An overhang hollow shaft carries a 900 mm diameter pulley, whose centre is 250 mm from the centre of the nearest bearing. The weight of the pulley is 600 N and the angle of lap is 180°. The pulley is driven by a motor vertically below it. If permissible tension in the belt is 2650 N and if coefficient of friction between the belt and pulley surface is 0.3, estimate, diameters of shaft, when the internal diameter is 0.6 of the external. Neglect centrifugal tension and assume permissible tensile and shear stresses in the shaft as 84 MPa and 68 MPa respectively.

Answer:

The distance and average speed are 54.79 m and 10.85 m.

Explanation:

Given that,

Speed = 21.7 m/s

Time = 5.05 s

(a). We need to calculate the distance

Firstly we will find the acceleration

Using equation of motion

[tex]v = u+at[/tex]

[tex]a = \dfrac{v-u}{t}[/tex]

Where, v = final velocity

u = initial velocity

t = time

Put the value in the equation

[tex]a = \dfrac{21.7-0}{5.05}[/tex]

[tex]a = 4.297 m/s^2[/tex]

Now, using equation of motion again

For distance,

[tex]s = ut+\dfrac{1}{2}at^2[/tex]

[tex]s = 0+\dfrac{1}{2}\times4.3\times(5.05)^2[/tex]

[tex]s=54.79\ m[/tex]

The distance is 54.79 m.

(b). We need to calculate the average speed during this time

[tex]v_{avg}=\dfrac{D}{T}[/tex]

Where, D = total distance

T = time

Put the value into the formula

[tex]v_{avg}=\dfrac{54.79}{5.05}[/tex]

[tex]v_{avg}=10.85\ m/s[/tex]

Hence, The distance and average speed are 54.79 m and 10.85 m.

To find the magnitude of the net force acting on the kayak, we will use the following kinematic equation derived from Newton's second law of motion:

v2 = u2 + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance over which the acceleration occurs.
We know that the kayak starts from rest, so u = 0.
The final velocity v is 0.680 m/s and the distance s is 0.428 m. Rearranging the equation for acceleration a, we get:

a = (v2 - u2) / (2s)

By plugging in the given values, we find that the acceleration a is:
a = (0.6802 - 02) / (2  imes 0.428) = 0.6802 / 0.856 = 0.541 m/s2

Now, using Newton's second law (Force = mass x acceleration), we find the net force F:
F = mass x a
F = 82.7 kg x 0.541 m/s2 = 44.7397 N

The magnitude of the net force acting on the kayak is approximately 44.74 N.

Answer:

Weight of gasoline = 2039.93 kN

Volume = 80783.80 gallon

Explanation:

Volume = Base area x Depth

[tex]\texttt{Base area = }\frac{\pi d^2}{4}=\frac{\pi \times 25^2}{4}=490.88ft^2[/tex]

Depth = 22 ft

Volume = 490.88 x 22 = 10799.22 ft³ = 10799.22 x 0.0283168 = 305.8 m³

Density of gasoline = 680 kg/m³

Mass of gasoline = 305.8 x 680 = 207943.65 kg

Weight of gasoline = 207943.65 x 9.81 = 2039.93 kN

We have 1 m³ = 264.172 gallon

            305.8 m³ = 305.8 x 264.172 = 80783.80 gallon

Explanation:

Mass = density × volume

m = (0.7 g/mL) × (40 mL)

m = 28 g

Rounding to 1 significant figure, the student should weigh out 30 grams of diethylamine.

By using the formula Mass = Density x Volume, we find that the student needs to weigh out 28 grams of diethylamine for 40mL volume, as the density of diethylamine is 0.7g/mL.

The question asks for the mass of diethylamine the student should weigh out to obtain 40mL of diethylamine. Diethylamine has a density of 0.7 g/mL, according to the CRC Handbook of Chemistry and Physics. The formula to find mass when you have volume and density is:Mass = Density x Volume. Therefore, to find the mass of the diethylamine, we multiply the volume (40 mL) by the density (0.7 g/mL), which equals 28 grams. So, the student should weigh out 28 grams of diethylamine.

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Answer:

option (a)

Explanation:

To make a galvanometer into voltmeter, we have to connect a high resistance in series combination.

The voltmeter is connected in parallel combination with teh resistor to find the voltage drop across it.

An ideal voltmeter has very high resistance that means it has a resistance as infinity.

Answer:

A. In series Fint

Explanation:

If you connect it to the resistor, the voltmeter's ideal internal resistance be  In series Fint.

Explanation:

It is given that,

Mass of bullet, m₁ = 0.0207 kg

Mass of wooden block, m₂ = 28 kg

Wooden block is initially at rest, u₂ = 0 m/s

After the collision the, system has speed of, v = 1.180 m/s

We need to find the initial speed (u₁) of the bullet. It can be calculated using the conservation of linear momentum as :

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

[tex]0.0207\ kg\times u_1+0=(0.0207\ kg+28\ kg)\times 1.180\ m/s[/tex]

[tex]0.0207\times u_1=33.06[/tex]

[tex]u_1=1597.10\ m/s[/tex]

So, the initial speed of the bullet is 1597.10 m/s. Hence, this is the required solution.

Getting a promotion is NOT an example of retaliation.

C. Getting A Promotion

the action of harming someone because they have harmed oneself; revenge.

An example of to retaliate is for a person to punch someone who has hit him. (intransitive) To do something harmful or negative to get revenge for some harm; to fight back or respond in kind to damage or affront. Johny insulted Peter to retaliate for Peter's acid remark earlier.

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Answer:

sin(C) = 1/n = 1/2.42

C = 24.4 deg

The critical angle for the diamond-air interface is 24 degrees.

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The light emerging from a red filter is red and less intense than the original white light. When passed through blue plastic, it would be very dim or blocked, as red light cannot pass through a blue filter.

When a beam of white light passes through a red piece of plastic, the color of the light that emerges is red. This color is seen because the red plastic acts as a filter, absorbing other colors and only allowing red light to pass through. The emerging light is less intense than the white light because some of the light energy has been absorbed by the plastic.

If the light that has already passed through a red filter is then passed through a blue piece of plastic, the emerging light would likely be very dim or completely blocked. This is because the red light, lacking blue components, will be mostly absorbed by the blue plastic, which only allows blue light to pass through.

Answer:

The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]

Explanation:

Given that,

Charge [tex]q= 2.00 \mu C[/tex]

Electric field E = 100 N/C i

Radius R = 10.0 cm

We need to find the flux. It can be calculate using Gauss's law

The flux of the net electric field

[tex]\phi=\dfrac{q}{\epsilon_{0}}[/tex]

[tex]\phi=\dfrac{2.00\times10^{-6}}{8.85\times10^{-12}}[/tex]

[tex]\phi=2.26\times10^{5}\ Nm^2/C[/tex]

Hence, The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]

The velocity of the fish relative to water is approximately 10.44 m/s when it hits the water. The fish inherits the eagle's horizontal velocity at the start and then accelerates downward due to gravity.

The eagle's speed is given as 3.00 m/s horizontally. Therefore, when the fish falls, it also initially has a horizontal velocity of 3.00 m/s because it was brought to that speed by the eagle. The vertical component of the fish's velocity can be calculated using the second equation of motion: vf = vi + a*t, which simplifies to vf = g*t in the absence of initial vertical velocity. The time 't' it takes for the fish to hit the water can be calculated using the equation d = 0.5*a*t2, which gives t = √(2d/g). Substituting 5.00 m for 'd' and 9.81 m/s2 (approx gravitational acceleration) for 'g', we get t ≈ 1.017 seconds. Replacing 't' in the velocity equation with 1.017 seconds, which results in a vertical velocity of 9.965 m/s. The resultant velocity of the fish relative to the water when it hits would then be calculated using Pythagoras theorem since the horizontal and vertical movements are independent and at right angles to each other, giving a result of √((3.00 m/s)2 + (9.965 m/s)2) = 10.44 m/s.

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The velocity of the fish relative to the water when it hits the water is approximately 10.36 m/s at an angle of about 73.74° below the horizontal.

To calculate the velocity of the fish relative to the water when it hits the water, we need to analyze both the vertical and horizontal components of its motion separately.

Given Data:

Horizontal speed of the eagle (and the fish when dropped): [tex]v_{horizontal} = 3.00 \, \text{m/s}[/tex]

Height from which the fish falls: [tex]h = 5.00 \, \text{m}[/tex]

Step 1: Calculate the time it takes for the fish to fall 5m.

To find the time of fall, we can use the kinematic equation for vertical motion:

[tex]h = \frac{1}{2} g t^2[/tex]

where:

[tex]h[/tex] is the height (5.00 m),

[tex]g[/tex] is the acceleration due to gravity (approximately [tex]9.81 \, \text{m/s}^2[/tex]),

[tex]t[/tex] is the time in seconds.

Rearranging the equation to solve for [tex]t[/tex], we have:

[tex]t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5.00}{9.81}} \approx \sqrt{1.02} \approx 1.01 \, \text{s}[/tex]

Step 2: Calculate the vertical velocity of the fish just before hitting the water.

The vertical velocity can be calculated using:

[tex]v_{vertical} = g t[/tex]

Substituting the time we found:

[tex]v_{vertical} = 9.81 \, \text{m/s}^2 \times 1.01 \, \text{s} \approx 9.91 \, \text{m/s}[/tex]

Step 3: Combine the horizontal and vertical velocities to find the resultant velocity.

The horizontal velocity is constant at 3.00 m/s, and the vertical velocity just before impact is approximately 9.91 m/s. We can use the Pythagorean theorem to find the magnitude of the resultant velocity:

[tex]v_{resultant} = \sqrt{v_{horizontal}^2 + v_{vertical}^2}[/tex]

Substituting the values:

[tex]v_{resultant} = \sqrt{(3.00)^2 + (9.91)^2} = \sqrt{9 + 98.20} = \sqrt{107.20} \approx 10.36 \, \text{m/s}[/tex]

Step 4: Find the direction of the velocity relative to the horizontal using tangent.

The angle [tex]\theta[/tex] can be calculated using:

[tex]\theta = \tan^{-1}\left(\frac{v_{vertical}}{v_{horizontal}}\right)[/tex]

Substituting our values:

[tex]\theta = \tan^{-1}\left(\frac{9.91}{3.00}\right) \approx \tan^{-1}(3.30) \approx 73.74^\circ[/tex]

Answer:

Option 2 is the correct answer.

Explanation:

We have equation of motion s = ut + 0.5at²

Here u = 0 m/s

So, s = 0.5at²

Distance traveled in first second = 0.5 x a x 1² = 0.5 a

Distance traveled in second second = 0.5 x a x 2² - 0.5 x a x 1²= 1.5 a

Distance traveled in third second = 0.5 x a x 3² - 0.5 x a x 2²= 2.5 a

Distance traveled in fourth second = 0.5 x a x 4² - 0.5 x a x 3²= 3.5 a

The percentage increase in its displacement during the 4th second compared to its displacement in the 3rd second

                  [tex]=\frac{3.5a-2.5a}{2.5a}=0.4=40\%[/tex]

Option 2 is the correct answer.

Final answer:

In Physics, work is done when a force causes a displacement. Since the bag of groceries is moved horizontally at constant speed and there's no vertical displacement or horizontal force in the direction of motion, the work done on the bag is 0 joules.

Explanation:

The question involves the concept of work in Physics, specifically related to the work-energy principle. In this scenario, the bag of groceries is being carried across the room at a constant speed and constant height. According to the scientific definition of work, work is done when a force causes a displacement in the direction of the force. The formula for work is W = force x displacement. Here, the only forces doing work would be if the bag was lifted or if it was accelerated. Since the bag is being moved at a constant speed and not being lifted any further, there is no work done in the direction of motion, and the vertical height does not change. Therefore, despite the effort you feel you are exerting, the work done on the bag with regard to Physics is 0 joules

Answer choice: (B) 0.00J.

Answer:

sned me short notes of Physics of all branches (post graduation level) thanks  

+923466867221

Explanation:

Answer:

The length of the rod should be

[tex]\frac{L}{4} \\ [/tex]

Explanation:

Period of simple pendulum is given by

[tex]T=2\pi\sqrt{\frac{l}{g}} \\ [/tex]

We have

[tex]\frac{T_1^2}{T_2^2}=\frac{l_1}{l_2}\\\\\frac{2^2}{1^2}=\frac{L}{l_2}\\\\l_2=\frac{L}{4} \\ [/tex]

The length of the rod should be

[tex]\frac{L}{4} \\ [/tex]

To solve this problem, we need to utilize the relationship between linear and angular motion.

Firstly, we know the speed of point A is vA = 2.7 ft/sec. In a rotating cylinder, the linear speed v of a point at distance r from the center of rotation is given by v = ω * r, where ω is the angular speed. This would allow us to find ω by dividing the speed at point A by its distance from the center.

However, in this problem, we don't know the exact distance of point A from the center. So let's denote the unknown distance as r ft. Then, the angular speed ω = vA / r.

Secondly, it's given that the magnitude of the acceleration of the point A is aA = 20.6 ft/sec². Angular acceleration α is related to linear acceleration a by the factor of the radius, i.e., a = α * r.

We don't know r itself, but we know vA and aA, so we are able to compute the ratio of α to ω or α/ω = aA/vA, which equals to 20.6/2.7. This calculation enables us to conclude that the angular velocity and angular acceleration are proportional to each other, regardless of the values of radius r and angle θ

Taken together, from this reasoning, we can conclude that the ratio of angular to linear velocity/acceleration is approximately 7.63.

Finally, considering the issue of whether the knowledge of the angle θ is necessary, we can see that the angle θ does not appear in our ratio, and it does not affect the ratios of angular to linear velocities and accelerations. Therefore, the knowledge of the angle θ is not necessary for this calculation.

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Answer:

0.035 (3.5 %)

Explanation:

The thermodynamic efficiency is given by:

[tex]\eta = 1 - \frac{T_C}{T_H}[/tex]

where

[tex]T_C[/tex] is the cold temperature

[tex]T_H[/tex] is the hot temperature

In this problem we have

[tex]T_C = 5 ^{\circ}C+ 273 = 278 KT_H = 15^{\circ}C+273 = 288 K[/tex]

So the efficiency is

[tex]\eta = 1 - \frac{278 K}{288 K}=0.035[/tex]

The maximum thermodynamic efficiency of an engine operating with a hot reservoir at 288 K and a cold reservoir at 278 K, according to the Carnot efficiency formula, is approximately 3.47%.

The student has asked about the thermodynamic efficiency of an engine that uses the temperature gradient of the ocean. To calculate this, we can use the Carnot efficiency formula:

Carnot Efficiency Formula

η = 1 - Tc/Th, where η represents the efficiency, Tc is the cold temperature, and Th is the hot temperature. It's important to remember these temperatures need to be in Kelvin.

First, convert the given temperatures from degrees Fahrenheit to Kelvin:

Now, apply the Carnot efficiency formula:

η = 1 - (278 K / 288 K) = 1 - 0.9653 = 0.0347 or 3.47%

Therefore, the maximum thermodynamic efficiency of this oceanic temperature gradient engine would be approximately 3.47%.

Answer:

[tex]\sqrt{\frac{4}{9}}[/tex]

Explanation:

The frequency of a simple pendulum is given by:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}[/tex]

where

g is the acceleration of gravity

L is the length of the pendulum

Calling [tex]L_1[/tex] the length of the first pendulum and [tex]g_1[/tex] the acceleration of gravity at the location of the first pendulum, the frequency of the first pendulum is

[tex]f_1=\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}[/tex]

The length of the second pendulum is 0.4 times the length of the first pendulum, so

[tex]L_2 = 0.4 L_1[/tex]

while the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum, so

[tex]g_2 = 0.9 g_1[/tex]

So the frequency of the second pendulum is

[tex]f_2=\frac{1}{2\pi}\sqrt{\frac{g_2}{L_2}}=\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}[/tex]

Therefore the ratio between the two frequencies is

[tex]\frac{f_1}{f_2}=\frac{\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}}{\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}}=\sqrt{\frac{0.4}{0.9}}=\sqrt{\frac{4}{9}}[/tex]

Final answer:

The frequency of the first pendulum compared to the second pendulum is in the ratio of 2/3, since the length of the second pendulum is 0.4 times the first and gravity on the second is 0.9 times the first, impacting their respective periods and inversely their frequencies.

Explanation:

The frequency of a simple pendulum is dependent on its length and the acceleration due to gravity it experiences. The formula to find the frequency (f) of a pendulum is given by:

f = 1 / T

Where T is the period of the pendulum, which can be calculated by:

T = 2π√(L/g)

Let's denote L1 and g1 as the length and the acceleration due to gravity for the first pendulum, and L2 and g2 for the second pendulum. Since the length of the second pendulum is 0.4 times that of the first, we have L2 = 0.4L1. And because the acceleration of gravity is 0.9 times for the second pendulum, g2 = 0.9g1.

Using these proportions:

T1 = 2π√(L1/g1)
T2 = 2π√(L2/g2) = 2π√((0.4L1)/(0.9g1))

We can simplify T2 by putting it in terms of T1:

T2 = T1√(0.4/0.9)

Now, since frequency is the inverse of period:

f1 = 1/T1
f2 = 1/T2

So the ratio of their frequencies (f1/f2) will be the inverse of the ratio of their periods (T2/T1):

f1/f2 = T2/T1

Plugging in the ratio:

f1/f2 = √(0.4/0.9) = √(4/9) = 2/3

Therefore, the correct answer is (a) 2/3.

Answer:

64.2 m/s

Explanation:

In the x direction:

x = x₀ + v₀ₓ t + ½ at²

400 m = 0 m + v₀ cos (π/5) t + ½ (0 m/s²) t²

t = 400 / (v₀ cos (π/5))

In the y direction:

y = y₀ + v₀ᵧ t + ½ gt²

0 m = 0 m + v₀ sin (π/5) t + ½ (-9.8 m/s²) t²

0 = v₀ sin (π/5) - 4.9 t

t = v₀ sin (π/5) / 4.9

Therefore:

400 / (v₀ cos (π/5)) = v₀ sin (π/5) / 4.9

1960 = v₀² sin (π/5) cos(π/5)

1960 = ½ v₀² sin(2π/5)

3920 / sin(2π/5) = v₀²

v₀ = 64.2 m/s

Answer:

Initial velocity = 423.08m/s

Explanation:

Using formular for Range of projectile

R =V^2 sin2theta/g

Given:

Range=0.4km= 400m

Theta=3.142/5

g= 9.8m/s^2

400= V^2×sin(2×3.142/5)/9.8

400×9.8= V^2Sin 1.26

3920= 0.0219V^2

V^2= 3920/0.0219

V^2= 178995.43

V=sqrt 178995.43

V= 423.08m/s

Answer:

40.3 min

Explanation:

First of all, let's convert every quantity into SI units:

[tex]v_1 = 92.5 km/h = 25.7 m/s[/tex] (speed in the first part of the trip)

[tex]t_2 = 28.0 min = 1680 s[/tex] time during which the person has stopped

[tex]v=72.2 km/h = 20.1 m/s[/tex] (average speed of the whole trip)

The average speed is the ratio between the total distance covered, d, and the total time taken, t:

[tex]v=\frac{d}{t}[/tex] (1)

The total distance covered is simply

[tex]d = v_1 t_1[/tex]

where [tex]t_1[/tex] is the time during which the person has moved at 92.5 km/h.

The total time taken is

[tex]t= t_1 + t_2[/tex]

So (1) becomes

[tex]v=\frac{v_1 t_1}{t_1 + t_2}[/tex]

Solving for [tex]t_1[/tex]:

[tex]v t_1 + v t_2 = v_1 t_1\\vt_2 = (v_1+v)t_1\\t_1 = \frac{v t_2}{v_1+v}=\frac{(20.1 m/s)(1680 s)}{25.7 m/s + 20.1 m/s}=737.3 s[/tex]

which corresponds to

[tex]t_2 = 737.3 s = 12.3 min[/tex]

So the total time of the trip is

[tex]t = 28.0 min + 12.3 min = 40.3 min[/tex]

The orbital period of a satellite at an altitude of 350 km above Earth's surface is approximately 1.93 hours or around 115.8 minutes. Given the options, the closest one is option D, with a period of 91 minutes.

To answer the question about the satellite orbiting Earth at an altitude of 350 km, we can utilize some physics concepts related to orbital mechanics. Note that a significant factor in determining the orbital period of a satellite (time required for a satellite to complete one orbit) is not the satellite's mass but rather its altitude or, more accurately, the total distance from the center of the Earth, considering the Earth's radius plus the satellite's altitude.

In general, the closer a satellite is to Earth, the shorter its orbital period. The information given indicates that a satellite at an altitude of 350 km will likely have an orbital period around 1.93 hours, which is roughly equivalent to 115.8 minutes. Given that, none of the available options exactly match this result. But the value closest to it would be 91 minutes, which is represented by option D.

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Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, [tex]q=1.60218\times 10^{-19}\ C[/tex]

Mass of a proton, [tex]m=1.67262 \times 10^{-27}\ kg[/tex]

The cyclotron frequency is given by :

[tex]f=\dfrac{qB}{2\pi m}[/tex]

[tex]f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}[/tex]

f = 2286785.40 Hz

or

[tex]\omega=14368296.44\ rad/s[/tex]

[tex]\omega=1.43\times 10^7 rad/s[/tex]

Hence, this is the required solution.

Answer:

The work done by the friction is 55.8 J.

Explanation:

Given that,

Height h = 7 m

Mass of the object = 3 kg

Angle = 30°

Speed = 10 m/s

The initial potential energy of the object is converted in to the kinetic energy at the bottom of the incline and  the work done by the friction

[tex]mgh=\dfrac{1}{2}mv^2+W[/tex]

Where, m = mass of the object

v = final velocity

g = acceleration due to gravity

h = height

Put the value in the equation

[tex]3\times9.8\times7=\dfrac{1}{2}\times3\times(10)^2+W[/tex]

[tex]W = (205.8-150)\ J[/tex]

[tex]W=55.8\ J[/tex]

Hence, The work done by the friction is 55.8 J.

Answer:

Explanation:

The escape velocity on a planet is directly proportional to the square root of acceleration due to gravity and the radius of the planet.

According to the question acceleration due to gravity that means g is same but there is no idea about the radius.

So if radius is change then escape velocity at that planet may be more or less depending on the radius.

If the radius is also same then the escape velocity is same.

Answer:

1.03 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

0 = 7.63 + (-2.40) t + ½ (-9.8) t²

0 = 7.63 - 2.40 t - 4.9 t²

Solve with quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 2.40 ± √(2.40² - 4(-4.9)(7.63)) ] / -9.8

t = -1.52, 1.03

Since t can't be negative here, the sandbag hits the ground after 1.03 seconds.

The sandbag is dropped from the hot air balloon with the balloon's existing speed. Using the second equation of motion and the known values for initial velocity, acceleration due to gravity, and distance, we can solve for the time taken for the sandbag to hit the ground.

In the given question, a hot-Air balloon is drifting straight downward with a constant speed of 2.40 m/s. When the balloon is 7.63 m above the ground, a ballast sandbag is dropped.

Given the bag is dropped from rest relative to the balloon (i.e., the initial velocity (u) of the sandbag is -2.4 m/s, negative because the direction is downward), and the acceleration due to gravity (a) is -9.8 m/s² (negative due to the downward direction), we can use the second equation of motion (s = ut + 0.5at²) where s is the distance covered, which is 7.63 m, to find time (t).

Setting this up, -7.63 = -2.4t + 0.5(-9.8)t². Solving this quadratic equation for time should give us the answer. The result will be the time taken by the sandbag to hit the ground after it is dropped from the hot air balloon.

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Answer:

the Answer is c: E/2

Explanation:

see attachment

Answer:

Magnitude of the force exerted on the handle = 66 N

Explanation:

Power is the ratio of work and time.

        [tex]P=\frac{W}{t}\\\\W=Pt=84\times 1.1=92.4J[/tex]

We have work done by rower = 92.4 J

We also have

            Work = Force x Displacement

            92.4 = Force x 1.4

            Force = 66 N

Magnitude of the force exerted on the handle = 66 N

Answer:

81.1 m

Explanation:

X = 237 m 27.2 east of north 237 (Sin27.2 i + Cos 27.2 j)

X = 108.33 i + 210.8 j

Y = 335 east = 335 i

Displacement = X + Y = 108.33 i + 210.8 j + 335 i = 443.33 i + 210.8 j

(magnitude of displacement)^2 = 443.33^2 + 210.8^2

magnitude of displacement = 490.9 m

Distance traveled = 237 + 335 = 572 m

Difference in the magnitude of displacement and distance = 572 - 490.9

                                                                                                   = 81.1 m

Answer:

137 J

Explanation:

Momentum is conserved, so:

(2.10)(13.0) + (6.50)(-4.10) = (2.10)(-8.20) + (6.50) v

v = 2.75 m/s

Energy before collision:

E = 1/2 (2.10) (13.0)² + 1/2 (6.50) (4.10)²

E = 232 J

Energy after collision:

E = 1/2 (2.10) (8.20)² + 1/2 (6.50) (2.75)²

E = 95.2 J

Energy lost in collision:

232 J - 95.2 J = 137 J

Final answer:

A total of 161.488 Joules of kinetic energy is lost in this collision.

Explanation:

The subject question involves a head-on inelastic collision between two balls of different masses traveling towards each other at different speeds and asks for the amount of kinetic energy lost in the collision. To find the kinetic energy lost, we first calculate the initial and final kinetic energies and then take their difference.

Initial kinetic energy (Eki) is the sum of the kinetic energies of the two balls before the collision:

Kinetic energy of the 2.10 kg ball: [tex](1/2) * 2.10 kg * (13.0 m/s)^2[/tex]

Kinetic energy of the 6.50 kg ball: [tex](1/2) * 6.50 kg * (4.1 m/s)^2[/tex]

Final kinetic energy (Ekf) involves only the kinetic energy of the 2.10 kg ball, as the final velocity of the 6.50 kg ball is not given:

Kinetic energy of the 2.10 kg ball: (1/2) * 2.10 kg * [tex](8.20 m/s)^2[/tex]

Kinetic energy lost during the collision is calculated as:

Elost = Eki - Ekf

After calculation:

Initial kinetic energy, Eki = (1/2) * 2.10 kg * [tex](13.0 m/s)^2[/tex] + (1/2) * 6.50 kg * [tex](4.1 m/s)^2[/tex]

Final kinetic energy, Ekf = (1/2) * 2.10 kg * [tex](8.20 m/s)^2[/tex]

Kinetic energy lost, Elost = Eki - Ekf

Plugging the numbers in:

Eki = 0.5 * 2.10 kg * [tex](13.0 m/s)^2[/tex] + 0.5 * 6.50 kg * [tex](4.1 m/s)^2[/tex]
= (0.5 * 2.10 * 169) + (0.5 * 6.50 * 16.81)
= 177.45 J + 54.64 J
= 232.09 J

Ekf = 0.5 * 2.10 kg * [tex](8.20 m/s)^2[/tex]
= (0.5 * 2.10 * 67.24)
= 70.602 J

Therefore, kinetic energy lost, Elost = 232.09 J - 70.602 J = 161.488 J

A total of 161.488 Joules of kinetic energy is lost in this collision.

Answer:

Charge, q = 1.15 C

Explanation:

It is given that,

Mass of sphere, m = 441 g = 0.441 kg

Electric field, E = 5 N/C

Due to this field, the sphere suddenly acquires a horizontal acceleration of 13.0 m/s² such that,

[tex]ma=qE[/tex]

[tex]q=\dfrac{ma}{E}[/tex]

[tex]q=\dfrac{0.441\ kg\times 13\ m/s^2}{5\ N/C}[/tex]

q = 1.146 C

or

q = 1.15 C

So, the charge carried by the small sphere is 1.15 C. Hence, this is the required solution.

The electric field accounts for the unbalanced motion of the sphere. The electrostatic forces balance the linear force, for which the charge acquired by the sphere is 1.146 C.

The region or space where a charged particle experiences the electrostatic force of attraction or repulsion, is known as an electric field.

Given data:

The mass of the small sphere is, m = 441 g = 0.441 kg.

The magnitude of the electric field is, E = 5.00 N/C.

The magnitude of horizontal acceleration is, a = 13.0 m/s².

Clearly, under the influence of an electric field, the electrostatic force balances the linear force. Then,

Fe = FL

[tex]qE = ma[/tex]

Here,

q is the magnitude of charge.

Solving as,

[tex]q \times 5.0 = 0.441 \times 13\\\\q = \dfrac{0.441 \times 13}{5.0}\\\\q = 1.146 \;\rm C[/tex]

Thus, we can conclude that the magnitude of charge carried by the sphere is 1.146 C.

Learn more about the electrostatic force here:

https://brainly.com/question/9774180