You are a passenger on a jetliner that is flying at constant velocity. You get up from your seat and walk toward the front of the plane. Because of this action, your forward momentum increases. What happens to the forward momentum of the plane itself?

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

[tex]Q = ms\Delta T + mL[/tex]

here we have

[tex]Q = 40(950)(680 - 32) + 40(450 \times 10^3)[/tex]

[tex]Q = 24624 kJ + 18000 kJ[/tex]

[tex]Q = 42624 kJ[/tex]

Since this is the amount of aluminium per hour

so power required to melt is given by

[tex]P = \frac{Q}{t}[/tex]

[tex]P = \frac{42624}{3600} kW[/tex]

[tex]P = 11.84 kW[/tex]

Since the efficiency is 85% so actual power required will be

[tex]P = \frac{11.84}{0.85} = 13.93 kW[/tex]

Part B)

Total energy consumed by the furnace for 30 hours

[tex]Energy = power \times time[/tex]

[tex]Energy = 13.93 kW\times 30 h[/tex]

[tex]Energy = 417.9 kWh[/tex]

now the total cost of energy consumption is given as

[tex]R = P \times 20 \frac{Cents}{kWh}[/tex]

[tex]R = 417.9 kWh\times  20 \frac{cents}{kWh}[/tex]

[tex]R = 8357.6 Cents[/tex]

Answer:

0.36 kg

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum at the beginning must be equal to the total momentum after the skater has thrown the ball.

Before the launch, the skater and the snowball are at rest, so the initial total momentum is zero:

[tex]p_i = 0[/tex]

After the launch, the total momentum is:

[tex]p_f = M V + m v[/tex]

where

M = 53.4 kg is the mass of the ice skater

v = -0.100 m/s is the velocity of the ice skater (here we assumed that east is the positive direction)

m is the mass of the snowball

v = +14.8 m/s is the velocity of the snowball

Since momentum must be conserved,

[tex]p_i = p_f\\0 = MV +mv[/tex]

so we can find m:

[tex]m=-\frac{MV}{v}=-\frac{(53.4 kg)(-0.100 m/s)}{14.8 m/s}=0.36 kg[/tex]

The question pertains to the principle of conservation of momentum. The mass of the snowball is found by equating the momentum of an ice skater with the snowball, resulting in a mass of about 0.36 kg for the snowball.

This question represents a physics concept known as conservation of momentum, stating that the total initial momentum of a system equals the total final momentum in the absence of external forces. In this scenario, the initial momentum of the system is zero because both the ice skater and the snowball are initially at rest.

In the final state, the 53.4 kg ice skater moves west at 0.100 m/s, and the snowball moves east at 14.8 m/s. The eastward momentum equates to the westward momentum.

To find the mass of the snowball, we can set the two momenta to be equal, resulting in the equation: (Mass of Skater) * (Velocity of Skater) = (Mass of Snowball) * (Velocity of Snowball), which simplifies to (53.4 kg) * (0.100 m/s) = (Mass of Snowball) * (14.8 m/s). Solving for the mass of the snowball gives approximately 0.36 kg.

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Answer:

catch the ball

Explanation:

if you push the ball back, that equal amount of force will applied to you. however if you catch it, you absorb less of the directional energy

Answer:

The distance is 54.6 m

Explanation:

Given that,

Mass = 2.0 kg

Frictional coefficient = 0.21

Initial velocity = 15 m/s

We need to calculate the acceleration

Using formula of frictional force  

[tex]F = \mu mg[/tex]

[tex]F=0.21\times2.0\times9.8[/tex]

[tex]F = 4.12\ N[/tex]

We need to calculate the acceleration

[tex]F = ma[/tex]

[tex]a = \dfrac{F}{m}[/tex]

[tex]a =\dfrac{4.12}{2.0}[/tex]

[tex]a=2.06\ m/s^2[/tex]

We need to calculate the initial velocity

Using equation of motion

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

Put the value  

[tex]0=15^2-2\times2.06\times s[/tex]

[tex]s = \dfrac{15^2}{2\times2.06}[/tex]

[tex]s=54.6\ m[/tex]

Hence, The distance will be 54.6 m.

Answer:

Explanation:

When a body is moving in a liquid, it experiences a backward dragging force. this backward dragging force is given by the formula

F = 6 π η r v

where, η is called the coefficient of viscosity, r be the radius, v be the velocity of object.

As we know that the coefficient of friction for oil is more than water so the dragging force in oil is more than water.

So, the balls in water comes first.

Answer:

9680 cal

Explanation:

A = cross-sectional area of the bar = 30 cm² = 30 x 10⁻⁴ m²

L = length of the bar = 1.5 m

T₁ = Temperature at one end of the bar = 25 °C

T₂ = Temperature at other end of the bar = 300 °C

k = Thermal conductivity of Aluminum = 1.76 x 10⁴ Cal cm /(m² ⁰C)

Q = amount of heat conducted per second

Amount of heat conducted per second is given as

[tex]Q = \frac{k A (T_{2} - T_{1})}{L}[/tex]

Q = (1.76 x 10⁴) (30 x 10⁻⁴) (300 - 25)/1.5

Q = 9680 cal

Answer:

When the angle is 56° the work done is 117.43 J

Explanation:

Work = F . s = Fscosθ

We have

   W1 = 210 J, θ = 0°

Substituting

      210 = F x s x cos 0 = Fs

Now we have to find W2 when angle θ = 56°

Substituting

      W2 = F x s x cos 56 = 210 cos56= 117.43 J

When the angle is 56° the work done is 117.43 J

Explanation:

Centripetal force is mass times centripetal acceleration:

F = m v² / r

If force is doubled while mass and radius are held constant, then velocity will increase.

2F = m u² / r

2 m v² / r = m u² / r

2 v² = u²

u = v√2

So the velocity increases by a factor of √2.

Doubling the centripetal force acting on an object in circular motion, while maintaining the same radius, necessitates an increase in the object's velocity, leading to faster circular motion.

The question explores how doubling the centripetal force acting on an object in uniform circular motion, while maintaining the same radius, affects its motion. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula for centripetal force is given as Fc = m(v2/r), where m is mass, v is velocity, and r is the radius of the circular path. Doubling the centripetal force while keeping the radius constant means that for the force to remain balanced and the object to stay in circular motion, the velocity of the object must increase. This is because the square of the velocity (v2) is directly proportional to the force applied. Therefore, the object's speed around the circular path will increase, resulting in a faster circular motion.

The power output of a 1.5 g fly moving upwards at a speed of 2.4 cm/s can be calculated by applying the physics formula for power. The resultant power output is ~0.036 Watts.

The power output of any moving body can be calculated using the formula P = mgh/t where 'P' is power, 'm' is mass, 'g' is acceleration due to gravity, 'h' is height and 't' is time. In order to calculate the power output of the fly, we need to convert the variables to the appropriate units.

So first, convert the mass of the fly to kg, which gives 0.0015 kg. Then, convert the speed from cm/s to m/s, giving us 0.024 m/s. The acceleration due to gravity is approximately 9.8 m/s2. We want to find the power as the fly walks 1 meter.

Substituting the values into the power equation, we get P = (0.0015 kg * 9.8 m/s2 * 1 m) / (1 m / 0.024 m/s) = 0.036 W, which when rounded off to two significant figures is 0.036 Watts.

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To calculate the power output of a fly walking up a windowpane, we need to use the equation Power = force x velocity. By converting the mass of the fly into kilograms, calculating the force exerted by the fly using the equation Force = mass x gravity, and multiplying the force by the velocity, we find that the power output of the fly is approximately 0.00004 Watts or 4 x 10^-5 Watts.

To calculate the power output of the fly, we can use the equation: Power = force x velocity.

First, let's convert the mass of the fly from grams to kilograms. 1.5 g = 0.0015 kg. The force exerted by the fly is equal to its weight, which is given by the equation: Force = mass x gravity.

Assuming the acceleration due to gravity is 9.8 m/s^2, the force exerted by the fly is: Force = 0.0015 kg x 9.8 m/s^2. Finally, we can calculate the power output of the fly by multiplying the force by the velocity: Power = (0.0015 kg x 9.8 m/s^2) x 0.024 m/s. This gives us a power output of approximately 0.00004 Watts or 4 x 10^-5 Watts.

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C) Staring At Someone's Body, Giving Them The “Once Over”

All of the following are examples of verbal sexual harassment EXCEPT:

c) Staring At Someone's Body, Giving Them The “Once Over”

Verbal harassment is considered any conscious and repeated attempt to humiliate, demean, insult, or criticize someone with words. Verbal sexual  harassment  is harassing someone by saying something or verbally disrespecting someone in a way that you are discomforting them . It is a type of non - physical harassment that  makes someone humiliated or threatened

hence ,a) Telling “Dirty" Jokes Or Stories b) Making Kissing Sounds, Howling, Or Smacking Lips d)Making Repeated Requests For A Date all are example of verbal sexual harassment all three are non-physical and only words are being used in harassing someone

c) Staring At Someone's Body, Giving Them The “Once Over” is an example of visual sexual harassment as staring will not be considered as verbal , it means that with your vision you are discomforting someone

correct option c)Staring At Someone's Body, Giving Them The “Once Over”

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Answer:

Radius = 0.11 m

Explanation:

To find the speed of the proton we know that

[tex]KE = PE[/tex]

here we have

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

now we have

[tex]v = \sqrt{\frac{2qV}{m}}[/tex]

now we have

[tex]v = \sqrt{\frac{2(1.60 \times 10^{-19})(1000)}{(1.67\times 10^{-27})}}[/tex]

[tex]v = 4.38 \times 10^5 m/s[/tex]

Now for the radius of the circular motion of charge we know

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

[tex]R = \frac{mv}{qB}[/tex]

[tex]R = \frac{(1.67\times 10^{-27})(4.38 \times 10^5)}{(1.60\times 10^{-19})(0.040)}[/tex]

[tex]R = 0.11 m[/tex]

The radius of the proton orbit is 0.114m

When the proton travels through the potential V it gains kinetic energy given below:

[tex]\frac{1}{2}mv^2=qV\\\\v=\sqrt[]{\frac{2qV}{m} }\\\\v=\sqrt{\frac{2\times (1.6\times10^{-19})\times10^3}{1.67\times10^{-27}}[/tex]

[tex]v=4.37\times10^5m/s[/tex]

Now, a moving charge under magnetic field B undergoes circular motion due to the magnetic force being perpendicular to the velocity of the charge, given by:

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

[tex]r=\frac{mv}{qB} \\\\r=\frac{1.67\times10^{-27}\times4.37\times10^5}{1.6\times10^{-19}\times0.040} m\\\\r=0.114m[/tex]

the radius of the orbit is 0.114m

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Answer:

200 KJ

Explanation:

h = 20 m, m = 1000 kg, g = 10 m/s^2

As the rock is sliding and it starts from rest. it falls from a f=height of 20 m

So, by using the energy conservation law

Potential energy at the top of hill = Kinetic energy at the bottom of hill

So, Kinetic energy at the bottom of hill = Potential energy at the top of hill

K.E = m g h

K. E = 1000 x 10 x 20 = 200,000 J

K.E = 200 KJ

The resulting current density in the  section of silver of uniform cross section be 1.54 × 10^11  A/m^ 2.

Electrical resistivity is a measurement of a material's degree of resistance to current flow. The SI unit for electrical resistivity is the ohm metre (m). It is frequently represented by the Greek letter rho.

Materials that easily transmit current and have a low resistance are called conductors. Insulators have a high resistance and do not conduct electricity. Resistivity is defined as the size of the electric field across it that results in a particular current density.

Resistivity at any temperature T is given by:  ρ = ρ₀[1 + α(T-T₀)]  

Given:

ρ₀ = 1.59x10^-8 ohms*m

reference temperature (T₀) = 20°C.

T =  3°C and α = 0.0038/°C.

So, resistivity  ρ =   1.59x10^-8[ 1 +  0.0038×( 3 -20)] ohm.m

= 7.39 × 10^-9 ohm.m.

Again: electric field : E = 1139 V/m

So,  resulting current density  = E/ρ = 1139/ 7.39 × 10^-9  A/m^2

=1.54 × 10^11  A/m^ 2.

Hence, The resulting current density in the  section of silver of uniform cross section be 1.54 × 10^11  A/m^ 2.

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To determine the current density in the silver specimen at 3 C, you must calculate its resistivity at that temperature using the temperature coefficient. Then, find the conductivity and use Ohm's law with the given electric field to calculate the current density.

To find the resulting current density (J), we need to first adjust the resistivity ( ) of silver for the given temperature of 3 C. The resistivity at this temperature can be found using the formula (T) = 0[1 + ( T - 20 C)], where (T) is the resistivity at temperature T, 0 is the resistivity at 20 C, and is the temperature coefficient of the material.

Substituting the given values into the formula, we get: (3 C) = 1.59 10^-8 [ 1 + 0.0038( 3 C - 20 C) ] = 1.59 ₓ 10⁻⁸ [ 1-0.0646] = 1.59 ₓ 10⁻⁸ (0.9354 m) = 1.49 ₓ 10⁻⁸ .

Now that we have the resistivity at 3 C, we can find the conductivity ( s) using s = 1/ which equals 1/(1.49 ₓ 10⁻⁸ ) = 6.71 ₓ 10⁷ S/m.

The current density J can then be calculated using Ohm's law, J = sE, where E is the electric field intensity. Using the provided electric field of 1139 V/m, J = 6.71 ₓ10⁷ S/m 1139 V/m = 7.64ₓ 10¹⁰ A/m.

The acceleration that car B appears to have to an observer in car A is equal to [tex]-16\;km/h/s[/tex].

Given the following data:

Relative acceleration can be defined as the acceleration of an observer B with rest to the rest frame of another observer A.

Mathematically, relative acceleration is given by this formula:

[tex]A_{BA}=A_B-A_A[/tex]

Substituting the given parameters into the formula, we have;

[tex]A_{BA}=-8-8\\\\A_{BA}=-16\;km/h/s[/tex]

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The acceleration that car A appears to have to an observer in car A is -8 km/h per second.

To determine the acceleration that car A appears to have to an observer in car A, we need to consider the relative motion between the two cars.

Since car A has a constant velocity of 100 km/h, the observer in car A would perceive car B's motion as a decrease in velocity with a rate of 8 km/h each second. This means that car B's acceleration as perceived by the observer in car A would be -8 km/h per second.

Therefore, the acceleration that car A appears to have to an observer in car A is -8 km/h per second.

Answer:

2666.67 m

Explanation:

Changing the 8km/s into SI units we 8000m/s.

Speed of a wave = λf where λ is the wavelength and f he frequency of the wave.

v=λf

Therefore making wavelength the subject of the formula we get

λ= v/f

Substituting with the values in the question we get:

λ= (8000m/s)/3Hz

=2666.667 m

Answer:

2.62 m

Explanation:

Let the small child sit at a distance x from the pivot.

The distance of big child from the pivot is 4 - x .

By using the concept of moments.

Clockwise moments = anticlockwise moments

27 x = 51 ( 4 - x )

27 x = 204 - 51 x

78 x = 204

x = 2.62 m

The frequency of the wave is 4921.57 Hz. The angular frequency is 30937.48 rad/s. The period of oscillation is 0.000203 s.

To find the frequency of the wave, we can use the formula:

v = λ × f

Plugging in the values given, we have:

251 = 0.051 × f

solving for f, we get:

f = 251 / 0.051 = 4921.57 Hz

The angular frequency (represented as ω) can be calculated using the formula:

ω = 2π × f

Substituting the value of f we found, we get:

ω = 2×3.1416×4921.57 = 30937.48 rad/s

The period of oscillation (represented as T) is the reciprocal of the frequency.

T = 1 / f = 1 / 4921.57 = 0.000203 s

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Explanation:

a) Calculate the z-score.

z = (x − μ) / σ

z = (1050 − 1100) / 93

z = -0.54

From a z-score table:

P(z<-0.54) = 0.2946

Therefore:

P(z>-0.54) = 1 - 0.2946

P(z>-0.54) = 0.7054

70.54% of steers are over 1050 pounds.

b) Calculate the z-score.

z = (x − μ) / σ

z = (1300 − 1100) / 93

z = 2.15

From a z-score table:

P(z<2.15) = 0.9842

98.42% of steers are under 1300 pounds.

c) Calculate the z-scores.

z = (x − μ) / σ

z = (1150 − 1100) / 93

z = 0.54

z = (1250 − 1100) / 93

z = 1.61

From a z-score table:

P(z<0.54) = 0.7054

P(z<1.61) = 0.9463

Therefore:

P(0.54<z<1.61) = P(z<1.61) - P(z<0.54)

P(0.54<z<1.61) = 0.9463 - 0.7054

P(0.54<z<1.61) = 0.2409

24.09% of steers weigh between 1150 pounds and 1250 pounds.

To calculate the proportion of steers in different weight categories, change the weights to z-scores, and use a standard normal distribution table to find the percentile ranks. For steers over 1050 pounds, about 70.43% meet this category. For steers under 1300 pounds, about 98.41% meet that mark. To find the proportion weighing between 1150 and 1250 pounds, find the percentile ranks for both weights and subtract.

This is a question related to the concept of Normal Distribution in statistics. First, you need to convert the weights to z-scores, which are measures of how many standard deviations an element is from the mean.

(a) To find the percent of steers weighing over 1050 pounds, first calculate the z-score: (1050-1100)/93= -0.5376. Using a standard normal distribution table, you find that the percentile rank for -0.5376 is approximately 0.2957 or 29.57%. However, this represents the proportion of steers that weigh less than 1050 pounds. To find how many weigh more, subtract this from 1. So about 70.43% of steers weigh more than 1050 pounds.

(b) Similarly, for under 1300 pounds, calculate the z-score: (1300-1100)/93= 2.1505. The percentile rank for 2.1505 is approximately 0.9841 or 98.41%, indicating 98.41% of steers will weigh less than 1300 pounds.

(c) To find the proportion weighing between 1150 and 1250 pounds, find the z-scores for both weights and their respective percentile ranks. Then subtract the smaller percentile from the larger one.

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The speed of the electron following a helical path under a uniform magnetic field of 0.340 T is approximately 3.39 x 107 m/s, derived from the relationship of the magnetic force on the electron, the strength of the magnetic field, and the charge of the electron.

The speed of an electron in a uniform magnetic field moving along a helical path can be determined by understanding the magnetic force on the electron and the pitch of its path. Known values for the magnetic field (B), force (F), and charge of an electron (e) can be used, relating F = e*v*B where v is the velocity (speed) of the electron.

From the question, we know that the force F = 1.85 x 10-15N and the magnetic field B = 0.340 T. We also know that the charge of an electron e = 1.6 x 10-19 C. Re-arranging the formula to solve for the velocity v, we get v = F / (e*B). Substituting the known values, we obtain v = 1.85 x 10-15N / (1.6 x 10-19 C * 0.340 T). This results in an electron speed of approximately 3.39 x 107 m/s.

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Answer:

The ball is at its maximum height.

Explanation:

As we know that when ball is thrown at some angle with the horizontal then the component of its velocity is given as

[tex]v_x = vcos\theta[/tex]

[tex]v_y = vsin\theta[/tex]

now here vertical velocity is the velocity in y direction

so it is given as

[tex]v_y = 3.0 sin30 = 1.5 m/s[/tex]

now as the velocity of ball in vertical direction becomes zero

then in that case

[tex]v_f - v_i = at[/tex]

[tex]0 - 1.5 = (-9.8)t[/tex]

[tex]t = 0.15 s[/tex]

since at this position the vertical component of the velocity is zero

so this is the position of ball when its height is maximum

Final answer:

The vertical velocity of a ball thrown at an angle becomes zero at its maximum height. This is the point in its trajectory where gravity has completely counteracted its initial upward velocity. At no other time during its flight will the ball's vertical velocity be zero.

Explanation:

The motion of a projectile launched at an angle involves two components of motion – horizontal and vertical. When a ball is thrown with a velocity of 3.0 m/s at an angle of 30° above horizontal, it will have both horizontal and vertical components of velocity. The vertical velocity of the ball becomes zero when the ball reaches its maximum height, as there is no upward velocity to counteract gravity at this point.

To find the vertical component of initial velocity (Vy), we use the formula Vy = V × sin(θ), where V is the initial velocity and θ is the launch angle. So in this case, the vertical velocity component is Vy = 3.0 m/s × sin(30°).

At maximum height, gravity has slowed the vertical velocity to zero. This is the only point during the ball's trajectory where the vertical component of velocity is zero. Just before the ball hits the ground, its vertical velocity is not zero but is equal in magnitude and opposite in direction to its vertical velocity on launch. When the ball changes direction horizontally, this does not affect the vertical velocity component. Therefore, the ball's vertical velocity is zero only when it reaches its highest point.

Answer:

28

Explanation:

X-13=15

X=15+13

X=28

You can report sexual harassment to all of the above.

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where we have

[tex]\alpha = -80.0 rad/s^2[/tex] is the angular acceleration (negative since the rotor is slowing down)

[tex]\omega_f [/tex] is the final angular speed

[tex]\omega_i = 2000 rad/s[/tex] is the initial angular speed

t = 10.0 s is the time interval

Solving for [tex]\omega_f[/tex], we find the final angular speed after 10.0 s:

[tex]\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s[/tex]

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

If we re-arrange it for t, we get:

[tex]t = \frac{\omega_f - \omega_i}{\alpha}[/tex]

where here we have

[tex]\omega_i = 2000 rad/s[/tex] is the initial angular speed

[tex]\omega_f=0[/tex] is the final angular speed

[tex]\alpha = -80.0 rad/s^2[/tex] is the angular acceleration

Solving the equation,

[tex]t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s[/tex]

Answer:

(bruh moment)

Explanation:

The electric force is a force that is inversely proportional to the square of the distance. This can be proved by Coulomb's Law, which states:  

"The electrostatic force [tex]F_{E}[/tex] between two point charges [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is proportional to the product of the charges and inversely proportional to the square of the distance [tex]d[/tex] that separates them, and has the direction of the line that joins them"  

Mathematically this law is written as:  

[tex]F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}[/tex]

Where [tex]K[/tex] is a proportionality constant.

So, if we have two electrons, this means we have two charges with the same sign, hence the electric force between them will be repulsive.

Final answer:

Explaining the attractive electrostatic force and gravitational force between two electrons at atomic scales.

Explanation:

The electrostatic force between two electrons separated by 10^-10 m is attractive. The magnitude of this force can be calculated using Coulomb's law, and for two electrons with charge e = 1.6 x 10^-19 C, the force is strong at atomic distances.

The gravitational force between the electrons is always attractive and weaker than the electrostatic force, with a magnitude of 5.54 x 10^-51 N. This force is much smaller compared to the electrostatic force at atomic scales.

The balance between attractive and repulsive forces is crucial in understanding the stability of atomic structures and the concept of bond length and bond energy in molecules.

Answer:

48 m/s

Explanation:

m = mass of the object = 28 kg

F = magnitude of net force acting on the object = 232 N

acceleration of the object is given as

a = F/m

a = 232/28

a = 8.3 m/s²

v₀ = initial velocity of the object = 5 m/s

v = final velocity of the object

t = time interval = 5.2 s

using the kinematics equation

v = v₀ + a t

v = 5 + (8.3) (5.2)

v = 48 m/s

Answer:

61.8 N

Explanation:

Given data

We can find the magnitude of the electric force (F) between the two protons using Coulomb's law.

[tex]F=k.\frac{q_{1}q_{2}}{d^{2} } \\F=8.99 \times 10^{9} N.m^{2} .C^{-2} .\frac{(1.60 \times 10^{-19}C)^{2} }{(1.93 \times 10^{-15}m)^{2} } \\F=61.8N[/tex]

The magnitude of the electric force is 61.8 N.

The magnitude of the electric force between two protons at this distance is mathematically given as

F=61.8N

Question Parameter(s):

The protons in a nucleus are approximately 2 ✕ 10^−15 m apart.

a distance d = 1.93 ✕ 10^−15 m apart.

Generally, the equation for the electric force (F)   is mathematically given as

[tex]F=k.\frac{q_{1}q_{2}}{d^{2} }[/tex]

Therefore

[tex]F=8.99* 10^{9} N.m^{2} .C^{-2} .\frac{(1.60 * 10^{-19}C)^{2} }{(1.93* 10^{-15}m)^{2} }[/tex]

F=61.8N

In conclusion, The force is

F=61.8N

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Answer:

Part a)

a = 1.37 m/s/s

Part b)

since the force of gravity is more than the thrust force of rocket so it will not lift off the surface of Earth

Explanation:

Part a)

Net force on the Module due to thrust of engine is given as

[tex]F = 30,000 N[/tex]

now net force while is ejected out of the surface of moon is given as

[tex]F_{net} = F - F_g[/tex]

here we know that

[tex]F_g = mg_{moon}[/tex]

where we have

[tex]g_{moon} = \frac{g}{6}[/tex]

[tex]F_{net} = 30,000 - (10000)(\frac{9.8}{6})[/tex]

[tex]F_{net} = 13666.67 N[/tex]

now the acceleration of the module on the moon is

[tex]a = \frac{F_{net}}{m}[/tex]

[tex]a = \frac{13666.67}{10,000} = 1.37 m/s^2[/tex]

Part b)

Now on the surface of earth the force of gravity on the module is given as

[tex]F_g = mg [/tex]

[tex]F_g = 10,000 \times 9.8[/tex]

[tex]F_g = 98000 N[/tex]

since the force of gravity is more than the thrust force of rocket so it will not lift off the surface of Earth

Acceleration is the rate of change of velocity. The acceleration of the module is 1.365 m/s².

Given to us

Mass of the Module = 10,000 kg

The thrust of the engine, Fₓ = 30,000 N

A.)

We know in order to lift the module the thrust produced by the engine must be greater than the gravitational pull by the module. therefore,

[tex]F_{N} = F_x - W[/tex]

where W is the weight of the moon,

[tex]F_{N} = F_x - (m\times g_{moon})[/tex]

Also, the acceleration on the moon is 1/6 of the acceleration on the earth,

[tex]F_{N} = 30,000- (10,000\times \dfrac{9.81}{6})\\F_N = 13,650\ N[/tex]

We know that force is the product of mass and acceleration, therefore,

[tex]F_N = m \times a\\\\13,650 = 10,000 \times a\\\\a = 1.365 m/s^2[/tex]

Hence, the acceleration of the module is 1.365 m/s².

B.)

If we need to lift the module on earth, we need a thrust that is greater than the weight of the module,

[tex]Weight = mass \times acceleration\\W = 10,000 \times 9.81\\W = 98,100\ N[/tex]

As the weight of the module is greater than the thrust produced by engines. therefore, the module can not take off from the earth.

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1. Maximum Height:

The maximum height (H) can be calculated using the vertical motion equation for projectile motion:

[tex]\[ H = \dfrac{v_{0y}^2}{2g}, \][/tex]

Here:

[tex]\(v_{0y}\)[/tex] is the initial vertical component of velocity (initial velocity [tex]\(v_0\)[/tex] multiplied by the sine of the launch angle),

g is the acceleration due to gravity ([tex]\(9.81 \, \text{m/s}^2\)[/tex] near the surface of the Earth).

Substitute the given values:

[tex]\[ H = \dfrac{(20.0 \, \text{m/s} \cdot \sin(37.0^\circ))^2}{2 \cdot 9.81 \, \text{m/s}^2}\\\\ \approx 10.4 \, \text{m}. \][/tex]

2. Time of Travel:

The total time of flight ([tex]\(t_{\text{total}}\)[/tex]) can be found using the equation for vertical motion:

[tex]\[ t_{\text{total}} = \dfrac{2v_{0y}}{g}. \][/tex]

Substitute the given values:

[tex]\[ t_{\text{total}} = \dfrac{2 \cdot 20.0 \, \text{m/s} \cdot \sin(37.0^\circ)}{9.81 \, \text{m/s}^2}\\\\ \approx 3.22 \, \text{s}. \][/tex]

3. Horizontal Distance:

The horizontal distance (d) can be calculated using the horizontal motion equation:

[tex]\[ d = v_{0x} \cdot t_{\text{total}}, \][/tex]

Here:

[tex]\(v_{0x}\)[/tex] is the initial horizontal component of velocity (initial velocity [tex]\(v_0\)[/tex] multiplied by the cosine of the launch angle),

[tex]\(t_{\text{total}}\)[/tex] is the total time of flight calculated in the previous step.

Substitute the given values:

[tex]\[ d = 20.0 \, \text{m/s} \cdot \cos(37.0^\circ) \cdot 3.22 \, \text{s}\\\\ \approx 54.3 \, \text{m}. \][/tex]

Thus, Maximum height is approximately 10.4 meters, time of travel is around 3.22 seconds, and horizontal distance is about 54.3 meters.

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The football kicked at an angle of 37.0° with a velocity of 20.0 m/s will reach a maximum height of 7.32 meters. It will stay in the air for a total of 2.44 seconds and land 39 meters away.

The subject of your question is projectile motion, which is a part of Physics. Given the initial angle of 37.0° and a velocity of 20.0 m/s, we can calculate the maximum height, time of travel, and horizontal distance traveled using the equations of motion.

First, we need to find the components of the initial velocity. The vertical component (Voy) is found using Voy = Vo*sin(θ) = 20.0 m/s * sin(37.0°) = 12.01 m/s. The horizontal component (Vox) is Vox = Vo*cos(θ) = 20.0 m/s * cos(37.0°) = 15.97 m/s.

The maximum height (H), also known as the apex of the trajectory, can be calculated using the formula H = Voy^2 / (2*g) where g is the acceleration due to gravity (9.81 m/s²). So, H = (12.01 m/s)^2 / (2*9.81 m/s²) = 7.32 meters.

The total time of travel (t), from launch to when the ball returns back to the ground, is t = 2*Voy/g = 2*(12.01 m/s) / (9.81 m/s²) = 2.44 seconds.

Finally, the horizontal distance traveled, known as the range (R), is R = Vox * t = 15.97 m/s * 2.44 seconds = 39 meters.

So the maximum height the football reaches is 7.32 m, the time it stays in the air is 2.44 seconds, and it hits the ground 39 meters away.

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