When you ride a bicycle, in what direction is the angular velocity of the wheels? When you ride a bicycle, in what direction is the angular velocity of the wheels? to your right forwards up to your left backwards g

Answers

Answer 1

When you ride a bicycle, the direction of the angular velocity of the wheels is; Option A; to your left

Complete question is;

When you ride a bicycle, in what direction is the angular velocity of the wheels? A) to your left B) to your right C) forwards D) backwards

While an object rotates, each particle will have a different velocity:

the 'Speed' component will vary with radius while the 'Direction' component will vary with angle.

Now, all of the velocity vectors are aligned in the same plane and as such we can be solve this by choosing a single vector normal to ALL of the possible velocity vectors of the rotating object in that plane.

The convention that will be used to answer this question is known as "Right-hand rule". The angular velocity vector points along the wheel's axle.

For instance, if you Imagine wrapping your right hand around the axle so that your fingers point in the direction of rotation, with your thumb sticking out. You will notice that your thumb points to the left.

Thus;

In conclusion, by right-hand rule, a wheel rotating on a forward - moving bicycle has an angular velocity vector pointing to the rider's left.

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Related Questions

The battery on your car has a rating stated in ampere minute which permit you to estimate the length of time a fully charged battery could deliver any particular current before discharge. Approximately how much energy is stored by a 50 ampere minute 12 volt battery

Answers

Answer:

Thus, the energy stored by a 50 Ampere minute battery is found to be  36 KJ.

Explanation:

The power delivered by a battery is given by the formula:

P = VI

where,

P = Power Delivered by battery in 1 second

V = Voltage of battery = 12 volt

I = Current stored in battery

But, if we multiply both sides of equation by time (t), then:

Pt = VIt

where,

Pt = Power x Time = E = Energy Stored = ?

It = Rating of Battery = (50 A.min)(60 sec/min) = 3000 A.sec

Therefore,

E = (12 volt)(3000 A.sec)

E = 36000 J = 36 KJ

Thus, the energy stored by a 50 Ampere minute battery is found to be  36 KJ.

Astronaut Flo wishes to travel to a star 20 light years away and return. Her husband Malcolm, who was the same age as Flo when she departs, stays home (baking cookies). If Flo travels at a constand speed of 80% of the speed of light (except for a short time to turn around), how much younger than Malcolm will Flo be when she returns? How long does Malcolm sit around baking cookies? How far is the distance to Flo?

Answers

Answer:

a. about 20 years younger

b. Malcolm sits around for 49.94 years

c. 2.268x[tex]10^{17}[/tex] m

Explanation:

light travels 3x[tex]10^{8}[/tex] m in one seconds

in 20 years that will be 3x[tex]10^{8}[/tex] x 20 x 60 x 60 x 24 x 365 = 1.89x[tex]10^{17}[/tex] m

for the to and fro journey, total distance covered will be 2 x 1.89x[tex]10^{17}[/tex]  = 3.78x[tex]10^{17}[/tex] m

Flo's speed = 80% of speed of light = 0.8 x 3x[tex]10^{8}[/tex]  = 2.4x[tex]10^{8}[/tex]  m/s

time that will pass for Malcolm will be  distance/speed = 3.78x[tex]10^{17}[/tex] /2.4x[tex]10^{8}[/tex]  

= 1575000000 s = 49.94 years

the relativistic time t' will be

t' = t x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]

t' = 49.94 x [tex]\sqrt{1 - 0.8^{2} }[/tex]

t' = 49.94 x 0.6 = 29.96 years       this is the time that has passed for Flo

this means that Flo will be about 20 years younger than Malcolm when she returns

relativistic distance is

d' = d x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]

d' = 3.78x[tex]10^{17}[/tex] x [tex]\sqrt{1 - 0.8^{2} }[/tex]

d' = 3.78x[tex]10^{17}[/tex] x 0.6

d' = 2.268x[tex]10^{17}[/tex] m     this is how far it is to Flo

A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of refraction of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its normal)?
A. 26
B. 35
C. 42
D. 22
E. 48

Answers

Answer:

35

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

Use the Lab screen to expand your ideas about what affects the landing location and path of a projectile. List any discoveries you made to identify additional things that affect the landing site of a projectile and/or path of a projectile. Next to each item, briefly explain why you think the motion of the projectile is affected..

Answers

Answer:

* air resistance.

*the direction of the rotation of the Earth

rotation of the thrown body

Explanation:

The projectile launch is described by the expressions

x-axis         x = v₀ₓ t

y-axis         y = [tex]v_{oy}[/tex] t - ½ gt²

When the things that affect this movement are analyzed, in order of importance we have:

* air resistance. This significantly changes the body's horizontal position, so it introduces a horizontal acceleration that is not contained in the equations.

* air resistance. At the height that the body reaches, since air resistance has the same direction as the gravity of gravity and therefore the relationship is more challenging.

* to a lesser extent the direction of launch, in the direction of the rotation of the Earth against. Since this creates an operational on the x and y axis that changes the initial assumption

* The possible rotation of the thrown body, since this rotation creates a lift that is not taken in the equations, this value is more noticeable the lighter the body, this effect has to keep the body longer in the air achieving more reach and height

When an electromagnetic wave falls on a white, perfectly reflectingsurface, it exerts a force F on that surface. If the surfaceis now painted a perfectly absorbing black, the force that the samewave would exert on the surface is:___________.
A) F
B) F/2
C) F/4
D) 2F
E) 4F

Answers

Answer:

B. F/2

Explanation:

The radiation force per unit area (radiation pressure Prad) exerted by an electromagnetic wave on a perfectly absorbing body has been found by experiment to be equal to the energy density of the wave

i.e Prad = u

For a reflecting body, this force exerted per unit area has been found to be twice the energy density of the wave.

i.e Prad = 2u.

Therefore, if the force exerted on a perfectly reflective body is F, then the force exerted on a perfectly absorbing body will be F/2

Archimedes and Heron are playing on a seesaw. Archimedes weighs 75 kg and Heron weighs 150 kg. If Heron is sitting 2 meters from the fulcrum, how many meters does Archimedes need to sit from the fulcrum?

Answers

Answer:

4metres

Explanation:

Using the principle of moment to solve the problem. The principle states that the sum of clockwise moments is equal to the sum of anticlockwise moment.

Moment = force *perpendicular distance

Moment of Archimedes about the fulcrum = 75 * x  ... 1

x is the distance of Archimedes from the fulcrum

Moment of Heron about the fulcrum = 150 * 2 = 300kgm... 2

Equation 1 and 2 according to principle of moment to get x we have;

75x = 300

x = 300/75

x = 4metres

Archimedes need to sit 4m from the fulcrum

Constants Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. The one goose is flying at 100 km/h relative to the air but a 44 km/h wind is blowing from west to east.
1. At what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground?2. How long will it take the bird to cover a ground distance of 450 from north to south?

Answers

Answer:

a. 63.89°  in the north-southward manner

b.  2.2 sec

Explanation:

The goose is flying at 100 km/h

Air from east to west is 44 km/h

angle relative to the north-south direction for the bird to travel south will be

cos∅ = 44/100 = 0.44

∅ = [tex]cos^{-1}[/tex]0.44 = 63.89°  in the north-southward manner

Speed south relative to the ground will be v

Tan 63.89 = v/100

2.04 = v/100

v = 2.04 x 100 = 204 km/hr

to cover a distance of 450 m from north to south at this speed time will be

t = d/v = 450/204 = 2.2 sec

A certain freely falling object, released from rest, requires 1.85 s to travel the last 26.5 m before it hits the ground. (a) Find the velocity of the object when it is 26.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.) -2.70 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the total distance the object travels during the fall.

Answers

Answer:

  a) -5.26 m/s

  b) 27.91 m

Explanation:

a) The acceleration due to gravity makes the velocity increase in magnitude in a linear way. The average velocity over the interval will be equal to the actual velocity halfway through the interval. The velocity at the beginning of the interval will be higher (less negative) by the amount velocity changes in the first half of the interval.

  average velocity = (0 -(26.5 m))/(1.85 s) ≈ -14.324 m/s

The change in velocity in the first half of the interval is ...

  Δv = (Δt/2)×(-9.8 m/s²) = (1.85 s)(-4.9 m/s²) = -9.065 m/s

So, the initial velocity (at the beginning of the last 1.85 s interval) is ...

  v1 = (average velocity) -Δv = (-14.324 m/s) -(-9.065 m/s)

  v1 = -5.259 m/s

__

b) The velocity when the object hits the ground is ...

  v2 = average velocity +Δv = -14.324 m/s -9.065 m/s = -23.389 m/s

This is related to the distance traveled by ...

  v² = 2dg . . . . . where g is the acceleration and d is the distance traveled

  d = v²/(2g) = 23.389²/(2·9.8) = 27.911 . . . . meters

The object travels a total distance of about 27.911 meters.

_____

The attached graph shows height vs. time.

EXPLANATION ⛔

A 20 gram mass is suspended from meter rod at 20cm. The meter rod is balanced on 40cm mark. Weight of meter rod is

A. 0.4N
B. 0.6N
C. 6N
D. 60N​

Answers

Answer:b

Explanation:I’m just trynna get more money dude

A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) . Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm

Answers

Complete Question

A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .

Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm

Part B

If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?

Answer:

Part A  

    [tex]R = 1081 \ J[/tex]

Part B  

     [tex]h = 169.7 \ m[/tex]

Explanation:

From the question we are told that

  The mass of the blade is  [tex]m_b = 0.65 \ kg[/tex]

   The length is  [tex]l = 0.55 \ m[/tex]

   The angular speed is  [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]

Generally the moment of inertia of the of this mower is mathematically evaluated as

         [tex]I = \frac{m_b * l^2 }{12}[/tex]

substituting values

         [tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]

         [tex]I = 0.016 \ kg m^2[/tex]

Generally the rotational kinetic energy of the bland is  

        [tex]R = \frac{1}{2} * I * w^2[/tex]

substituting values

       [tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]

     [tex]R = 1081 \ J[/tex]

At point where the gravitational potential energy is equal to the rotational kinetic energy  we have that

       [tex]P = R = m_b * h * g[/tex]

Where P is the  gravitational potential energy

substituting values

          [tex]1081 = 0.65 * 9.8 * h[/tex]

=>       [tex]h = 169.7 \ m[/tex]

       

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 3.5 mV/m. At what rate is the magnetic field changing?

Answers

Answer

The rate at which the magnetic field is changing is  [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]

Explanation

From the question we are told that

   The electric field strength is [tex]E = 3.5mV/m = 3.5 *10^{-3} \ V/m[/tex]

    The radius is  [tex]r = 1.5 \ m[/tex]

The rate of change of the  magnetic  field  is mathematically represented as

        [tex]\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}[/tex]

Where [tex]dl[/tex] is change of a unit length

     [tex]\frac{d \phi}{dt} = A * \frac{dB}{dt}[/tex]

Where A is the area which is mathematically represented as

     [tex]A = \pi r^2[/tex]

    So

    [tex]E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

  [tex]E L = ( \pi r^2) (\frac{dB}{dt} )[/tex]  

where L is the circumference of the circle which is mathematically represented as

     [tex]L = 2 \pi r[/tex]

So

     [tex]E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ][/tex]

      [tex]E = \frac{r}{2} [\frac{dB}{dt} ][/tex]

       [tex][\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }[/tex]

substituting values

      [tex][\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }[/tex]

      [tex][\frac{dB}{dt} ] = 0.000467 T/s[/tex]    

You are moving a desk that has a mass of 36 kg; its acceleration is 0.5 m / s 2. What is the force being applied

Answers

Answer:

18 N

Explanation:

Force can be found using the following formula.

f= m*a

where m is the mass and a is the acceleration.

We know the desk has a mass of 36 kilograms. We also know that its acceleration is 0.5 m/s^2.

m= 36 kg

a= 0.5 m/s^2

Substitute these values into the formula.

f= 36 kg * 0.5 m/s^2

Multiply 36 and 0.5

f=18 kg m/s^2

1 kg m/s^2 is equivalent to 1 Newton, or N.

f= 18 Newtons

The force being applied is 18 kg m/s^2, Newtons, or N

5.

The solar system coalesced due to rotational forces and

gravity.

heat.

radioactivity.

solar wind.

Answers

Answer:

Gravity

Explanation:

The solar system is held together by rotational forces and gravity. This can be seen from billions of years ago when the solar system was a cloud of dust and gas. This cloud of dust and gas is known as the solar nebula. All of these dust and gas were brought together by the rotational movement as well as the action of gravity which brought all the particles together to form a larger one. This alone brought about the sun's formation in the center of the nebula as well the formation of other planetary bodies, etc.

Cheers.

a steel ball is dropped from a diving platform use the approximate value of g as 10 m/s^2 to solve the following problem what is the velocity of the ball 0.9 seconds after its released

Answers

Answer:

The final speed of the ball is 9 m/s.

Explanation:

We have,

A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :

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

u = 0 (at rest), a = g

[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]

So, the final speed of the ball is 9 m/s.

The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?

Answers

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position =  5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

5000 m - 1900 m = a(1800 s²)

a(1800 s²) = 3100 m

a = 3100 m/1800 s²

a = 1.72 m/s²

A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x

Answers

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m  (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.

What is electric field?

The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.

Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.

The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)

The horizontal distance traveled by particle is

x = ucosθt

t = x/ucosθ

The force in an electric field is F = qE...................(1)

where, q is charge , E is the strength of electric field

From, newton 2nd law of motion, Force F = ma.................(2)

Equating both the equations, we get

ma = qE

a = qE/m..................(3)

The vertical distance, y =usinθt - 1/2at²

From equation 3, we have

y = usinθt  -  1/2 (qE/m) t²

if y = 0, t = 2musinθ/(qE) = x / (ucosθ)

The electric field is represented as

Also, E = 2mu²×sinθ×cosθ/(xq)

Plug the values, we get

E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)

E = -4556.18 N/m

Thus, the electric field of the particle is  -4556.18 N/m.

Learn more about electric field.

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A 1,269 kg rocket is traveling at 413 m/s with 2,660 kg of fuel on board. If the rocket fuel travels at 1,614 m/s relative to the rocket, what is the rockets final velocity after it uses half of its fuel?

Answers

Answer:

About 2104m/s

Explanation:

[tex]F=ma \\\\F=\dfrac{2660kg}{2}\cdot 1614m/s=2,146,620N \\\\2,146,620N=1,269kg\cdot a \\\\a\approx 1691m/s \\\\v_f=v_o+at=413m/s+1691m/s=2104m/s[/tex]

Hope this helps!

A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Answers

Answer:

(a) q = 2.357 x 10⁻ C

(b) Φ = 2.66 x 10 N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a)  Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10 N.m²/C

Assuming 100% efficient energy conversion, how much water stored behind a 50 centimetre high hydroelectric dam would be required to charge battery

Answers

Complete question is;

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes

Answer:

Amount of water required to charge the battery = 7.35 m³

Explanation:

The formula for Potential energy of the water at that height = mgh

Where;

m = mass of the water

g = acceleration due to gravity = 9.8 m/s²

H = height of water = 50 cm = 0.5 m

We know that in density, m = ρV

Where;

ρ = density of water = 1000 kg/m³

V = volume of water

So, potential energy is now given as;

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Now, formula for energy of the battery is given as;

E = qV

We are given;

q = 50 A.min = 50 × 60 = 3,000 C

V = 12 V

Thus;

qV = 3,000 × 12 = 36,000 J

E = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery.

Thus;

(4900V) = (36,000)

4900V = 36,000

V = 36,000/4900

V = 7.35 m³

wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, how many more revolutions will it rotate through in the next 5.00 s?

Answers

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Nuclear fusion in our Sun happens when


- hydrogen atoms combine to make helium atoms and release energy

- uranium atoms break apart and release energy

- hydrogen atoms are burned and release energy

- helium atoms break apart and release energy

Answers

Answer:

A

Explanation:

Fussion occurs when elements of lower atomic mass combines to form that of a larger atomic mass, releasing energy in the process .

Hydrogen has a lower atomic mass than Helium.

Carbon is added to iron to make steel. Steel is stronger than either carbon or iron by itself.


What does this example show?

Answers

Answer:

This example shows that alloys are stronger than either of it's parent materials by themselves.

Explanation:

Since carbon is added to iron to make steel, it means steel is an alloy of iron and carbon.

This is because an alloy is a mixture of two or more elements, where at least one element is a metal.

Now, steel is stronger than either carbon or or iron by itself because Steel contains atoms of other elements including carbon and iron. These atoms have different sizes to iron carbon atoms, so they distort the layers of atoms in the pure iron and carbon. This means that a greater force is required for the layers to slide over each other in steel, so steel is harder than pure iron.

The “turning effect of a force” (T = F * r) is:
(a) determined as the product of force and the moment of inertia.
(b) generated by concentric forces.
(c) equivalent to the angular momentum.
(d) determined as a product of torque and moment arm.
(e) called “moment” or “torque”.

Answers

Answer:

b and e

Explanation:

r x F is the formula for torque.

The "turning effect" or torque happens when concentric forces rotate an object along said center.

a) False because T = Fr = Ia (a = angular acceleration)

b) True

c) False. L = Iw (w = angular velocity), which does not equal Ia

d) False. It is torque, not the product of torque and something else

e) True.

A woman weighs 129 lb. If she is standing on a spring scale in an elevator that is traveling downward, but slowing up, the scale will read:___________.
A) more than 129 lb
B) 129 lb
C) less than 129 lb
D) It is impossible to answer this question without knowing the acceleration of the elevator.

Answers

Answer:

C) less than 129 lb.

Explanation:

Let the elevator be slowing up with magnitude of a . That means it is accelerating downwards  with magnitude a .

If R be the reaction force

For the elevator is going downwards with acceleration a

mg - R = ma

R = mg - ma

R measures its apparent weight . Spring scale will measure his apparent weight.

So its apparent weight is less than 129 lb .

Question 9 of 10
2 Powie
You are riding a bicycle. You apply a forward force of 100 N, and you and the
bicycle have a combined mass of 80 kg. What is the acceleration of the
bicycle?
A. 125 m/s
B. 1.5 m/s2
c. 1.8 m/s?
D. 0.8 m/s​

Answers

Answer:

1.25 m/s^2

Explanation:

F = m*a ...... force = mass * acceleration

force = 100 N, mass = 80 kg

100 = 80 * a

100/80 = a = 1.25 m/s^2

Answer:

The acceleration is 1.25m/s².

Explanation:

You have to apply Newton's Second Law which is F = m×a where F represents force, m is mass and a is acceleratipn. Then you have to substitute the following values into the formula :

[tex]f = m \times a[/tex]

Let F = 100,

Let m = 80,

[tex]100 = 80 \times a[/tex]

[tex]100 = 80a[/tex]

[tex]a = 100 \div 80[/tex]

[tex]a = 10 \div 8[/tex]

[tex]a = 1.25[/tex]

A block of mass 15.0 kg slides down a ramp inclined at 28.0∘ above the horizontal. As it slides, a kinetic friction force of 30.0 N parallel to the ramp acts on it. If the block slides for 5.50 m along the ramp, find the work done on the block by friction.

Answers

Answer:

Work is done by friction = -165 J

Explanation:

Given:

Mass of block (m) = 15 kg

Ramp inclined = 28°

Friction force (f) = 30 N

Distance (d) = 5.5 m

Find:

Work is done by friction.

Computation:

Work is done by friction = -Fd

Work is done by friction = -(30)(5.5)

Work is done by friction = -165 J

Explain why it is necessary to have a high voltage​

Answers

Answer:

SO THAT

EACH APPLIANCE CAN GET SUFFICIENT POTENTIAL DIFF. TO RUN

Four long wires are each carrying 6.0 A. The wires are located
at the 4 corners of a square with side length 9.0 cm. All of
these wires are carrying current out of the page. The
magnetic field (in T) at one corner of the square is:

Answers

Answer:

[tex]B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

[tex]B=\frac{\mu_oI}{2\pi r}[/tex]

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i)  and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

[tex]B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}][/tex]

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

[tex]r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m[/tex]

Then you have:

[tex]B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]

The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its
A.climate.
b.weather.
C. global warming
d. seasons

Answers

Answer:

a. global warming

Explanation:

that's the definitain of global warming

Answer:

A climate

Explanation:

Assuming 100% efficient energy conversion, how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery?

Answers

Complete question is;

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes.

Answer:

Amount of water required to charge the battery = 7.35 m³

Explanation:

The formula for Potential energy of the water at that height = mgh

Where;

m = mass of the water

g = acceleration due to gravity = 9.8 m/s²

h = height of water = 50 cm = 0.5 m

We know that in density, m = ρV

Where;

ρ = density of water = 1000 kg/m³

V = volume of water

So, potential energy is now given as;

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Now, formula for energy of the battery is given as;

E = qV

We are given;

q = 50 A.min = 50 × 60 = 3,000 C

V = 12 V

Thus;

qV = 3,000 × 12 = 36,000 J

E = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery.

Thus;

(4900V) = (36,000)

4900V = 36,000

V = 36,000/4900

V = 7.35 m³

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