A bicycle wheel, of radius 0.300 m and mass 2.07 kg (concentrated on the rim), is rotating at 4.00 rev/s. After 59.0 s the wheel comes to a stop because of friction. What is the magnitude of the average torque due to frictional forces?

Answers

Answer 1

ANSWER:

0.079 J

STEP-BY-STEP EXPLANATION:

The average torque exerted due to friction force is:

[tex]\tau=I\alpha[/tex]

Here, I is the moment of inertia and α is the angular acceleration.

The mass of wheel is concentrated at the rim. Therefore, the moment of inertia of wheel is,

[tex]I=m\cdot r^2[/tex]

Here, m is the mass of wheel and r is the radius of wheel.

replacing:

[tex]\begin{gathered} I=2.07\cdot0.3^2 \\ I=0.1863\text{ kg}\cdot s^2 \end{gathered}[/tex]

The angular speed is decreased from 4 rev/s to 0 rev/s in 59 sec. Therefore, the angular acceleration is calculated as:

[tex]\begin{gathered} \alpha=\frac{0-4}{59} \\ \alpha=0.068\cdot\frac{rev}{\sec}\cdot\frac{2\pi\text{ rad/rev}}{1rev/sec^2} \\ \alpha=-0.426rad/sec^2 \end{gathered}[/tex]

The negative sign represents that the wheel is de-accelerating.

Replacing:

[tex]\begin{gathered} \tau=I\alpha \\ \tau=0.1863\cdot-0.426 \\ \tau=-0.079\text{J} \end{gathered}[/tex]

The negative sign represents the direction of torque i.e. clockwise direction. Thus, the magnitude of average torque is 0.079 J.


Related Questions

An element with an atomic number of 90 and an atomic mass of 230 would have what atomic number after Alpha decay?88909186

Answers

Given data:

The atomic number of

In the calculation of the area of a rectangle with dimensions of 4.282 m by 0.050 m, which of the following answers has the correct number of significant figures?

Answers

We have

[tex]A=l\times w[/tex]

in our case

l=4.282 m

w=0.050m

we substitute

[tex]A=(4.282)(0.050)=0.2141m^2=0.21m^2[/tex]

ANSWER

0.21m^2

hello I need help with question 7 a and b please

Answers

Given data:

* The mass of the child is 40kg.

* The reading of the scale is 30 N.

Solution:

(a). The free body diagram of the given system is,

(b). The weight of the child is,

[tex]\begin{gathered} W=40\times9.8 \\ W=392\text{ N} \end{gathered}[/tex]

The Scale reading is,

[tex]N=30\text{ Newton}[/tex]

The net force acting on the box in terms of the weight of the box is,

[tex]\begin{gathered} F_{\text{net}}=N-mg \\ F_{\text{net}}=30-392 \\ F_{\text{net}}=-362\text{ N} \end{gathered}[/tex]

According to the Newton's second law,

[tex]undefined[/tex]

I need help on number 5. We need to use one of the four kinematics equations.

Answers

ANSWER

[tex]35.02m[/tex]

EXPLANATION

Parameters given:

Initial velocity, u = 26.2 m/s

When the vase reaches its maximum height, its velocity becomes 0 m/s. That is the final velocity.

We can now apply one of Newton's equations of motion to find the height:

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

where a = g = acceleration due to gravity = 9.8 m/s²

Therefore, we have that:

[tex]\begin{gathered} 0=26.2^2-2(9.8)s \\ \Rightarrow19.6s=686.44 \\ s=\frac{686.44}{19.6} \\ s=35.02m \end{gathered}[/tex]

That is the height that the vase will reach.

A wheel of radius 30.0 cm is rotating at a rate of 2.20 revolutions every 0.0910 s. What is the linear speed of a point on the wheels rim?

Answers

Explanation:

First, we need to calculate the frequency of the wheel, so if it is rotating at a rate of 2.20 revolutions every 0.0910 seconds, the frequency is:

[tex]f=\frac{2.20\text{ revolutions}}{0.0910\text{ seconds}}=24.16\text{ Hz}[/tex]

Then, the magnitude of the angular velocity is equal to:

[tex]w=2\pi f=2(3.14)(24.16)=151.9\text{ rad/s}[/tex]

Finally, we can calculate the linear speed as the angular velocity times the radius, so:

[tex]v=w\cdot r=(151.9\text{ rad/s)(30 cm) = 4557.04 m/s}[/tex]

Therefore, the linear speed is 45.47 m/s

What is the wavelength of the electromagnetic wave emitted and absorbed by a phone?

Answers

The wavelengths of the electromagnetic waves emitted and absorbed by a phone is approximately 10-1000 m. Radio waves are those waves that are used for the cell phones to communicate with the towers.

What is the momentum of a 7.30 kg bowling ball going down the alley with a speed of 20.0 m/s?

Answers

Given data

*The given mass of the bowling ball is m = 7.30 kg

*The given speed is v = 20.0 m/s

The formula for the momentum of a 7.30 kg bowling ball is given as

[tex]p=mv[/tex]

Substitute the known values in the above expression as

[tex]\begin{gathered} p=(7.30)\times(20.0) \\ =146\text{ kg.m/s} \end{gathered}[/tex]

Hence, the momentum of a 7.30 kg bowling ball is p = 146 kg.m/s

How to calculate electric field on a charge

Answers

Let the charge be denoted by q.

Let the total electric force on the charge be denoted by F.

We have to the electric field on a charge.

Electric field on a charge is given by the formula

[tex]E=\text{ }\frac{F}{q}[/tex]

2. A parallel-plate capacitor has an area of 2.0 cm², and the plates are separated by 2.0 mm. a. What is the capacitance? b. How much charge does this capacitor store when connected to a 6.0 V battery?

Answers

Given data:

* The area of the parallel plate capacitor is,

[tex]\begin{gathered} A=2cm^2 \\ A=2\times10^{-4}m^2^{} \end{gathered}[/tex]

* The distance between the plates is,

[tex]\begin{gathered} d=2\text{ mm} \\ d=2\times10^{-3}\text{ m} \end{gathered}[/tex]

Solution:

(a). The capacitance of the capacitor in terms of area and distance between the plates is,

[tex]C=\frac{\epsilon_{\circ}A}{d}[/tex][tex]\text{where }\epsilon_{\circ}\text{ is the electrical permittivity of the fr}ee\text{ spaces}[/tex]

Substituting the known values,

[tex]\begin{gathered} C=8.85\times10^{-12}\times\frac{2\times10^{-4}}{2\times10^{-3}} \\ C=8.85\times10^{-13}\text{ F} \end{gathered}[/tex]

Thus, the value of the capacitanc is 8.85 times 10 power -13 Farad.

(b). The voltage across the battery is,

[tex]V=6\text{ Volts}[/tex]

The charge stored in the capacitor in terms of the voltage and the capacitance is,

[tex]\begin{gathered} C=\frac{Q}{V} \\ Q=CV \end{gathered}[/tex]

where Q is the charge stored in the capacitor

Substituting the known values,

[tex]\begin{gathered} Q=8.85\times10^{-13}\times6 \\ Q=53.1\times10^{-13}\text{ Coulomb} \end{gathered}[/tex]

Thus, the charge stored in the parallel plate capacitor is 53.1 times 10 power -13 coulomb.

A grocery shopper tosses a 6 kg bag of rice into a stationary 27 kg grocery cart. The bag hits the cart with a horizontal speed of 8.0 m/s toward the front of the cart. What is the final speed of the cart and the bag? Round to the hundredths place

Answers

Given,

The mass of the bag, m=6 kg

The mass of the cart, M=27 kg

The initial velocity of the cart, u=0 m/s

The initial velocity of the bag, v=8.0 m/s

From the law of conservation of momentum, the sum momentum of the bag and the momentum of the cart before the bag hits the cart must be equal to the momentum of the bag and the cart after the bag hits the cart.

Thus,

[tex]mv+Mu=(M+m)V[/tex]

Where V is the velocity of the bag and the cart after the bag hits the cart.

On rearranging,

[tex]V=\frac{mv+Mu}{(M+m)}[/tex]

On substituting the known values,

[tex]\begin{gathered} V=\frac{6\times8.0+0}{(27+6)} \\ =\frac{48}{33} \\ =1.45\text{ m/s} \end{gathered}[/tex]

Thus the final speed of the cart and the bag is 1.45 m/s

2. Connect an ammeter and voltmeter in the circuit below.

Answers

ANSWER and EXPLANATION

We want to connect a voltmeter and an ammeter in the given circuit.

In a circuit, a voltmeter is connected in parallel with a device in order to measure its voltage. Hence, the voltmeter will be connected in parallel with the resistor.

The ammeter is connected in a circuit in series in order to measure its current.

Let us connect the meters below:

That is the answer.

A car starts from rest and accelerates uniformly for a distance of 137 m over an 9.6-second time interval. The car's acceleration ism/s².

Answers

In order to solve this question, we will need to use kinematics

Let's see what is given to us:

Distance traveled is 137 meters

Time elapsed is 9.6 seconds

and inital velocity is 0 m/s

Since we are trying to find acceleration, we can use this formula

[tex]\Delta x=v_0t+\frac{1}{2}at^2[/tex]

Where Δx is the distance traveled, v0 is the inital velocity, t is time, and a is acceleration

Plugging in what we have, we get

137 = 0(9.6) + 1/2(a)(9.6)^2

Solving for a, we get 2.97 m/s^2

Explain what inertia is. What causes inertia in an object that is at rest? In an object that is moving?

Answers

Inertia is the ability of the body to remain in the state or in uniform motion unless some external force acts on the body.

The inertia of the body is direly proportional to the mass of the body.

Thus, the greater the mass of the body greater is its inertia.

Three types of inertia:

In the state of rest, the inertia of the body resists the cause of motion. This inertia is known as resting inertia.

In the state of uniform motion, the inertia resists the cause of changing of motion. This inertia is known as motion inertia.

Some external force is applied to the body to change the direction of the motion of the body, the inertia that opposes this change in direction is known as directional inertia.

a cannon sitting on top of a 40 m mound shoots a projectile with an initial velocity of 4.47 m/s parallel to ground. how far away did it land?

Answers

Given that,

The initial height of the projectile, y₀=40 m

The initial velocity of the projectile, v₀=4.47 m/s

The direction of the initial velocity is parallel to the ground. Therefore the x-component of the initial velocity v₀x=4.47 m/s.

And the y-component of the initial velocity is v₀y=0 m/s

From the equation of the motion we have,

[tex]y=y_0+v_{0y}t+\frac{1}{2}gt^2[/tex]

Where y is the final height of the projectile which is zero as it finally hits the ground. And g =-9.8m/s² is the acceleration due to gravity. And t is the time interval of the flight of the projectile.

On substituting the known values in the above equation,

[tex]\begin{gathered} 0=40+0+\frac{1}{2}\times-9.8\times t^2 \\ =40-4.9t^2 \end{gathered}[/tex]

On rearranging the above equation and simplifying it,

[tex]\begin{gathered} t=\sqrt{\frac{40}{4.9}} \\ =2.86\text{ s} \end{gathered}[/tex]

The x-component of the velocity remains constant as there is no acceleration in that direction.

The horizontal distance travelled or the range of the projectile can be calculated using the formula,

[tex]R=v_{0x}t[/tex]

On substituting the known values in the above equation,

[tex]R=4.47\times2.86=12.78\text{ m}[/tex]

Therefore the projectile will land 12.78 meters away

An airplane flying horizontally at a velocity of 138 m/s and at an altitude of 1500 meters when one of its wheels falls off.What horizontal distance (in meters) will the wheel travel before it strikes the ground?

Answers

ANSWER:

2415 meters

STEP-BY-STEP EXPLANATION:

Given:

Initial horizontal velocity (ux) = 138 m/s

Initial vertical velocity (uy) = 0 m/s

Height (h) = 1500 meters

The first thing is to calculate the time it takes for the airplane to reach the ground, just like this:

[tex]\begin{gathered} h=u_yt+\frac{1}{2}at^2 \\ \\ \text{ we replacing} \\ \\ 1500=0\cdot t+\frac{1}{2}(9.8)t^2 \\ \\ 4.9t^2=1500 \\ \\ t^2=\frac{1500}{4.9} \\ \\ t=\sqrt{\frac{1500}{4.9}} \\ \\ t=17.5\text{ sec} \end{gathered}[/tex]

Therefore, the horizontal distance would be:

[tex]\begin{gathered} x=u_x\cdot t \\ \\ \text{ we replacing} \\ \\ x=138\cdot17.5 \\ \\ x=2415\text{ m} \end{gathered}[/tex]

Therefore, the horizontal distance is 2415 meters

Find the magnitude of the vector sum A→+B→+C→ . Each grid square is 2.00 N on a side. If the vector sum is to the west, enter a negative value. If the vector sum is to the east, enter a positive value.

Answers

Given data:

* The magnitude of the vector C is,

[tex]\begin{gathered} C=2\times2 \\ C=4\text{ N} \end{gathered}[/tex]

* The magnitude of the vector B is,

[tex]\begin{gathered} B=4\times2 \\ B=8\text{ N} \end{gathered}[/tex]

Solution:

From the given diagram, the magnitude of the vector A is,

[tex]\begin{gathered} A=\sqrt[]{Base^2+Perpendicular^2} \\ A=\sqrt[]{(3\times2)^2+(4\times2)^2} \\ A=\sqrt[]{6^2+8^2} \\ A=\sqrt[]{36+64} \\ A=\sqrt[]{100} \\ A=10\text{ N} \end{gathered}[/tex]

The count of square grid in the hypotenuse is,

[tex]\begin{gathered} n=\frac{A}{2} \\ n=\frac{10}{2} \\ n=5 \end{gathered}[/tex]

The angle of the vector A with the x-axis is,

[tex]\begin{gathered} cos(\theta)=\frac{Base}{\text{Hypotenuse}} \\ \cos (\theta)=\frac{3}{5} \\ \theta=53.13^{\circ} \end{gathered}[/tex]

Thus, the value of vector A is,

[tex]\begin{gathered} \vec{A}=A\cos (53.13^{\circ})+A\sin (53.13^{\circ}) \\ \vec{A}=10\times\cos (53.13^{\circ})i+10\times\sin (53.13^{\circ})j \\ \vec{A}=6\text{ i + 8 j} \end{gathered}[/tex]

The value of vector B is,

[tex]\vec{B}=-8\text{ j}[/tex]

The value of vector C is,

[tex]\vec{C}=-4\text{ i}[/tex]

Thus, the sum of the vectors is,

[tex]\begin{gathered} \vec{A}+\vec{B}+\vec{C}=6\text{ i+8 j-8 j-4 i} \\ \vec{A}+\vec{B}+\vec{C}=2\text{ i} \\ |\vec{A}+\vec{B}+\vec{C}|=\sqrt[]{2^2} \\ |\vec{A}+\vec{B}+\vec{C}|=\text{ 2 N} \end{gathered}[/tex]

Thus, the magnitude of the sum of three given vectors is 2 N towards the east (positive of the x-axis).

Romeo and Juliet are sitting on a balcony 1.5 meters apart. If Romeo has a mass of 61.6 Kg and Juliet has a mass of 48.8 kg. What is the attractive force between them?

Answers

[tex]\begin{gathered} m_{Romeo}=61.6\text{ kg} \\ m_{Juliet}=48.8\text{ kg} \\ r=1.5\text{ m} \\ G=6.673x10^{-11}Nm^2/kg^2 \\ F=G*\frac{m_{Juliet}m_{Romeo}}{r^2} \\ \\ F=\left(6.673x10^{-11}Nm^2/kg^2\right)\frac{\left(48.8\text{ kg}\right)\left(61.6\text{ kg}\right)}{\left(1.5\text{ m}\right)^2} \\ \\ F=8.91x10^{-8}N \\ The\text{ attractive force is 8.91x10}^{-8}\text{N} \end{gathered}[/tex]

If a plane covers a distance of 700 miles in 1 hour and 43 minutes, what is its speed?   A)   Give your answer in mph.   B)   Give your answer in ft/second

Answers

Given:

Distance covered by the plane, d = 700 miles

Time, t = 1 hour and 43 minutes.

Let's find the speed of the plane.

To find the speed, apply the formula:

[tex]\text{Speed = }\frac{dis\tan ce}{time}[/tex]

Where:

distance = 700 miles

time = 1 hour 43 minutes

(A) speed in mph.

mph is miles per hour.

Where:

60 minutes = 1 hour

Thus, to find the speed the time is to be in hours.

We have:

[tex]t=1\frac{43\text{ minutes}}{60\text{ minutes}}=1\frac{43}{60}=1.716666\text{ hrs}\approx1.72\text{ hrs}[/tex]

Thus, to find the speed in mph, we have:

[tex]\text{Speed}=\frac{\text{distance}}{\text{time}}=\frac{700}{1.72}=407.8\text{ mph}[/tex]

Therefore, the speed of the plane in mph is 407.8 mph

(B) To find the speed in ft/second

Let's first convert the distance from miles to feet

Where:

1 mile = 5280 feet

700 miles = 700 x 5280 = 3696000 feet

Also convert the time to seconds.

Where:

1 hour = 60 minutes x 60 seconds = 3600 seconds

1.7166 hours = 1.7166 x 3600 = 6180 seconds

Thus, we have:

Distance = 3696000 feet

Time = 6180 seconds

[tex]\text{Speed = }\frac{dis\tan ce}{\text{time}}=\frac{3696000}{6180}=598.06\text{ ft/second}[/tex]

Therefore, the speed in ft/second is 598.06 ft/second

ANSWER:

(A) 407.8 mph

(B) 598.06 ft/second

When an ice cube melts in your hand, it changes from a solid to a liquid.

H2O (s) → H2O (l)

This is an example of ________________ reaction, because heat is _____________ the system.

Answers

The reaction is endothermic as heat is taken into the system.

What is an endothermic reaction?

We know that in an endothermic reaction. Heat is taking in by the reaction. The heat that is taken into the system is used to break apart the chemical bonds that are in the system and this would facilitate the conversion of the reactants to the products.

In this case, there is the intake of heat by the solid water and this would lead to the breakage of all the rigid hydrogen bonds that hold the molecules of the water in the solid state and when that occurs we can now see that we would have liquid water.

Thee main point here is that heat is taken into the system thus heat is a reactant and ought to appear at the left hand side of the reaction equation and leads to the formation of the liquid water product.

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Suppose a 1.90 N force can rupture an eardrum having an area of 1.14 cm².(a) Calculate the maximum tolerable gauge pressure inside the eardrum (in the middle ear) in N/m². (Pressures in themiddle ear may rise when an infection causes a fluid buildup. Use 13.6 x 10³ kg/m³ as the density of mercury.) submit answer in N/m²(a) part 2: Convert this value to mm Hg.mm Hg(b) At what depth in fresh water would this person's eardrum rupture, assuming the gauge pressure in the middle ear iszero?Submit Answer in m

Answers

Given:

The force is

[tex]F=1.90\text{ N}[/tex]

The area of the eardrum is

[tex]\begin{gathered} A=1.14\text{ cm}^2 \\ =1.14\times10^{-4}\text{ m}^2 \end{gathered}[/tex]

To find:

The maximum tolerable gauge pressure inside the eardrum

a) the pressure in mm of Hg

b) At what depth in freshwater would this person's eardrum rupture

Explanation:

The pressure at the eardrum is

[tex]\begin{gathered} P=\frac{F}{A} \\ =\frac{1.90}{1.14\times10^{-4}} \\ =16.67\times10^3\text{ N/m}^2 \end{gathered}[/tex]

Hence, the pressure is

[tex]16.67\times10^3\text{ N/m}^2[/tex]

a)

We know,

[tex]1\text{ N/m}^2=0.0075\text{ mm of Hg}[/tex]

So,

[tex]\begin{gathered} 16.67\times10^3\text{ N/m}^2=0.0075\times16.67\times10^3\text{ mm of Hg} \\ =125.02\text{ mm of Hg} \end{gathered}[/tex]

Hence, the pressure is 125.02 mm of Hg.

b)

The depth of fresh water is,

[tex]\begin{gathered} h=\frac{P}{dg} \\ Here,\text{ d=1000 kg/m}^3 \\ g=9.8\text{ m/s}^2 \end{gathered}[/tex]

So,

[tex]\begin{gathered} h=\frac{16.67\times10^3}{1000\times9.8} \\ =1.70\text{ m} \end{gathered}[/tex]

Hence, the depth of water is 1.70 m.

What is the centripetal force exerted on a 1,600-kg car that rounds a 100-m curve at 12 m/s?Question 18 (2 points) What is the centripetal force exerted on a 1,600-kg car that rounds a 100-m radi curve at 12 m/s?

Answers

2304 Newton  is the centripetal force exerted on a 1,600-kg car that rounds a 100-m curve at 12 m/s

centripetal force=mv^2/r

m=1,600-kg

v= 12 m/s

r= 100-m

centripetal force=mv^2/r

centripetal force=(1,600-kg×12 m/s×12 m/s)÷100-m

centripetal force=230400/100

centripetal force=2304 N

Any motion along a curved road is accelerated, necessitating the application of force to the path's center of curvature. Centripetal, which means "center seeking," is the name of this force.

The Centripetal Force Formula states that the centripetal force equals the product of mass (in kg) and tangential velocity (in meters per second) squared, divided by the radius (in meters). This formula predicts that the centripetal force will double with a doubling of tangential velocity.

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URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!

Answers

Answer:

LAST ONE!!!!

Explanation:

Answer:

the Temperature in each room will become equal over time

If you have a concave mirror whose focal length is 100.0 cm, and you want an image that is upright and 10.0 times as tall as the object, where should you place the object?

Answers

given

[tex]\begin{gathered} focal\text{ length=100 cm} \\ let\text{ object distance is u form the pole of mirror } \\ so\text{ u=-u} \\ and\text{ v=?} \\ here\text{ v is the distance of image from the pole of mirror } \\ let\text{ height of the object h} \\ so\text{ h=+h} \\ according\text{ to the question image of the object is } \\ \text{ h}^{^{\prime}}=+10h \\ \end{gathered}[/tex]

since the image is in the upright direction so the image is virtual.

and in the case of the mirror, an image can be virtual if and only if

the image is placed between the pole and the focus. so the object

must be placed between the pole and the focus.

[tex]\begin{gathered} the\text{ formula for magnification is given by;} \\ m=-\frac{v}{u}=\text{ }\frac{h^{^{\prime}}}{h} \\ and\text{ the mirror formula is given by the following} \\ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \end{gathered}[/tex][tex]\begin{gathered} -\frac{v}{u}=\frac{h^{^{\prime}}}{h} \\ by\text{ putting all the values} \\ -\frac{v}{u}=\frac{10h}{h} \\ v=-10u \end{gathered}[/tex][tex]undefined[/tex]

URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!

Answers

B.

Hot air is transferred through heat being radiated from a thermal burner.

It can't be A. because the food isn't touching the burner, it can't be C. because food can't be completely cooked that way and that's not how ovens work, and it can't be D. because cool air doesn't rise.

B is the answer because it accurately describes how a oven is supposed to work.

In a bike race, a sport person starts his bike from rest and maintains a constant acceleration of 3 m/s2 for 8 seconds, and then a constant acceleration of 2 m/s2 for another 8 seconds. Determine the acquired velocity of the bike.

Answers

Answer:

40 m/s

Explanation:

First, we need to calculate the velocity after 8 seconds, so we will use the following equation:

v₂ = v₁ + at

Where v₁ is the initial velocity, so v₁ = 0 m/s, a is the acceleration a = 3 m/s² and t is the time, so t = 8s. Replacing the values, we get:

v₂ = 0 m/s + (3 m/s²)(8s)

v₂ = 0 m/s + 24 m/s

v₂ = 24 m/s

Then, it accelerates for another 8 seconds with a = 2 m/s², so the acquired velocity v₃ is equal to:

v₃ = v₂ + at

v₃ = 24 m/s + (2 m/s²)(8 s)

v₃ = 24 m/s + 16 m/s

v₃ = 40 m/s

Therefore, the acquired velocity of the bike is 40 m/s

2. Two charges are repelled by a force of 9.0 N. If the distance between them triples, what is the force between the charges?

Answers

F = k q1q2 / d^2

F= 9 N

The force is inversely related to the square of the distance. If d is 3 times larger, F is 9 times smaller.

F= 1 N

I need help with all three questions for this problem

Answers

Answer:

Explanation:

The given force is:

[tex]F=(2.2N)x+(1.0N)y[/tex]

Part A

If the displacement of the dog is:

d = (0.29m)x

The work done is the dot product of the force and displacement

[tex]\begin{gathered} W=F.d \\ \\ W=\text{ \lparen2.2\rparen\lparen0.29\rparen+\lparen1.0\rparen\lparen0\rparen} \\ \\ W=0.638Joules \end{gathered}[/tex]

Part B

If the displacement of the dog is:

d = (0.29m)y

The workdone will be calculated as:

[tex]\begin{gathered} W=F.d \\ \\ W=(2.2)(0)+(1.0)(0.29) \\ \\ W=0.29\text{ Joules} \end{gathered}[/tex]

Identify each setup and describe the difference in the magnetic field created by each.

Answers

In A. we have a magnetic field created by a coil in physics, the term coil refers to a long, thin loop of wire, which produces a magnetic field when an electric current is passed through it. in this case, the core inside the coil is the air

In B. We have a magnetic field created by a magnet that is a piece of iron or other magnetic material that has its component with atoms so ordered that the material have properties of magnetism, such as attracting other iron-containing objects or aligning itself in an external magnetic field

In C. Also we have a coil but this time with a core of metal material, a core can increase the magnetic field to thousands of times the strength of the field of the coil alone.

E. O 58.456 S48. A bullet is fired from the ground making an angle of 20 degwith the horizontal with a speed of 1500 m/s.Calculate the maximum height reached? (1 point)A. O20831.097 mB. 3642.875 mC. O8232.474 mD. O15803.894 mE. 13428.572 m9. A bullet is fired from the ground making an angle of 20 degwith the horizontal with a speed of 1500 m/s.How far (horizontally) will it fall? (1 point)

Answers

The maximum height of a projectile is given by:

[tex]h=\frac{v_0^2\sin^2\theta}{2g}[/tex]

In this case, the initial velocity is 1500 m/s, the angle is 20°; then we have:

[tex]\begin{gathered} h=\frac{(1500^)^2(\sin20)^2}{(2)(9.8)} \\ h=13428.572 \end{gathered}[/tex]

Therefore, the maximum height is 13428.572 m

Define vector and state two examples.

Answers

Vector is a mathematical entity that has a magnitude and a direction.

It is graphically represented by an arrow. The size of the arrow indicates the magnitude, and the tip of the arrow indicates the direction.

For example, we have the vector below:

Some forms of writing this vector are:

[tex]\begin{gathered} \vec{v}=ai+bj=4i+3j\\ \\ \vec{v}=<4,3> \end{gathered}[/tex]

Now, let's write a second example of a vector with 6 units in horizontal direction and -4 units in vertical direction, starting at the origin:

The magnitude and angle of a vector are given by:

[tex]\begin{gathered} \vec{v}=ai+bj\\ \\ magnitude:\\ \\ |\vec{v}|=\sqrt{a^2+b^2}\\ \\ angle:\\ \\ \theta=\tan^{-1}(\frac{b}{a}) \end{gathered}[/tex]

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