A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

a) What is the magnitude of the force of the wall on ladder?b) What is the magnitude of the normal force of the ground on ladder?c) What is the minimum coefficient of friction so the ladder does not slip?d) What is the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder?

The velocity of the boat relative to the earth is 7.2 m/s, 32.0° south of east. It will take the boat 0.118 seconds to cross the river. The boat will reach a point 0.236 meters south of the starting point.

To find the velocity of the boat relative to the earth, we need to find the resultant of the boat's velocity relative to the water and the river's velocity relative to the earth. Using vector addition, we can find that the magnitude of the total velocity is 7.2 m/s and the direction is 32.0° south of east.

To find the time required to cross the river, we can use the formula t = d/v, where d is the width of the river and v is the horizontal component of the boat's velocity relative to the earth. Plugging in the values, we find that the time required is 0.118 seconds.

To find how far south of the starting point the boat will reach the opposite bank, we can use the formula d = v*t, where v is the vertical component of the boat's velocity relative to the earth and t is the time. Plugging in the values, we find that the boat will reach a point 0.236 meters south of the starting point.

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We use vector addition to find the magnitude of 4.65 m/s at an angle of 25.4° south of east. The time required to cross the river is approximately 119.05 seconds, and the boat will reach a point around 238.1 meters south of its starting point.

The question is about calculating the motion of a boat in a river flowing in a specific direction. Let's solve each part step-by-step.

(a) The boat's velocity relative to the water is 4.2 m/s due east, and the river flows due south at 2.0 m/s. We can use the Pythagorean theorem to find the magnitude:

[tex]Magnitude = \sqrt{(velocity_x^2 + velocity_y^2)}\\Magnitude = \sqrt{(4.22 + 2.02)}\\Magnitude = \sqrt{(17.64 + 4.00)}\\Magnitude = \sqrt{21.64}\\Magnitude = 4.65 m/s[/tex]

To find the direction, we need to calculate the angle θ relative to the east:

[tex]tan(\theta) = velocity_y / velocity_x = 2.0 / 4.2\\\theta = tan-1(2.0 / 4.2)\\\theta \approx 25.4\textdegree south\right \left of\right \left east[/tex]

(b) Given the river's width is 500 m and the boat's velocity relative to the water is 4.2 m/s, we use the formula:

Time = Distance / Velocity
Time = 500 m / 4.2 m/s
Time ≈ 119.05 s

(c) To find the southward distance, we use the river's flow velocity and the time calculated in part (b):

[tex]Distance_{south} = Velocity_{river} \times Time\\Distance_{south} = 2.0 m/s \times 119.05 s\\Distance_{south} \approx 238.1 m[/tex]

To produce a second tone or the first overtone in a tube closed at one end, the length of the tube required is three times the length used for the fundamental frequency, resulting in a length of 1.008 m.

To understand the length required to produce a second tone or the first overtone in a tube closed at one end, it's essential to grasp the concept of harmonics in sound resonance. In such a tube, the resonant frequencies occur in odd multiples of the fundamental frequency. The first resonance the students observed, with the fundamental frequency of 256 Hz at a length of 0.336 m, corresponds to a quarter wavelength of the sound wave in the tube.

For the first overtone (second resonance), the air column in the tube must accommodate three-quarters of a wavelength, meaning the effective length will be three times larger than that of the fundamental. Thus, if the fundamental resonance occurs at a length of 0.336 m, the length for the second resonance will be:

This calculation is based on the understanding that the second tone or first overtone in a closed tube happens at three times the length necessary for the fundamental frequency, leading to the described increase in the length of the air column.

To find the length of tube for the second resonance, halve the initial length where the first resonance occurred at a fundamental frequency of 256 Hz.

The length required to produce a second tone under the same experimental conditions can be calculated based on the concept of resonance in a closed tube.

To find the length for the second resonance (first overtone), we know that the first resonance occurs at 0.336m for a fundamental frequency of 256 Hz. The second resonance, in this case, would occur at half the wavelength of the fundamental frequency, so the length would be half of the initial length: 0.168m.

Answer:

Option (2)

Explanation:

The Cryosphere refers to the frozen water bodies on earth. This includes the glaciers, icebergs, ice sheets and the frozen water surrounding the Arctic as well as Antarctica.

The Hydrosphere refers to all the water bodies on earth including the rivers, streams, lakes, and ponds.

The given condition is based on the interaction between the cryosphere and the hydrosphere.

The frozen ice in the Antarctic and Arctic is melting rapidly due to the increase in the global warming effect. This declining ice in the polar region results in the rise in the global sea level. This can be catastrophic as many of the big cities will be flooded because of this increasing height of sea level.

Thus, the correct answer is option (2).

The decline of Arctic sea ice and its impact on water levels is an interaction between two Earth system spheres: the cryosphere and hydrosphere.

The interaction of Arctic sea ice decline and increasing water levels involves the cryosphere and hydrosphere spheres. The cryosphere refers to the frozen components of the Earth system, including ice caps, glaciers, and sea ice. The hydrosphere encompasses all the water on Earth, including oceans, lakes, and rivers.

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To calculate the magnitude of the total acceleration of the car 11 seconds after it starts, consider both tangential and centripetal accelerations. The tangential acceleration is 2.457 m/s², and the centripetal acceleration is 8.63 m/s², resulting in a total acceleration of 8.97 m/s².

To determine the magnitude of the total acceleration of the car 11 seconds after the start, we need to consider both the tangential acceleration (uniform acceleration as it speeds up) and the centripetal acceleration (due to the car's circular motion). To calculate the tangential acceleration ([tex]a_t[/tex]), we use the final speed (v) the car reaches in 13 seconds, which is 115 km/h, and convert it to meters per second:

v = 115 km/h = (115 * 1000 m) / (3600 s) = 31.94 m/s

Since the car accelerates uniformly from rest, the tangential acceleration is given by:

[tex]a_t[/tex] = v / t = 31.94 m/s / 13 s = 2.457 m/s²

After 11 seconds, the speed of the car (v₁₁) is:

v₁₁ = at * 11 s = 2.457 m/s² * 11 s = 27.03 m/s

The centripetal acceleration ([tex]a_c[/tex]) is calculated using the formula:

[tex]a_c[/tex] = v₁₁² / r = (27.03 m/s)² / 85 m = 8.63 m/s²

To find the total acceleration, we use the Pythagorean theorem because the tangential and centripetal accelerations are perpendicular to each other:

a = √([tex]a_t[/tex]² + [tex]a_c[/tex]²) = √((2.457 m/s²)² + (8.63 m/s²)²) = √(6.036 + 74.516) = √(80.552) = 8.97 m/s².

Final answer:

To find the car's acceleration, use the kinematic equation v = u + at. The car's acceleration is 2.5 m/s^2. To find the car's speed as it passes the 0.14 km post, plug the values into the kinematic equation v = u + at. The car's speed is 20 m/s.

Explanation:

To find the car's acceleration, we can use the kinematic equation:
v = u + at
where v is the final velocity, u is the initial velocity (which is 0 m/s since the car starts from rest), a is the acceleration, and t is the time. We are given that the car passes the 0.14 km post at 8.0 s after starting, which means the car travels a distance of 0.14 km in 8.0 s. Converting 0.14 km to meters gives us 140 m. Plugging the values into the equation, we have:
20 = 0 + a * 8.0
Simplifying, we find that the car's acceleration is 2.5 m/s^2.

To find the car's speed as it passes the 0.14 km post, we can use the kinematic equation:
v = u + at
Since the car starts from rest (u = 0 m/s) and the car's acceleration is 2.5 m/s^2 (which we just found), we can plug these values into the equation along with the time (8.0 s) to find the car's speed:
v = 0 + 2.5 * 8.0
Simplifying, we find that the car's speed as it passes the 0.14 km post is 20 m/s.

Answer:

maximum width of the doorway = 35.77ft

Explanation:

The detailed calculation and derivation from first principle is as shown in the attachment

Answer:

the maximum width is x= 4√2 ft = 5.656 ft

Explanation:

for the parabola

y= a*x² + b*x + c

where y= height and x= width

an aircraft hangar should be symmetric with respect to the y axis , then

y(-x)=y(x) → a*x² + b*x + c = a*x² - b*x + c →-2*b*x =0 → b=0

it also should be pointing downwards → a is negative

, then the parabola would be

y= c- a*x²

since c= maximum height = 18 ft

then for y=0 , x= 48 ft/2 = 24 ft  →  0 = 18 ft - a*(24 ft)² → a= 1/32 ft⁻¹

then

y= 18 ft- 1/32 ft⁻¹ *x²

since the doorway cannot go beyond the parabola , the maximum possible doorway is obtained when the doorway touches the parabola.

then for a height y= 8 ft

8 ft = 18 ft- 1/32 ft⁻¹ *x²

x= 4√2 ft = 5.656 ft

When an object is dropped from a certain height with no air resistance, it takes a specific time to reach the ground. The time it takes to reach the ground is proportional to the square root of the height. If the object is dropped from three times the original height, it will take T * sqrt(3) time to reach the ground.

When an object is dropped from a certain height with no air resistance, it takes a specific time (T) to reach the ground. If the object is dropped from three times that height, it will take a longer time to reach the ground. The relationship between the height and the time taken is linear, assuming no air resistance.

To determine the time it would take to reach the ground from three times the original height, we need to consider that the time is proportional to the square root of the height. Let's represent the original time as T and the original height as H. Therefore, the time it would take to reach the ground from three times the original height (3H) would be:

t = T * sqrt(3)

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Answer:

Reduced by 49 times

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

[tex]F_G = G\frac{M_1M_2}{R^2}[/tex]

where G is the gravitational constant. [tex]M = M_1 = M_2[/tex] are the masses of the 2 objects. and R is the distance between them.

Since R squared is in the denominator of the formula, if we make it 7 times as large with no change in mass, gravitational force would be dropped by 7*7 = 49 times

To solve the problem we should know about Newton's Law of gravity.

According to Newton's law of gravity, there is an attractive force between any two-particle carrying mass, such that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

[tex]F \propto m_1m_2\\\\F \propto \dfrac{1}{R^2}[/tex]

[tex]F = G\dfrac{m_1m_2}{R^2}[/tex]

Where G is the proportionality constant and the value of G is 6.67 x 10-11 N m² / kg².

The force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.

Given to us,

Assumption

Let's assume that the distance between the moon and the planet is d.

Values

As it is given that there is no change in the mass of the moon or the planet, therefore,

Also, it is given that the distance between them changes to 7 times, therefore,

Substitute the value Newton's Law of gravity,

[tex]F = G\dfrac{m_1m_2}{(7d)^2}\\\\\\F = G\dfrac{m_1m_2}{49d^2}[/tex]

Thus, the force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.

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Answer:

Explanation:

V = Deltax/Deltat

V = 15.0 m/s

Displacement:

(a) Vf = Vi + adeltat

Vf = 15.0m/s - 9.8m/s^2 x 0.500s = 10.1m/s

Displacement = (15.0m/s x 0.500s) - (0.5)(9.8m/s^2)(0.500s)^2 = 6.275m

(b) Vf = 15.0m/s - 9.8m/s^2 x 1.00s = 5.2m/s

Displacement = (15.0m/s x 1.00s) - (0.5)(9.8m/s^2)(1s)^2 = 10.1m

(c) Vf = 15.0m/s - 9.8m/s^2 x 1.50s = 14.7m/s

Displacement = (15.0m/s x 1.50s) - (0.5)(9.8m/s^2)(1.5s)^2 = 11.475m

(d) Vf = 15.0m/s - 9.8m/s^2 x 2.00s =  19.6m/s

Displacement = (15.0m/s x 2.00s) - (0.5)(9.8m/s^2)(2s)^2 = 10.4m

To calculate the molecular weight of a polyethylene molecule with n=750, multiply the molecular weight of the ethylene unit by n.

To calculate the molecular weight of a polyethylene molecule with n=750, we need to know the molecular formula and the atomic weights of the elements present in the molecule. Polyethylene is made up of repeating ethylene (C2H4) units, so we can calculate the molecular weight of the polyethylene molecule by multiplying the molecular weight of the ethylene unit (28.05 g/mol) by the value of n (750).

Calculation:
Molecular weight of polyethylene = Molecular weight of ethylene unit × n = 28.05 g/mol × 750 = 21,037.5 g/mol

Therefore, the molecular weight of the polyethylene molecule with n=750 is 21,037.5 g/mol, rounded to three significant figures.

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Answer:

1. True

2. True

3. False

Explanation:

According to Ohm's law, V = IR where V is the voltage, I is the current and R is the resistance. So if V is constant, R is inversely proportional to I

1. The resistivity of metal wires would depend on the temperature, this is true.

2. When length L of a wire decreases, the resistance R decreases as well, this would make I increase.

3. When the resistivity increases, resistance R increases as well, this would make I decrease.

The resistivity of a wire is affected by temperature, with most metals increasing in resistivity with higher temperatures. Current increases when the length of a wire decreases at constant voltage. However, current decreases when resistivity increases at constant voltage.

When examining the resistance of a wire, there are several factors to consider that impact how the resistance changes under various conditions. The statements about resistance of a wire need to be evaluated for their truthfulness.

Final Answer:

The speed at which the car hits the water is approximately 75.2 feet per second.

Explanation:

To find the speed at which the car hits the water, we can use one of the kinematic equations that relates the initial velocity, acceleration due to gravity, the height it fell from, and the final velocity. The kinematic equation that we need is:

[tex]\[ v^2 = u^2 + 2gh \][/tex]

Where:
-  v  is the final velocity,
-  u  is the initial velocity,
-  g  is the acceleration due to gravity (which we will use  [tex]\( 32.174 \, \text{ft/s}^2 \)[/tex] for since we are dealing with feet),
-  h  is the height.

Here, we are given:
-  [tex]\( u = 8 \, \text{ft/s} \)[/tex] (initial velocity)
- [tex]\( h = 87 \, \text{ft} \)[/tex] (height)
- [tex]\( g = 32.174 \, \text{ft/s}^2 \)[/tex] (acceleration due to gravity)

Let's find the final velocity \( v \) using these values.

[tex]\[ v^2 = u^2 + 2gh \][/tex]

[tex]\[ v^2 = (8 \, \text{ft/s})^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]

[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]
[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 5591.148 \, \text{ft}^2/\text{s}^2 \][/tex]
[tex]\[ v^2 = 5655.148 \, \text{ft}^2/\text{s}^2 \][/tex]

Now we take the square root of both sides to solve for the final velocity \( v \):

[tex]\[ v = \sqrt{5655.148} \, \text{ft/s} \][/tex]

Performing the square root calculation, we get:

[tex]\[ v \approx 75.2 \, \text{ft/s} \][/tex]

So, the speed at which the car hits the water is approximately 75.2 feet per second.

Answer: Asteroids, meteoroids, and comets are remnants of the early solar system. This Statement is TRUE.

Explanation:

METEOROID: these are small rocky or metallic objects found in outer space.

ASTEROIDS: these are also known as minor planets of the inner solar system. They are irregularly shaped object in space that orbits the Sun.

COMETS: these are dusty chunk of ice, that moves in a highly elliptical orbit about the sun.

Asteroids, meteoroids, and comets as remnants of the early solar system was further proved in nebular hypothesis

initially proposed in the eighteenth century by German philosopher Immanuel Kant and French mathematician Pierre-Simon Laplace. (The word nebula means a gaseous cloud.) According to the modern version of the theory, about 4.5 to 5 billion years ago the solar system developed out of a huge cloud of gases and dust floating through space. These materials were at first very thin and highly dispersed.

Answer:

The y-component of the car's position vector is 670m/s.

The x-component of the acceleration vector is -3, and the y-component is 40.

Explanation:

The displacement vector of the car with velocity

[tex]\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s[/tex]

is the integral of the velocity.

Integrating [tex]\boldsymbol{v}[/tex] we get the displacement vector [tex]\boldsymbol{d}[/tex]:

[tex]\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j} )[/tex]

Now if the initial position if the car is

[tex]\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})[/tex]

then the displacement of the car at time [tex]t[/tex] is

[tex]\boldsymbol{d(t)}= \boldsymbol{r+d}[/tex]

[tex]\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j} )}[/tex]

Now at [tex]t=10s[/tex], we have

[tex]\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j} )}m[/tex]

The y-component of the car's position vector is 670m/s.

The acceleration vector is the derivative of the velocity vector:

[tex]\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})[/tex]

and at [tex]t=10s[/tex] it is

[tex]\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2[/tex]

The x-component of the acceleration vector is -3, and the y-component is 40.

The x and y components of the car's position at 10 s are -147.0 m and 668.67 m, respectively. The x and y components of the car's acceleration at 10 s are -3 m/s² and 40 m/s², respectively.

The problem involves determining the position and acceleration components of a remote-controlled car from given velocity functions over time.

1.) X Component of Car's Position at 10 s:

Given the velocity component, vx = -3t m/s, we need to integrate it with respect to time to find the position (x). The initial position x0 is 3.0 m.

When t = 10 s:

2.) Y Component of Car's Position at 10 s:

Given the velocity component, vy = 2t² m/s, integrating it with respect to time gives the position (y). The initial position y0 is 2.0 m.

When t = 10 s:

3.) X Component of Car's Acceleration at 10 s:

Given the velocity component, vx = -3t m/s, the acceleration is the time derivative of velocity.

Hence, at t = 10 s:

4.) Y Component of Car's Acceleration at 10 s:

Given the velocity component, vy = 2t² m/s, the acceleration is the time derivative of velocity.

Hence, at t = 10 s:

Answer:

a)  x_{cm} = m₂/ (m₁ + m₂)   d , b)   x_{cm} = 52.97 pm

Explanation:

The expression for the center of mass is

                [tex]x_{cm}[/tex] = 1 / M  ∑ [tex]x_{i}[/tex] [tex]m_{i}[/tex]

Where M is the total masses, mI and xi are the mass and position of each element of the system.

Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon

              x_{cm} = 1 / (m₁ + m₂)   (0+ m₂ x₂)

Let's reduce the magnitudes to the SI system

             m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg

             m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg

             d = 128 pm = 128 10⁻¹² m

The equation for the center of mass is

               x_{cm} = m₂/ (m₁ + m₂)   d

b) let's calculate the value

            x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷    128 10-12

            x_{cm} = 52.97 10⁻¹² m

            x_{cm} = 52.97 pm

(a) The expression for the center mass of these two atoms relative to oxygen atom is  [tex]X_{cm} = \frac{m_1 d_0 \ +\ m_2d}{m_1 + m_2}[/tex]

(b) The numeric value for the center of mass of carbon monoxide is 53 pm.

The given parameters;

The center mass of these two atoms relative to oxygen atom is calculated as follows;

[tex]X_{cm} = \frac{m_1 d_0 \ +\ m_2d}{m_1 + m_2}[/tex]

where;

(b)

The numeric value for the center of mass of carbon monoxide in units of pm is calculated as follows;

[tex]X_{cm} = \frac{17u(0) \ +\ 12u(128 \ pm)}{(12u + 17u)}\\\\X_{cm} = \frac{(12 \times 128u) \ pm}{29u} \\\\X_{cm} = 53 \ pm[/tex]

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Answer:

The correct answer is:

d) A and B have the same magnitude flux

Explanation:

Electric flux is the property of electric field that measures the electric field lines, passing through a surface and electric flux is also directly proportional to the number of electric field lines passing through a surface.  

The formula of electric flux is:

Φ = E A Cos θ

(where E is the electric field, A is the area of face and θ is the angle between the face and the electric field).

Since, faces A and B are perpendicular to the electric field and the electric field lines passing through face A also passes through face B therefore, both of these faces have larger and same magnitude of electric flux.

Since, face C is parallel to the electric field so, the electric flux is smaller at face C, because the magnitude of Cos 180 (when face is parallel) is smaller than the magnitude of Cos 90 (when face is perpendicular).

Face A, which is perpendicular to the uniform electric field, has the largest magnitude electric flux through it because the angle between the field lines and the normal to the surface is zero, maximizing the electric flux.

The question revolves around calculating the electric flux through different faces of a prism in a uniform electric field. Electric flux (Φ) is given by the equation Φ = E ⋅ A ⋅ cos(θ), where E is the magnitude of the electric field, A is the area through which the field lines pass, and θ is the angle between the field lines and the normal (perpendicular) to the surface.

Face A is perpendicular to the electric field, which means the angle θ is 0 degrees and cos(θ) is 1. Thus the flux through Face A is maximum. For Face B, the leaning face, θ is greater than 0 degrees but less than 90 degrees, thus cos(θ) will be less than 1. Hence, flux through Face B will be less than through Face A. Face C, being parallel to the electric field, has θ as 90 degrees, and cos(90) is 0, so the flux through Face C is zero. Therefore, in comparison, Face A has the largest magnitude electric flux through it.

Answer:

532.0725 m

102.17270893 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = g

H = Height of cliff

Distance traveled in 3 seconds

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m[/tex]

Distance traveled by sound = 2H-44.145 m

[tex]2H-44.145=ut+\dfrac{1}{2}at^2\\\Rightarrow 2H-44.145=340\times 3\\\Rightarrow H=\dfrac{340\times 3+44.145}{2}\\\Rightarrow H=532.0725\ m[/tex]

The height of the cliff is 532.0725 m

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 532.0725+0^2}\\\Rightarrow v=102.17270893\ m/s[/tex]

Her speed just before she hits the ground is 102.17270893 m/s

Answer:

12.9121148614 m/s

35.9393048982 m

Explanation:

t = Time taken

u = Initial velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 38=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{38\times 2}{9.81}}\\\Rightarrow t=2.78337865516\ s[/tex]

Time taken for the bag to fall is 2.78337865516 seconds

Time she has been jogging for

9+2.78337865516 = 11.78337865516 seconds

Total distance traveled by her

[tex]s=vt\\\Rightarrow s=3.05\times 11.78337865516=35.9393048982\ m[/tex]

Henrietta is 35.9393048982 m away

Velocity of throwing

[tex]\dfrac{35.9393048982}{2.78337865516}=12.9121148614\ m/s[/tex]

The velocity of throwing is 12.9121148614 m/s

Final answer:

Bruce must throw the bagels at an initial speed of 12.92 m/s for Henrietta to catch them, and Henrietta will be 35.93 m from the point directly below Bruce's window when she catches the bagels.

Explanation:

Projectile Motion and Kinematics Problem

To find the initial speed Bruce must throw the bagels, we need to consider two aspects of projectile motion: the horizontal motion, which is constant because air resistance is neglected, and the vertical motion, which is influenced by gravity.

Firstly, we need to calculate the time it takes for the bagels to fall from the window to the ground. Using the equation for free fall h =
1/2 g t², where h is the height (38.0 m), and g is the acceleration due to gravity (9.81 m/s²), we can solve for t, the time to fall:

38.0 m = 1/2 * 9.81 m/s² * t²

t = sqrt(2 * 38.0 m / 9.81 m/s²) = sqrt(7.74) ≈ 2.78 s

Bruce throws the bagels 9.00 s after Henrietta has passed below the window. In this time, Henrietta has jogged a distance of d = speed * time = 3.05 m/s * 9.00 s = 27.45 m horizontally.

Since Henrietta is already past the point directly below the window, we need to add the distance she will cover in the time it takes for the bagels to fall. This distance is additional distance = jogging speed * fall time = 3.05 m/s * 2.78 s ≈ 8.48 m.

Overall, Henrietta will be approximately 27.45 m + 8.48 m = 35.93 m from the point directly below the window when she catches the bagels.

To find the initial speed with which Bruce throws the bagels, we use the horizontal motion formula initial speed = distance / time, which gives us an initial speed of approximately 35.93 m / 2.78 s ≈ 12.92 m/s.

Bruce must throw the bagels horizontally at an initial speed of approximately 12.92 m/s for Henrietta to catch them just before they hit the ground, at a distance of approximately 35.93 m from the point directly below Bruce's window.

Explanation:

A nucleus that absorbs 11.45δ is less shielded than a nucleus that absorbs at 4.13δ.

the nucleus that absorbs at 11.45δ requires weaker applied field strength to come into resonance than the nucleus that absorbs at 4.13δ.

A nucleus absorbing at 4.13 δ is more shielded than one absorbing at 11.45 δ because it is in a weaker local magnetic field and resonates at a lower frequency.

In the context of nuclear magnetic resonance (NMR) spectroscopy, the chemical shift is given in units of delta (δ) which represents the resonance frequency of a nucleus relative to a standard reference compound. When a nucleus is surrounded by a dense cloud of electrons, it is considered to be ‘shielded’. A shielded nucleus is influenced by a smaller local magnetic field because the electrons repel some of the external magnetic field. As a result, shielded nuclei resonate at a lower frequency (higher δ values) when compared to de-shielded nuclei. Therefore, a nucleus that absorbs at 4.13 δ is more shielded than a nucleus that absorbs at 11.45 δ because the latter is in a stronger local magnetic field and is, hence, de-shielded.

Answer:

The angle θ between the direction of the current and the direction of the uniform magnetic field is 41° or 139°.

Explanation:

The force on a current carrying wire is given by the following equation:

[tex]\vec{F} = I\vec{L}\times \vec{B}[/tex]

The cross-product can be written with a sine term:

[tex]F = ILB\sin(\theta)\\0.025 = (3.4)(0.28)(0.04)\sin(\theta)\\\sin(\theta) = 0.6565[/tex]

Therefore, the angle θ is 41.03° and 138.97°

The angles can be calculated using the formula sin(θ) = F / (I L B), giving two symmetrical values about 90° in the first and second quadrants because the sine function is periodic.

The question is asking for the angles at which the force on a current-carrying wire in a magnetic field is a specific magnitude. The magnitude of the force exerted on a current-carrying wire placed in a magnetic field is given by the formula F = I L B sin(θ), where F is the force, I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the direction of the current and the direction of the magnetic field.

By rearranging for θ, we get the equation sin(θ) = F / (I L B). Plugging in the values from the question, we find sin(θ) = 0.0250 N / (3.40 A  imes 0.280 m  imes 0.0400 T). This gives us θ values that correspond to the sine of this ratio.

There are two angles that will produce the same sine value because sine is a periodic function, which are θ and 180°-θ. Therefore, the two angles between the direction of the current and the direction of the uniform magnetic field for which the force on the wire has a magnitude of 0.0250 N will be symmetrical about 90° in the first and second quadrants.

Answer:

Explanation:

Given

Initial velocity of ball [tex]u=10\ m/s[/tex]

height of window [tex]h=20\ m[/tex]

Using Equation of motion

[tex]y=ut+\frac{1}{2}at^2[/tex]

where u=initial velocity

t=time

a=acceleration

As ball is already is at a height of 20 m so

[tex]Y=ut+\frac{1}{2}at^2+20[/tex]

[tex]Y=10\times t+0.5\times (-9.8)t^2+20[/tex]

[tex]Y=-4.9t^2+10t+20[/tex]

(b)highest point is obtained at v=0

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

where

v=final velocity

u=initial velocity

a=acceleration

s=displacement

[tex](0)-10^2=2\times (-9.8)\times s[/tex]

[tex]s=\frac{100}{19.6}[/tex]

[tex]s=5.102\ m[/tex]

Highest Point will be [tex]s+20=25.102\ m[/tex]

(c)Time taken when the ball hit the ground i.e. at Y=0

[tex]-4.9t^2+10t+20=0[/tex]

[tex]t=3.28\ s[/tex]

impact velocity [tex]v=\sqrt{2\times 9.8\times 25.102}[/tex]

[tex]v=22.181\ m/s[/tex]

(a) The equation be "Y = -4.9t² + 10t + 20".

(b) The highest point be "25.102 m".

(c) The impact velocity be "22.181 m/s"

According to the question,

Ball's initial velocity, u = 10 m/s

Window's height, h = 20 m

(a) By using equation of motion,

Y = ut + [tex]\frac{1}{2}[/tex]at²

By substituting the values,

  = ut + [tex]\frac{1}{2}[/tex]at² + 20

  = 10 × t + 0.5 × (9.8)t² + 20

  = -4.9t² + 10t + 20

(b) We know that,

→ v² - u² = 2as

here, Final velocity, v = 0

0 - (10)² = 2 × (-9.8) × s

          s = [tex]\frac{100}{19.6}[/tex]

             = 5.102 m

(c) Time taken will be:

→ -4.9t² + 10t + 20 = 0

                            t = 3.28 s

hence,

The impact velocity,

v = [tex]\sqrt{2\times 9.8\times 25.102}[/tex]

  = 22.181 m/s

Thus the above response is correct.

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In a single phase, the addition or removal of energy changes Kinetic Energy not Potential Energy. However, during a phase change, this energy addition or subtraction results in a change in Potential Energy, not Kinetic Energy.

The subject of this question is energy and phase, particularly in the context of Potential Energy (PE) and Kinetic Energy (KE). When only one phase is present, adding or removing energy will mainly change the KE, not the PE. This is because the energy is utilized to speed up or slow down the particles, thus changing their kinetic energy. However, during a phase change, adding or removing energy changes PE but not KE as it alters the state rather than the speed of the particles. Statement B is the one that is accurate while only one phase is present, whereas the correct option for the phase change scenario is option D.

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112.11 V

The electromotive force, e.m.f, (E) of the generator is related to its terminal voltage(V) and the current (I) it delivers as follows;

E = V + (I x r) -------------------------(i)

Where;

r = the internal resistance of the generator

Now, as stated in the question;

at V = 108V, I = 10.2A

Substitute the values into equation (i) as follows;

E = 108 + (10.2 x r)

E = 108 + 10.2r       ------------------(ii)

Also, as stated in the question;

at V = 93V, I = 47.4A

Substitute these values also into equation (i) as follows;

E = 93 + (47.4 x r)

E = 93 + 47.4r      ------------------------(iii)

Now solve equations (ii) and (iii) simultaneously;

Subtract equation (iii) from (ii)

  E = 108 + 10.2r

_

  E = 93 + 47.4r

______________

  0 =  15 - 37.2r           ----------------(iv)

_______________

Solve for r in equation (iv)

37.2r = 15

r = 15 / 37.2

r = 0.403Ω

The internal resistance is therefore 0.403Ω

Substitute r = 0.403Ω into equation (ii);

E = 108 + 10.2(0.403)

E = 108 + 4.11

E = 112.11

Therefore, the emf is 112.11 V

The electromotive force (emf) produced in the generator is 112.1126 Volts.

Given the following data:

To calculate the electromotive force (emf) produced in the generator:

Mathematically, the electromotive force (emf) is given by the formula:

[tex]E = V + Ir[/tex]

For the first instance A:

[tex]E = 108 + 10.2r[/tex]   .....equation 1.

For the first instance B:

[tex]E = 93 + 47.4r[/tex]   .....equation 2.

Next, we would equate eqn 1 and eqn 2:

[tex]108 + 10.2r = 93 + 47.4r\\\\108 - 93 = 47.4r - 10.2r\\\\15 = 37.2r\\\\r = \frac{15}{37.2}[/tex]

Internal resistance, r = 0.4032 Ohms

Now, we can calculate the electromotive force (emf) produced in the generator:

[tex]E = 108 + 10.2r\\\\E = 108 + 10.2(0.4032)\\\\E = 108 + 4.1126[/tex]

Emf, E = 112.1126 Volts.

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The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.

Through these waves, you can know that there is beyond the clouds.

Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.

Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.

It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.

Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.

Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.

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To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:

[tex]v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})[/tex]

Here,

[tex]v_{o} =[/tex]  Orbital Velocity

[tex]\lambda_{max} =[/tex] Maximal Wavelength

[tex]\bar{\lambda}} =[/tex] Average Wavelength

c = Speed of light

Replacing with our values we have that,

[tex]v_{o} = (3*10^5) (\frac{3.00036-3}{3})[/tex]

Note that the average signal is 3.000000m

[tex]v_o = 36 km/s[/tex]

Now using the definition about centripetal acceleration we have,

[tex]a_c = \frac{v^2}{r}[/tex]

Here,

v = Orbit Velocity

r = Radius of Orbit

Replacing with our values,

[tex]a = \frac{(36km/s)^2}{100000km}[/tex]

[tex]a= 0.01296km/s^2[/tex]

[tex]a = 12.96m/s^2[/tex]

Applying Newton's equation for acceleration due to gravity,

[tex]a =\frac{GM}{r^2}[/tex]

Here,

G = Universal gravitational constant

M = Mass of the planet

r = Orbit

The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,

[tex]M = \frac{ar^2}{G}[/tex]

[tex]M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}[/tex]

[tex]M = 1.943028*10^{27}kg[/tex]

Therefore the mass of the planet is [tex]1.943028*10^{27}kg[/tex]

Answer:

31 m/s

Explanation:

As both the monkey and the darts are subjected to constant gravitational acceleration g = 9.8 m/s2 and both start from rest (vertically speaking). Their vertical position will always be the same. For the dart to hit the monkey, its horizontal position must be the same as the monkey's, which is unchanged before reaching the ground. Therefore, the time it takes for the dart to travel across 70 m must be less than the time it takes for the monkey to drop 25m to the ground. We can find it out using the following equation of motion

[tex]s_m = gt_m^2/2[/tex]

[tex]25 = 9.8t_m^2/2[/tex]

[tex]t_m^2 = 50/9.8 = 5.1[/tex]

[tex]t_m = \sqrt{5.1} = 2.26 s[/tex]

For the dart to takes less that 2.26 s to travel 70m, its horizontal speed must at least be 70 / 2.26 = 31 m/s

Answer:

Distance between them increase

Explanation:

The position S of the water droplet can be determined  using equation of motion

[tex]S=ut+\frac{1}{2} at^2[/tex]

where [tex]u[/tex] is the initial velocity which is zero here

[tex]t[/tex] is time taken, [tex]a[/tex] is acceleration due to gravity

the position of  first drop after time [tex]t[/tex] is given by

[tex]S_{1} =0 \times t+ \frac{1}{2} at^2=\frac{1}{2} at^2............(1)[/tex]

the position of  next drop at same time is

[tex]S_{2} =\frac{1}{2} a(t-1)^2 = \frac{1}{2} a(t^2+1-2t)............(2)[/tex]

distance between them is [tex]S_{1} -S_{2}[/tex]  is [tex]a(t-1)[/tex]

from the above the difference will increase with the time

As the water drops fall, their velocity increases due to the force of gravity, which causes the distance between each subsequent drop to increase.

The response to the student's question deals with the notion of acceleration due to gravity. As the water drops fall, they are accelerated by gravity, which means their velocity (speed) increases over time. If we consider two subsequent droplets, the second drop begins its descent 1.0 seconds after the first. Therefore, when the second drop begins to fall, the first drop has already accelerated for 1.0 seconds. This causes the distance between the two drops to increase as they fall.

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Answer:

Explanation:

False

Electric conductivity in metals is the result of the motion of charged particles called electrons. Atoms of metallic elements are distinguished by the availability of valence electrons. Valence electrons are the electrons in the outer shell of an atom and free to move around.

These free electrons are free to move in the lattice of metal and result in electric current when a voltage is applied.

Answer:

83816746.4254 m/s

Explanation:

m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

V = Voltage = [tex]2\times 10^4\ V[/tex]

The kinetic energy of the electron is

[tex]K=\dfrac{1}{2}mv^2[/tex]

Energy is given by

[tex]E=qV[/tex]

Balancing the energy

[tex]qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s[/tex]

The velocity of the electrons is 83816746.4254 m/s

Answer:

42.5 Hz.

Explanation:

The fundamental frequency of a closed pipe is given as

f₀ = v/4l....................... Equation 1

Where f₀ = lowest frequency, v = speed of sound in air, l = length of the organ pipe

Given: v = 340 m/s, l = 2.00 m.

Substitute into equation 1

f₀ = 340/(4×2)

f₀ = 340/8

f₀ = 42.5 Hz.

Hence the smallest frequency that will resonant in the organ pipe = 42.5 Hz.