An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/s2 and there is no atmosphere. How long does it take for the instrument to return to where it was thrown?

Answer:

28

Explanation:

X-13=15

X=15+13

X=28

(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:

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:

The change in internal energy is zero.

(a) is correct option

Explanation:

Given that,

Work done W = -400 J

The First law of thermodynamics is law conservation of energy

Law of conservation energy is defined the total amount of energy is constant in isolated system.

The energy can not be created and destroyed but the energy can be changed from one form to other form.

The First law of thermodynamics is given by

[tex]\Delta U=Q+W[/tex]

Where,  

W = work done  by the system or on the system

[tex]\Delta U[/tex] = Total internal energy

Q = heat

In isothermal process, the temperature is constant

The energy can not be change and the internal energy is the function of temperature.

So, The internal energy will be zero .

Hence, The change in internal energy is zero.

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:

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]

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.

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.

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:

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:

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.

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:

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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You can report sexual harassment to all of the above.

Answer:

power, P = 90 hp

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Initial velocity of car, u = 0

Final velocity of car, v = 25 m/s

Time taken, t = 7 s

We need to find the average power delivered by the engine. Work done divided by total time taken is called power delivered by the engine. It is given by :

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

According to work- energy theorem, the change in kinetic energy of the energy is equal to work done i.e.

[tex]W=\Delta E=\dfrac{1}{2}m(v^2-u^2)[/tex]

[tex]P=\dfrac{\dfrac{1}{2}m(v^2-u^2)}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times 1500\ kg\times (25\ m/s)^2}{7\ s}[/tex]

P = 66964.28 watts

Since, 1 hp = 746 W

So, P = 89.76 hp

or

P = 90 hp

So, the average power delivered by the engine is 90 hp. Hence, the correct option is (E) " 90 hp".                                                                                                  

The average power delivered by the engine of a 1500-kg car accelerating from 0 to 25 m/s in 7.0 s is about 90 horsepower (hp). This is calculated by determining the work done as the change in kinetic energy divided by time, and then converting watts to horsepower.

The kinetic energy (KE) of the car at 25 m/s, given that it started from rest, is:

KE = ½ m v^2

KE = ½ × 1500 kg × (25 m/s)^2

KE = 468750 J

The average power, P, delivered by the engine is the work done divided by the time:

P = Work / time = KE / time

P = 468750 J / 7 s

P = 66964.29 W

To convert the power to horsepower:

Power (hp) = Power (W) / 746

Power (hp) = 66964.29 W / 746

Power (hp) = 89.74 hp

Therefore, rounding to the nearest whole number, the average power delivered by the car’s engine is about 90 hp.

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:

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:

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

Answer:

(a) 0.139 N/m

(b) 2.83 Hz

Explanation:

(a) m = 0.22 g = 0.22 x 10^-3 kg

f = 4 Hz

The formula for the frequency of spring is given by

f = [tex]f = \frac{1}{2\pi }\times \sqrt{\frac{k}{m}}[/tex]

Where, k be the spring constant

[tex]4 = \frac{1}{2\times 3.14 }\times \sqrt{\frac{k}{0.22\times 10^{-3}}}[/tex]

k = 0.139 N/m

(b)

m = 0.44 g = 0.44 x 10^-3 kg,

k = 0.139 N/m

Let the frequency be f.

f = [tex]f = \frac{1}{2\pi }\times \sqrt{\frac{k}{m}}[/tex]

[tex]f = \frac{1}{2\times 3.14 }\times \sqrt{\frac{0.139}{0.44\times 10^{-3}}}[/tex]

f = 2.83 Hz

In a simple harmonic system, the spring stiffness constant 'k' and frequency of oscillations are directly linked by the formula f = 1/(2π) * √(k/m). In the given scenario, the spring stiffness constant of the web is calculated as 34.5 N/m. When the mass of the insect doubles, the expected oscillation frequency drops to around 2.8Hz.

In the study of physics, particularly mechanics, harmonic motion is usually demonstrated using a simple spring-mass system. The spring stiffness constant 'k', also known as the force constant, is a key aspect to consider.

In part (a) of your question, we can use the equation for the frequency of oscillations in a simple harmonic oscillator, which is formulated as f = 1/(2π) * √(k/m), where 'f' is the frequency, 'k' is the spring stiffness constant, and 'm' is the mass of the object causing the oscillations. Given the known values in your question, we would then rearrange this equation to solve for 'k': k = (f * 2π)² * m. Inserting the values given (f = 4.0Hz, m = 0.22g = 0.00022kg), we find that k = 34.5 N/m.

For part (b) of your question, if the mass of the insect trapped in the web is doubled (m = 0.44g = 0.00044kg), while the spring stiffness constant remains the same, we would expect the web to oscillate with a lower frequency. We can calculate this using the same formula as before, rearranged to solve for 'f': f = 1/(2π) * √(k/m), and find that the frequency would be approximately 2.8Hz.

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Answer:

A. because the Sun is aligned with the moon

Explanation:

There are two types of tides:

- Spring tides: spring tides occur when the Sun, the Earth and the Moon are aligned along the same line. When this occurs, the gravitational attractions exerted by the Sun and the Moon on the Earth are along the same line - as a result, the water of the oceans is pulled "stronger" along this line, causing a higher tide than normal.

- Neat tides: neat tides occur when the Sun and the Moon are placed at right angle with respect to the Earth - in this case, the tide level is lower, because the gravitational attractions exerted by the Sun and the Moon on the Earth are perpendicular to each other.

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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Answer:

The total resistance is 5000 N. So the net force left for acceleration is 7 000 N (12 000 - 5 000).

7000 N applied to 2x700 kg gives a maximum acceleration of

a = F/m = 7000/1400kg = 5 m/s^2

At 5m/s^2, it will take 8 s to reach 40 m/s. In 8 second the distance covered will be

d = 1/2 at^2 = 1/2 x 5 x 8^2 = 160 m

Because each glider has the same drag and inertia, the tension in the rope between the gliders will be exactly half of the tension between plane and the two gliders:

12 000/2 = 6 000N.

The second law of Newton and kinematics allows to find the answers for the distance and tension are:

A) The distance traveled is 127.2 m

B) The tension in the second glider is 6003 N

Kinematics studies the movement of bodies establishing relationships between the position, velocity and acceleration of bodies

               v² = [tex]v_o^2 + 2 a x[/tex]

Where v is the velocity, v₀ the initial velocity, a the acceleration and x the distance traveled

Newton's second law states that the net force is proportional to the mass and the acceleration of the body

         F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body

A) Ask the length needed for takes off

We look for the acceleration that the system has using Newton's second law,

The free body diagram is a representation of the system without the details of the bodies, in the adjoint we can see a free body diagram of the forces, let's write Newton's second law for each axis

Glider 1

x-axis

          T₁ - T₂ -fr = m a

y-axis  

         N₁ - W = 0

         N₁ = W

Glider 2

x-axis

         T₂ -fr = m a

y-axis  

         N₂ - W =

         N₂ = W

We write the system of equations

        T₁ - T₂ -fr  = m a

              T₂ - fr = m a

We solve the system

         T₁ - 2 fr = 2 m a

         a = [tex]\frac{T_1 - 2 fr}{2m}[/tex]

Indicates that the friction force for each glider is 1600 N

Calculate

         a = [tex]\frac{12000 - 2 \ 1600}{2 \ 700}[/tex]

         a = 6.29 m / s²

Taking the acceleration we can use the kinematics relationship to find the distance traveled

          v² = [tex]v_o^2+2ax[/tex]

As part of rest the initial velocity is zero

           v² = 0 + 2 ax

           x = [tex]\frac{v^2}{2a}[/tex]v² / 2 a

indicate that the speed of 40 { m/s} is required for takeoff

           x = [tex]\frac{40^2}{2 \ 6.29}[/tex]

           x = 127.2 m

B) The tension (T2) between the two gliders is requested

We write the second Newton law for the second glider, see attached

                  T₂ - fr = m a

                  T₂ = m a + fr

                  T₂ = 700 6.29 + 1600

                  T₂ = 6003 N

In conclusion using Newton's second law and kinematics we can find the answers for the distance and tension are;

A) The distance traveled is 127.2 m

B) The tension in the second glider is 6003 N

Learn more about Newton's second law and kinematics here:

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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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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:

3 N

Explanation:

According to the Coulomb's law in electrostatics, the Coulomb force acting between two charges is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

Case I :

F = 300 N, d = 33 cm

300 ∝ 1/ 3^2

300 ∝ 1/9    ..... (1)

Case II:

Let F be the force, d = 30 cm

F ∝ 1/30^2

F ∝ 1/900     ....(2)

Dividing equation (2) by equation (1), we get

F / 300 = 9 / 900

F = 300 / 100

F = 3 N

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

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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