(f)

The new level of sound β2 can be calculated using the formula β2 = β1 + 10 log10(f)

Here,β1 is the initial sound intensity levelβ2 is the new sound intensity level if is the factor by which the sound intensity level is increased by.

Sound intensity, also known as acoustic intensity, is defined as the power carried by sound waves per unit area in a direction perpendicular to that area.

The SI unit of intensity, which includes sound intensity, is the watt per square meter (W/m2)

Hence, The new level of sound β2 = β1 + 10 log10(f).

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The new level of the sound, β2, can be found by using the property of logarithms. The formula to calculate the new level is β2 = β1 + 10 log10(f). If one sound is twice as intense as another, it has a sound level about 3 dB higher.

The new level of the sound, β2, can be found by using the property of logarithms. When the intensity of a sound increases by a factor of f, the sound level increases by 10 times the logarithm (base 10) of f. So, in this case, if the initial sound has intensity β1, the new level of sound β2 can be obtained by:

β2 = β1 + 10 log10(f)

For example, if the initial sound has intensity β1 = 60 dB and it increases by a factor of f = 2, then:

β2 = 60 + 10 log10(2) = 60 + 3 = 63 dB

Answer:

[tex]\lambda'=\dfrac{1}{2}\times \lambda[/tex]

Explanation:

The number of oscillation per unit time is called frequency of a wave while the distance between two consecutive crests and the trough is called its wavelength.

The wavelength of a wave is given by :

[tex]\lambda=\dfrac{v}{f}[/tex]

If frequency is doubled, f' = 2f

[tex]\lambda'=\dfrac{v}{f'}[/tex]

[tex]\lambda'=\dfrac{v}{2f}[/tex]

[tex]\lambda'=\dfrac{1}{2}\times \dfrac{v}{f}[/tex]

[tex]\lambda'=\dfrac{1}{2}\times \lambda[/tex]

So, when the frequency is doubled the wavelength of a wave on the string becomes half.

Answer:

Vacuum lamp experiment

Explanation:

Edison began his work on electricity the year 1878, where he made use of platinum wire to make filaments for his lamp as the melting point of Platinum is.

However, later he concluded that Platinum pores absorbed air while heating thus reducing its melting point.

So he came up with the idea of vacuum lamp and developed quite a good vacuum lamp with the help of carbon.

Being highly resistive in nature, carbon would end up burning at quite a rapid rate.

Along with his staff, in the year 1879, Edison conducted his first fruitful experiment by using vacuum lamp with carbon filament.

The angular acceleration of a ceiling fan with a net torque of 2.1 N · m and blades having a moment of inertia of 0.19 kg · m² is approximately 11.05 rad/s².

The subject of this question is Physics, specifically rotation dynamics. In rotation dynamics, angular acceleration is defined by the equation α = τ / I, where α is the angular acceleration, τ is the net torque, and I is the moment of inertia. In the given problem, a net torque of 2.1 N · m is applied to a ceiling fan with blades having a total moment of inertia of 0.19 kg · m². Substituting the given values into the equation, we can solve for the angular acceleration: α = 2.1 N · m / 0.19 kg · m² = approximately 11.05 rad/s². Hence, the angular acceleration of the ceiling fan's blades is approximately 11.05 rad/s².

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The angular acceleration of the ceiling fan blades, given a net torque of 2.1 N·m and a moment of inertia of 0.19 kg·m², is calculated using Newton's second law for rotation. The resulting angular acceleration is 11.05 rad/s².

The student's question is related to the field of Physics, specifically rotational kinematics and dynamics. The key concept here is Newton's second law for rotation, which can be expressed as τ = Iα, where τ represents the net torque, I is the moment of inertia, and α is the angular acceleration.

Given that the net torque (τ) is 2.1 N·m and the moment of inertia (I) is 0.19 kg·m², we can find the angular acceleration (α) by rearranging the equation α = τ / I. Substituting the given values into this equation: α = 2.1 N·m / 0.19 kg·m² = 11.05 rad/s². Hence, the angular acceleration of the blades of the ceiling fan is 11.05 rad/s².

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The magnitude of the acceleration (ae) of the Earth due to the gravitational pull of the Moon can be calculated using Newton's law of universal gravitation.

The magnitude of the acceleration (ae) of the Earth due to the gravitational pull of the Moon can be calculated using Newton's law of universal gravitation. The formula is F = G × (m1 × m2) / r^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers. In this case, the Earth is the object experiencing the acceleration and the Moon is the object exerting the gravitational force.

Given the distance between the center of the Earth and the center of the Moon as r = 4700 km (or 4.7 x 10^6 m) and the gravitational constant F = G × (m1 × m2) / r^2[tex]F = G × (m1 × m2) / r^2[/tex], the formula becomes:

[tex]F = G × (m1 × m2) / r^2[/tex]

Since we know the mass of the Earth, we can calculate the acceleration by rearranging the formula to solve for ae:

ae = F / m1

Substituting the values, we get:

[tex]ae = (G × (m1 × m2) / r^2) / m1[/tex]

Simplifying the equation, we are left with:

[tex]ae = G × m2 / r^2[/tex]

Now, we can plug in the known values. The mass of the Moon (m2) is approximately 7.35 x 10^22 kg:

[tex]ae = (6.67 x 10^-11 N · m^2/kg^2) × (7.35 x 10^22 kg) / (4.7 x 10^6 m)^2[/tex]

Calculating this expression will give us the value for the magnitude of the acceleration of the Earth due to the gravitational pull of the Moon.

Answer:

new moon

Explanation:

A solar eclipse take place at new moon phase, when the moon passes between the earth and the sun and its shadows fall on the Earth's surface which by definition a solar eclipse.

The moon must be in the new moon phase for a solar eclipse to occur.

The moon must be in the new moon phase for a solar eclipse to occur. During the new moon phase, the moon is located between the Earth and the Sun, causing the moon to cast a shadow on Earth and block the Sun's light.

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Answer:

The  maximum kinetic energy of the ejected electron=[tex]3.53\times 10^{-10} J[/tex]

Explanation:

We are given that

Frequency of light source=[tex]5.32\times 10^{23}[/tex] Hz

Work function of arsenic=[tex]7.67\times 10[/tex] eV

We have to find the maximum kinetic energy of ejected electron.

We know that the maximum kinetic energy of ejected electron

[tex]K.E=h\nu-w_o[/tex]

Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]

[tex]\nu[/tex] =Frequency of light source

[tex]w_o[/tex]=Work function

Substitute the values in the given formula

Then, the maximum kinetic energy of ejected electron

[tex]K.E=6.63\times 10^{-34}\times 5.32\times 10^{23}-7.67\times 10\times 1.6\times 10^{-19}[/tex]

Because 1 e V=[tex]1.6\times 10^{-19} J[/tex]

[tex]K.E=35.2716\times 10^{-11}-12.272\times 10^{-18}[/tex]

[tex]K.E=3.53\times 10^{-10}[/tex] J

Hence, the maximum kinetic energy of the ejected electron=[tex]3.53\times 10^{-10} J[/tex]

Answer:

option E

Explanation:

given,

decibel level of a busy freeway = 100 dB

when flow is = 100 car/minutes

traffic flow= 20 car/minute

Intensity factor =[tex]\dfrac{I}{I_0}[/tex]

                         =[tex]\dfrac{20}{100}[/tex]

                         =[tex]\dfrac{1}{5}[/tex]

[tex]\Delta \beta = 10 log\dfrac{I}{I_0}[/tex]

[tex]\Delta \beta = 10 log\dfrac{1}{5}[/tex]

[tex]\Delta \beta = - 6.989\ dB[/tex]

The late night decibel level

       = 100 - 6.989

       = 93

The correct answer option E

Answer:

343m

Explanation:

From the question Sally hits one blow every second

However, the sound is delayed by that same amount of time (1 sec)

If the speed of sound is 343m/s

Speed = Distance/Time

Speed = 343m/s

Time = 1s

Distance = Speed x Time

= 343 x 1 = 343m

Answer:

L - h = 12.3672 in

Explanation:

Given

P = 41.0 lb/in² = 41 P.S.I

L = 16.8 in

A = 3.00 in²

h = ?

In order that air flows into the tire, the pressure in the pump must be more than the tire pressure,  41.0  PSI.

We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be  14.6959  P.S.I , we can use the ideal gas equation

P*V = n*R*T

As number of moles of air do not change during its compression in the pump, n*R*T of the gas equation is constant. Therefore we have

P₁*V₁ = P₂*V₂    ⇒    V₂ = P₁*V₁ / P₂

where  

1  and  2  are initial and final states respectively,

V₁ = A*L = (3.00 in²)*(16.8 in)   ⇒   V₁ = 50.4 in³

P₁ = 14.6959  P.S.I

P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²

Inserting various values we get

V₂ = (14.6959  P.S.I)*(50.4 in³) / (55.6959 lb/in²)

⇒  V₂ = 13.2985 in³

Length of pump, measured from bottom, this volume corresponds to is

h = V₂ / A  = (13.2985 in³) / (3.00 in²)

⇒  h = 4.4328 in

Piston must be pushed down by more than

L - h = 16.8 in - 4.4328 in = 12.3672 in

Answer:

Technical director.

Explanation:

The answer is the Technical director.

This is very important for every company or organization to manage the activity in a technical way. Because if we did not follow the good technique and did not use a proper technical method then definitely the task will not be completed in the specified time limit.

Answer:

c. the hole in the plate and the opening in the ring get larger.

Explanation:

The hole in the plate and the ring both of them get enlarged.

The thermal expansion is very much similar to the optical expansion of the shadow of an object.

The change in length of an object is given by the relation:

[tex]\Delta l=l.\alpha.\Delta T[/tex]

where:

l = original length of the object

[tex]\alpha=[/tex] coefficient of linear expansion

ΔT = change in temperature

When heated from 20 degrees C to 100 degrees C, the diameter of the hole in a steel plate and the inner diameter of a steel ring both increase due to thermal expansion. So the correct option is C.

The correct answer to the question about what happens to the diameter of a hole in a steel plate and the inner diameter of a steel ring when both are heated is that both will get larger. This is due to the thermal expansion of materials, which states that objects expand in all dimensions as their temperature increases. When the steel plate and ring are heated from 20 degrees C to 100 degrees C, the particles within the steel vibrate more and push each other farther apart, causing both the solid material and the empty space within, like the hole and the opening of the ring, to increase in size.

Final answer:

To find the spring constant of the car's springs, we use Hooke's Law and the given information about the car's mass, maximum amplitude, and speed. The spring constant is approximately 92638 N/m.

Explanation:

To find the spring constant of the car's springs, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is F = -kx, where F is the force, k is the spring constant, and x is the displacement.

In this case, the car bounces up and down with the maximum amplitude when traveling at 5.7 m/s. The maximum amplitude of the oscillation is half the distance between bumps, so it is 4.9 m/2 = 2.45 m. We can use this information to solve for the spring constant k using the formula:

k = mω²

where m is the mass of the car and ω is the angular frequency, which can be calculated using the formula:

ω = 2πf

where f is the frequency, which is the reciprocal of the period T.

Given that the bumps are spaced 4.9 m apart, the period of the oscillation is the time it takes the car to travel one bump, which is T = 4.9 m / 5.7 m/s = 0.859 s.

The frequency is the reciprocal of the period, so f = 1 / T = 1 / 0.859 s = 1.164 Hz.

Plugging the values into the formulas, we have:

ω = 2π(1.164) ≈ 7.294 rad/s

k = (1800 kg)(7.294 rad/s)² ≈ 92638 N/m

Therefore, the spring constant of the car's springs is approximately 92638 N/m.

Answer:

(a.) 4z

(b.) 4w

Explanation:

From the equation y=4zcos(8πwt), where z and w are positive constants.

Comparing this equation to the equation of a wave y = Acos(Wt), where A is the amplitude (largest distance from equilibrium) and W is the angular frequency (W=2πf)

(a.) Comparing our wave equation with the given equation, we see that A = 4z in this case (furthest distance of the mass from equilibrium)

(b.) Comparing similarly we can see from our given equation that angular frequency W =8πw we also know that W = 2πf from our wave equation, therefore 2πf = 8πw

Solving for f we have f = 8πw÷2π

f = 4w (Proves our second answer because the frequency is the number of oscillations completed per second)

Answer:

60 ft

Explanation:

The tree man and their shadow system forms two similar right angle triangles.

the figure is in the attachment.

let ∠BAC= θ

Therefore, in triangle ABC

tanθ = BC/AC= 35/75= 7/15

now in ΔAEF also

tanθ = EF/AE = 7/75-x= 7/15

solving we get x=60 ft

Now AC and FA are the shadows of the tree and the man respectively.

now FA =75-x=75-60= 15 ft

Therefore, the man must stand at a distance of 60 ft from the tree can you stand and still be completely in the shadow of the tree

A person 7 feet tall can stand 60 feet away from a 35-foot tree casting a 75-foot shadow to be completely in the shadow of the tree.

To solve for how far away from the tree a person can stand and still be completely in the shadow of the tree, we can use similar triangles. Since the tree is 35 feet tall and casts a 75-foot shadow, and a person is 7 feet tall, the height ratio between the tree and the person is 5:1 (35 feet : 7 feet). Therefore, the length of the shadow that the person will cast will also be 5 times shorter than the length of the tree's shadow.

We calculate the length of the person's shadow as follows:

Length of person's shadow = Length of tree's shadow / Height ratio

Length of person's shadow = 75 feet / 5

Length of person's shadow = 15 feet

The total distance from the tree at which the person can stand and still be completely in the shadow is the length of the tree's shadow minus the length of the person's shadow.

Total distance from the tree = Length of tree's shadow - Length of person's shadow

Total distance from the tree = 75 feet - 15 feet

Total distance from the tree = 60 feet

Answer:

Power, P = 560 W

Explanation:

Given that,

Weight of the student, F = 700 N

Distance, d = 8 m

Time taken, t = 10 s

To find,

The average power expended by the student.

Solution,

Let P is the power. The work done per unit time is called its power. It is given by :

[tex]P=\dfrac{W}{t}[/tex]

[tex]P=\dfrac{F\times d}{t}[/tex]

[tex]P=\dfrac{700\ N\times 8\ m}{10\ s}[/tex]

P = 560 W

So, the  average power expended by the student to overcome gravity is most nearly 5670 watts.

The average power expended by the  student to overcome gravity =

( 5.) P = 560 W

The formula for Power is given by the equation (1)  

[tex]Power = \dfrac{Work}{Time}[/tex]...........(1)

Also the formula for Work is given by equation (2)

[tex]Work = Force \times Distance[/tex]............(2)

(Considering magnitude of Displacement)

From equation (1) and (2) we can get

[tex]Power = \dfrac{Force \times Distance}{Time}[/tex].......(3)

Also   [tex]Speed= \dfrac{ Distance}{Time}[/tex]....(4)

From equation (2) , (3) and (4)

So [tex]Power= Force \times Speed[/tex]

Given

Weight of the Student = 700 N

Height of the rope  = 8 m

Time taken  to climb the rope = 10 s.

So from equation (3)  we can get

Average Power expended by the student to overcome gravity = (700)[tex]\times[/tex] (8)/(10) = 560 W.

Hence option 5 is correct.

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Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

[tex]a=(-2v^3)\ m/s^2[/tex]

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]\dfrac{dv}{dt}=-2v^3[/tex]

[tex]\dfrac{dv}{-v^3}=2dt[/tex]

On integrating

[tex]int{\dfrac{dv}{-v^3}}=\int{2dt}[/tex]

[tex]\dfrac{1}{2v^2}=2t+C[/tex]

[tex]v^2=\dfrac{1}{4t+2C}[/tex]....(I)

At t = 0, v = 10 m/s

[tex]10^2=\dfrac{1}{4\times0+2C}[/tex]

[tex]C=\dfrac{1}{200}[/tex]

Put the value of C in equation (I)

[tex]v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}[/tex]

[tex]v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}[/tex]

[tex]v=0.099\ m/s[/tex]

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

[tex]\dfrac{ds}{dt}=v[/tex]

On integrating

[tex]\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt[/tex]

[tex]s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'[/tex]

At t = 0, s = 15 m

[tex]15=\dfrac{200}{800}+C'[/tex]

[tex]C'=15-\dfrac{200}{800}[/tex]

[tex]C'=14.75[/tex]

Put the value in the equation

[tex]s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75[/tex]

[tex]s=19.75\ m[/tex]

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

The velocity of an object depends on its position.

Part A: The velocity of the moving particle at t = 25 s is 0.099 m/s.

Part B: The position of the particle at t = 25 s is 15.475 m.

Velocity is defined as the rate at which the position of an object changes.

Given that the deacceleration of the particle is [tex]a = (-2v^3)[/tex]. The velocity of the particle v = 10 m/s and a position s = 15 m when t = 0.

The acceleration of the particle is given below.

[tex]a = \dfrac {dv}{dt}[/tex]

[tex]-2v^3 = \dfrac {dv}{dt}[/tex]

[tex]- 2 dt = \dfrac {1}{v^3}dv[/tex]

Integrating the above equation, we get,

[tex]\int (-2 dt) = \int \dfrac {1}{v^3} dv[/tex]

[tex]2t + C = \dfrac {1}{2v^2}[/tex]

[tex]v^2 = \dfrac {1}{4t + 2C}[/tex]

Putting t=0 s, v = 10 m/s

[tex]100 = \dfrac {1}{4\times 0 + 2C}[/tex]

[tex]100 = \dfrac {1}{2C}[/tex]

[tex]C = 5 \times 10^{-3}[/tex]

Part A: Velocity

The velocity of the particle at t = 25 s is given as below.

[tex]v^2 = \dfrac { 1}{(4\times 25) + (2\times 5\times 10^{-3})}[/tex]

[tex]v^2 = \dfrac{1}{100.01}[/tex]

[tex]v = 0.099 \;\rm m/s[/tex]

Hence the velocity of the moving particle at t = 25 s is 0.099 m/s.

Part B:  Position

The velocity of the particle is given as,

[tex]v = \dfrac {ds}{dt}[/tex]

[tex]v dt = ds[/tex]

Integrating the above equation, we get,

[tex]\int vdt = \int ds[/tex]

[tex]\int \sqrt{\dfrac {1}{4t + (\dfrac {1}{100})}} dt = s[/tex]

[tex]\int\sqrt{ {\dfrac {100}{400 t + 1}}}dt = s[/tex]

[tex]s =\dfrac {\sqrt{100} \times 2 \sqrt{400 t + 1}}{400} + C'[/tex]

Put t = 0 s, and s = 10 m,

[tex]15=\dfrac {\sqrt{100} \times 2 \sqrt{400 \times 0 + 1}}{400} + C'[/tex]

[tex]15 = \dfrac {\sqrt{100}\times 2}{400} + C'[/tex]

[tex]C' = 14.95[/tex]

Now the position at t = 25 s is,

[tex]s =\dfrac {\sqrt{100} \times 2 \sqrt{400 \times 25 + 1}}{400} + 14.95[/tex]

[tex]s =15.475 \;\rm m[/tex]

Hence the position of the particle at t = 25 s is 15.475 m.

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The change in angular velocity of the bicycle tire is -5.40 rad/s, indicating a change in direction. The average angular acceleration during this time period was -2.84 rad/s2, also reflected by the negative sign representing a change in direction.

This question pertains to physics, specifically focusing on angular velocity and angular acceleration. For the bicycle tire spinning counterclockwise at 2.70 rad/s to stop and spin in the clockwise direction at the same angular velocity, its total change in angular velocity (∆ω) is a combination of the initial and final angular velocities, given by ∆ω = ω_final - ω_initial. Since the tire goes from spinning counterclockwise (considered positive) to clockwise (considered negative), ∆ω = (-2.70 rad/s) - (2.70 rad/s) = -5.40 rad/s.

Next, to find the average angular acceleration (∆av), we use the formula ∆av = ∆ω / ∆t. Substituting in the time given (1.90 s) along with our calculated ∆ω, ∆av = -5.40 rad/s / 1.90 s = -2.84 rad/s2. Note that the negative sign indicates a change in direction.

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Answer:Image will be located at 2F on the other side of the lens.

Explanation:

Answer:

2 f

Explanation:

Explanation:

Given that, the height of the tide measured at a seaside community varies according to the number of hours t after midnight. The height is given by the equation as :

[tex]h=-\dfrac{1}{2}t^2+6t-9[/tex]

When the tide first be at 6 ft, put h = 6 ft in above equation as :

[tex]-\dfrac{1}{2}t^2+6t-9=6[/tex]

[tex]-t^2+12t-18=0[/tex]

On solving the above equation to find the value of t. It is equal to :

t = 3.551 seconds

or

t = 8.449 seconds

So, the tide of 6 ft is at  3.551 seconds and 8.449 seconds. Hence, this is the required solution.

Answer:

574.25 °

Explanation:

To know this, we need to get the expressions to calculate °F and °K.

In the case of Kelvin:

°K = °C + 273

in fahrenheit:

°F = 9/5(°C + 32)

Now, in order to know the temperature which both Kelvin and Fahrenheit are the same, we need to use both above equations, and solve for °C.

°C = °K - 273 (1)

°C = 5/9(°F - 32) (2)

Using 1 and 2 into a same expression:

°K - 273 = 5/9(°F - 32)

With this, we need to know now the moment which K = F, so, all we need to do is replace the F for the K in the above expression. Doing this, we have:

°K - 273 = 5/9(°K - 32)

°K = 0.555(°K - 32) + 273

°K = 5/9K - 17.78 + 273

°K - 5/9K = 255.22

4/9°K = 255.22

°K = 255.22 * 9 / 4

°K = 574.25 °

And this is the temperature in which both Kelvin and Fahrenheit are the same.

Answer:

Minimum required energy

The correct orientation

Explanation:

A collision is described as successful or effective if the particles:

The success of a collision between reactant molecules in causing a reaction depends on the molecules colliding with enough energy to overcome the activation energy, proper orientation during the collision, and a higher collision frequency influenced by reactant concentration.

To answer your question about the factors that determine whether a collision between two reactant molecules results in a reaction, we need to consider several key factors. First, reactant molecules must collide with sufficient energy; this energy must be equal to or higher than the activation energy of the reaction. Second, the molecules must have proper orientation during the collision to allow the necessary rearrangement of electrons and atoms. Lastly, the collision frequency is also essential, which is influenced by the concentration of the reactants; higher concentration generally leads to more collisions and an increased likelihood of reaction.

The deformation of molecules during a collision can also be a contributing factor, as more violent collisions tend to lead to greater deformation and enhance the probability of reaction. Additionally, external factors such as temperature and the presence of catalysts can increase the kinetic energy of the molecules or reduce the activation energy, both of which can lead to a higher rate of successful collisions and reactions.

Answer:

Part a)

[tex]W = 2716.5 J[/tex]

Part B)

[tex]W = -2716.5 J[/tex]

Explanation:

Part A)

While escalator is moving up work done to move the person upwards is given as

[tex]W = mgh[/tex]

here we know that

m = 63 kg

h = 4.4 m

now we have

[tex]W = 63 \times 9.8 \times 4.4[/tex]

[tex]W = 2716.5 J[/tex]

Part B)

Work done while we move from up to down

So we have

[tex]W = - mgh[/tex]

so we have

[tex]W = -2716.5 J[/tex]

Answer:

mass of the baryon = [tex]\frac{(qRB)^{2} }{2K}[/tex]

speed of the baryon = [tex]\frac{2K}{qRB}[/tex]

Explanation:

For any body to move in a circular path, there must be a centripetal force which is directed towards the center of the circle. Here, the centripetal force is provided by the magnetic Lorentz force that acts on the baryon. Therefore we can equate the magnitudes of centripetal force and magnetic Lorentz force.

Magnitude of Centripetal force = [tex]\frac{mv^{2} }{R}[/tex]

Magnitude of Magnetic Lorentz force =  qvB

where,

m = mass of the baryon

v = velocity of the baryon

Thus,

[tex]\frac{mv^{2} }{R}[/tex] = qvB

[tex]\frac{mv}{R}[/tex] = [tex]qB[/tex] (cancelling v on both sides)

v = [tex]\frac{qRB}{m}[/tex]

We know, Kinetic energy, K = [tex]\frac{1}{2} mv^{2}[/tex]

Substituting v in the above equation we get,

K = [tex]\frac{1}{2} m(\frac{qRB}{m} )^{2}[/tex]

K = [tex]\frac{(qBR)^{2} }{2m}[/tex] (simplifying)

Thus,

m = [tex]\frac{(qRB)^{2} }{2K}[/tex]

We already got that

v = [tex]\frac{qRB}{m}[/tex]

substituting the value of m in this equation gives,

[tex]v = \frac{qRB}{\frac{(qRB)^{2} }{2K}}[/tex]

v = [tex]\frac{2K}{qRB}[/tex] (simplifying)

Thus,

mass of the baryon = [tex]\frac{(qRB)^{2} }{2K}[/tex]

velocity of the baryon = [tex]\frac{2K}{qRB}[/tex]

Answer:

The magnitude of the resultant is 30.4 N.

The resultant angle direction is 25.3°.

Explanation:

To find the resultant of the magnitude and direction for given forces “P” and “Q” are 20 N and 15 N respectively, the angle (θ) between them is 60°.

We know that from triangle law of forces,  

[tex]R=\sqrt{P^{2}+2 P Q \cos \theta+Q^{2}}[/tex]

Substitute the given values in the above formula,

[tex]R=\sqrt{20^{2}+2 (20)(15) Q \cos 60+15^{2}}[/tex]

[tex]R=\sqrt{400+600(0.5) + 225}[/tex]

[tex]R=\sqrt{400+300 + 225}[/tex]

[tex]R=\sqrt{925}[/tex]

R = 30.4 N

The magnitude of the resultant is 30.4 N.

To find the direction of the resultant we know that [tex]\text {Resultant angle}=\tan ^{-1} \frac{Q \sin \theta}{P+Q \cos \theta}[/tex]

Substitute the given values in the above formula,

[tex]\text {Resultant angle}=\tan ^{-1} \frac{15 \sin 60}{20+15 \cos 60}[/tex]

[tex]\text {Resultant angle}=\tan ^{-1} \frac{12.99}{20+7.5}[/tex]

[tex]\text {Resultant angle}=\tan ^{-1} \frac{12.99}{27.5}[/tex]

[tex]\text { Resultant angle }=\tan ^{-1} 0.472[/tex]

Resultant angle=25.3°

The resultant angle direction is 25.3°.

Answer:

68 db

Explanation:

Since now instead of one two dogs are barking simultaneously close to each other, therefore we take n =2.

Ignoring interference effects, the barking of two dogs result in a higher level of intensity which is given by,

β(db)=10×㏒(2)

=3 db

So, a reasonable estimate for the raised Intensity Level is: 65db+3db = 68db

The intensity level of two identical dogs barking together would be approximately 68 dB, as an increase in

sound intensity leads to an increase in the decibel level.

The intensity level of a sound is determined by its amplitude and is measured in decibels (dB). The relationship between intensity and decibel level is logarithmic, which means that an increase of 10 dB corresponds to a tenfold increase in intensity. Hence, if two identical dogs are barking very close to each other their combined sound would be twice as intense as one dog barking alone. However, this does not mean that the decibel level would double. In fact, when the intensity of a sound is doubled, the increase in decibel level is about 3 dB. Thus, if a single dog barking has an intensity level of 65 dB, two dogs barking together would have an estimated intensity level of 68 dB.

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Answer:

Cold front

Explanation:

The occurrence of cold front takes place when colder air at the surface pushes up the warmer into the atmosphere.

The movement of these cold fronts takes place at a faster pace as compared to the other fronts and are these are linked with the most violent types of weather such as severe thunderstorms are produced.

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

Final answer:

In subduction zones, the denser oceanic crust is pushed beneath another plate, which may be either oceanic or continental. This process results in the formation of ocean trenches and volcanic activity. Subduction zones play a crucial role in the cycle of plate tectonics.

Explanation:

The two kinds of crust involved in a subduction zone are the oceanic crust and the continental crust. In a subduction zone, one tectonic plate, usually the denser oceanic plate, is forced beneath another plate, which can be either oceanic or continental. Often it is the older and colder oceanic plate that subducts beneath the younger and warmer one. The subduction process is marked by features such as ocean trenches, volcanic island arcs, and is associated with powerful earthquakes and volcanic eruptions.

Subduction zones are integral to the Earth's plate tectonics cycle, balancing the creation of new crust at mid-ocean ridges by destroying the old crust as it descends into the mantle. Such areas of crust destruction are most commonly found around the edges of the Pacific Ocean. The descending plate undergoes changes due to high pressures and temperatures, leading to the release of water and the melting of the crust, forming magma that can result in volcanic activity.