A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 4.30 rev/s in 3.05 s . Part A What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 rev/s

Answers

Answer 1

Answer:

[tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

Explanation:

Angular acceleration

[tex]\begin{aligned}

\alpha &=\frac{\left(\omega_{f}-\omega_{i}\right)}{t} \\

\omega_{i} &=0 \\

\omega_{f} &=4.30 \mathrm{rev} / \mathrm{s} \\

&=4.30 \times 2 \pi \mathrm{rad} / \mathrm{s} \\

&=27.02 \mathrm{rad} / \mathrm{s} \\

\alpha &=\frac{(27.02-0)}{3.15} \\

&=8.57 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

a)Tangential acceleration

[tex]\begin{aligned}

a &=r \alpha \\

&=\frac{12}{2} \times 10^{-2} \times 8.57 \\

a &=0.51 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

The tangential acceleration of the disc is [tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

Answer 2

This question involves the concepts of the equations of motion for angular motion.

The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed reaches 2 rev/s will be "0.532 m/s²".

First, we will use the first equation of motion for the angular motion to find out the angular acceleration:

[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]

where,

[tex]\alpha[/tex] = angular acceleration = ?

[tex]\omega_f[/tex] = final angular speed = (4.3 rev/s)[tex](\frac{2\pi\ rad}{1\ rev})[/tex] = 27.02 rad/s

[tex]\omega_i[/tex] = initial angular speed = 0 rad/s

t = time taken = 3.05 s

Therefore,

[tex]\alpha =\frac{27.02\ rad/s-0\ rad/s}{3.05\ s}\\\\\alpha= 8.86\ rad/s^2[/tex]

Now, the tangential acceleration can be given as follows:

[tex]a=r\alpha\\a=(\frac{diameter}{2})(8.86\ rad/s^2)\\\\a=(\frac{0.12\ m}{2})(8.86\ rad/s^2)\\\\[/tex]

a = 0.532 m/s²

Learn more about the angular motion here:

brainly.com/question/14979994?referrer=searchResults

The attached picture shows the angular equations of motion.

A Compact Disk, Which Has A Diameter Of 12.0 Cm, Speeds Up Uniformly From Zero To 4.30 Rev/s In 3.05

Related Questions

A standing wave on a string that is fixed at both ends has frequency 80.0 Hz. The distance between adjacent antinodes of the standing wave is 12.0 cm. What is the speed of the waves on the string, in m/s

Answers

Answer:

v = 19.2 m/s

Explanation:

In order to find the speed of the string you use the following formula:

[tex]f=\frac{v}{2L}[/tex]          (1)

f: frequency of the string = 80.0Hz

v: speed of the wave = ?

L: length of the string = 12.0cm = 0.12m

The length of the string coincides with the wavelength of the wave for the fundamental mode.

Then, you solve for v in the equation (1), and replace the values of the other parameters:

[tex]v=2Lf=2(0.12m)(80.0Hz)=19.2\frac{m}{s}[/tex]

The speed of the wave is 19.2 m/s

While her kid brother is on a wooden horse at the edge of a merry-go-round, Sheila rides her bicycle parallel to its edge. The wooden horses have a tangential speed of 6 m/s. Sheila rides at 4 m/s. The radius of the merry-go-round is 8 m. At what time intervals does Sheila encounter her brother, if she rides opposite to the direction of rotation of the merry-go-round?
a. 5.03 s
b. 8.37 s
c. 12.6 s
d. 25.1 s
e. 50.2 s

Answers

Answer:

t = 5.03 s

Explanation:

To find the time interval when Sheila encounter her brother, you first calculate the angular speed of both Sheila and her brother.

You use the following formula:

[tex]\omega = \frac{v}{r}[/tex]

w: angular speed

v: tangential speed

r: radius of the trajectory = 8 m

For  you have:

[tex]\omega=\frac{4m/s}{8m}=0.5\frac{rad}{s}[/tex]

For her brother:

[tex]\omega'=\frac{6m/s}{8m}=0.75\frac{rad}{s}[/tex]

Next, they will encounter to each other when the angular distance of the Brother of sheila equals the angular distance of Sheila in the opposite direction. This can be written as follow:

[tex]\theta=\omega t\\\\\theta'=\omega ' t[/tex]

They encounter for θ = 2π-θ':

[tex]\omega t=2\pi-\omega' t[/tex]

You replace the values of the parameters in the previous equation and solve for t:

[tex]0.5t=2\pi-0.75t\\\\1.25t=2\pi\\\\t=5.026\approx5.03[/tex]

Hence, Sheila encounter her brother in 5.03 s

Mark Watney (Matt Damon in the Martian movie) and Marvin the Martian (Looney Tunes cartoon character) are having an argument on the surface of Mars (negligible air resistance). They are testing out their new potato launcher that fires projectiles at a constant speed. Mark launches his potato at an angle of 60◦ and Marvin launches his identical potato at an angle of 30◦ . Without any calculations try to answer the following questions, and justify each answer.

(A) Which potato lands farther away from the launcher (potatoes are launched from ground level)?

(B) Which potato spends more time in the air before hitting the ground

(C) Which potato has a greater speed just before it hits the ground?

Answers

Answer:

A) The two potatoes cover the same horizontal distance from the launcher.

B) Mark's potato spends more time in the air than Marvin's potato before hitting the ground.

C) Marvin's potato hits the ground with a greater speed than Mark's potato

Explanation:

A) For projectile motion, the final horizontal distance of the projectile from where it was initially launched (its range) is given as

R = (u² sin 2θ)/g

where

u = initial velocity of the projectile

θ = angle above the horizontal at which the projectile was launched = 30°, 60°

g = acceleration due to gravity on Mars

Since, u and g are the same for Mark and Marvin, sin 2θ would determine which range is higher.

Sin (2×60°) = sin 120°

Sin (2×30°) = sin 60°

Sin 120° = Sin 60°

Hence, the two potatoes cover the same horizontal distance from the launcher.

B) Time spent in the air for a projectile is given as

T = (2u sin θ)/g

Again, since u and g are the same for Mark and Marvin on Mars, sin θ will give the required idea of whose potato spends more time in the air.

Sin 60° = 0.866

Sin 30° = 0.50

Sin 60° > Sin 30°

Hence, Mark's potato spends more time in the air than Marvin's potato.

C) The horizontal velocity for projectile motion is constant all through the motion and is equal to u cos θ

u cos 60° < u cos 30°

And the initial vertical velocity is u sin θ

Final vertical velocity

= (initial vertical velocity) - gt

g = acceleration due to gravity on Mars

T = time of flight

For Mark,

initial vertical velocity = u sin 60°, greater than Marvin's u sin 30°

And Mark's potato's time of flight is greater as established in (B) above.

But for Marvin

initial vertical velocity = u sin 30°, less than Mark's u sin 60°

And Marvin's potato's time of flight is lesser as established in (B) above

So, at the end of the day, the final vertical velocity is almost the same for both Mark's and Marvin's potatoes.

Hence, the horizontal component of the final velocity edges the final speed of the potatoes just before hitting the ground in Marvin's favour.

Hope this Helps!!!

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.

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

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

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]

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

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]    

A car travels around an oval racetrack at constant speed. The car is accelerating:________.
A) at all points except B and D.
B) at all points except A, B, C, and D.
C) everywhere, including points A, B, C, and D.
D) nowhere, because it is traveling at constant speed.
2) A moving object on which no forces are acting will continue to move with constant:_________
A) Acceleration
B) speed
C) both of theseD) none of these

Answers

Answer:

1A,2D,3B

Explanation:

hope this helps

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

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.

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

510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectanglar prism with cross-sectional area of width 11.6 cm and length 23.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the gr

Answers

Answer:

The terminal velocity is [tex]v_t =17.5 \ m/s[/tex]

Explanation:

From the question we are told that

       The mass of the squirrel is  [tex]m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg[/tex]

      The surface area is   [tex]A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2[/tex]

       The height of fall is  h =4.8 m

        The length of the prism is [tex]l = 23.2 = 0.232 \ m[/tex]

          The width of the prism is [tex]w = 11.6 = 0.116 \ m[/tex]

 

The terminal velocity is mathematically represented as

       [tex]v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }[/tex]

Where [tex]\rho[/tex]  is the density of a rectangular prism with a constant values of [tex]\rho = 1.21 \ kg/m^3[/tex]

            [tex]C[/tex] is the drag coefficient for a horizontal skydiver with a value = 1

            A  is the area of the prism the squirrel is assumed to be which is mathematically represented as

      [tex]A = 0.116 * 0.232[/tex]

       [tex]A = 0.026912 \ m^2[/tex]

 substituting values

      [tex]v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }[/tex]

     [tex]v_t =17.5 \ m/s[/tex]

       

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) What is the cost of heating a hot tub containing 1440 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take heat loss to surroundings into account? The cost of electricity is 9.00¢/(kW · h) and the specific heat for water is 4184 J/(kg · °C). $ 67 Incorrect: Your answer is incorrect. How much heat is needed to raise the temperature of m kg of a substance? How many joules are in 1 kWh? (b) What current was used by the 220 V AC electric heater, if this took 3.45 h? 88.2 Correct: Your answer is correct. A

Answers

Answer:

a) [tex]E = 6.024\,USD[/tex], For m kilograms, it is 4184m J., 3600000 joules, b) [tex]i = 88.200\,A[/tex]

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

[tex]Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})[/tex]

Where:

[tex]m_{w}[/tex] - Mass of water, measured in kilograms.

[tex]c_{w}[/tex] - Specific heat of water, measured in [tex]\frac{J}{kg\cdot ^{\circ}C}[/tex].

[tex]T_{f}[/tex], [tex]T_{i}[/tex] - Initial and final temperatures, measured in [tex]^{\circ}C[/tex].

Then,

[tex]Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)[/tex]

[tex]Q_{needed} = 180748800\,J[/tex]

The energy needed in kilowatt-hours is:

[tex]Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)[/tex]

[tex]Q_{needed} = 50.208\,kWh[/tex]

The electric energy required to heat up the water is:

[tex]E = \frac{50.208\,kWh}{0.75}[/tex]

[tex]E = 66.944\,kWh[/tex]

Lastly, the cost of heating a hot tub is: (USD - US dollars)

[tex]E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)[/tex]

[tex]E = 6.024\,USD[/tex]

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

[tex]i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}[/tex]

[tex]i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]

[tex]i = 88.200\,A[/tex]

Two students are pushing their stalled car down the street. If the net force exerted on
the car by the students is 1000 N at an angle of 20° below horizontal, the horizontal
component of the force is:
(a) greater than 1000 N.
(b) less than 1000 N.
(c) sum of the pushing force and the weight of the students.
(d) (a) and (b)
(e) (a) and (c)

Answers

Answer:

B

Explanation:

Because the force has 2 components (horizontal and vertical), the horizontal component must be smaller than the total force. The Pythagorean theorem only adds positive values (because they're squared) so it makes sense. Using trigonometry, 100*cos(-20) yields a horizontal force of around 939.7N, which is less than 1000N.

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]

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

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

Someone please helpp me out thanks !

Answers

Answer:

Silver.

Explanation:

To determine the identity of the metal, we need to calculate the density of the metal. This is illustrated below:

Mass of metal (m) = 18.15g

Length (L)= 1.2cm

Volume (V) = L³ = 1.2³ = 1.728cm³

Density =.?

The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically, it is expressed as:

Density = Mass /volume

With the above formula, we can obtain the density of the metal as follow:

Mass = 18.15g

Volume = 1.728cm³

Density =.?

Density = Mass /volume

Density = 18.15g/1.728cm³

Density of the metal = 10.50g/cm³

Comparing the density of metal obtained with the densities given in the table above, we can see that the density of the metal is the same with that of silver.

Therefore, the metal is silver.

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

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:

Organ pipe A, with both ends open, has a fundamental frequency of 475 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. Use 343 m/s for the speed of sound in air. How long are (a) pipe A and (b) pipe B?

Answers

Answer:

The length of organ pipe A is [tex]L = 0.3611 \ m[/tex]

The length of organ pipe B is  [tex]L_b = 0.2708 \ m[/tex]

Explanation:

From the question we are told that

    The fundamental frequency is  [tex]f = 475 Hz[/tex]

     The speed of sound is  [tex]v_s = 343 \ m/s[/tex]

The fundamental frequency of the organ pipe A  is mathematically represented as

        [tex]f= \frac{v_s}{2 L}[/tex]

Where L is the length of  organ pipe

   Now  making L the subject

        [tex]L = \frac{v_s}{2f}[/tex]

substituting values

        [tex]L = \frac{343}{2 *475}[/tex]

        [tex]L = 0.3611 \ m[/tex]

The second harmonic frequency of the  organ pipe A is mathematically represented as

       [tex]f_2 = \frac{v_2}{L}[/tex]

The third harmonic frequency of the  organ pipe B is mathematically represented as      

      [tex]f_3 = \frac{3 v_s}{4 L_b }[/tex]

So from the question

       [tex]f_2 = f_3[/tex]

So

    [tex]\frac{v_2}{L} = \frac{3 v_s}{4 L_b }[/tex]

Making  [tex]L_b[/tex] the subject

     [tex]L_b = \frac{3}{4} L[/tex]

substituting values

    [tex]L_b = \frac{3}{4} (0.3611)[/tex]

    [tex]L_b = 0.2708 \ m[/tex]

Your electric drill rotates initially at 5.35 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.331 rad/s2 for 4.81 s. What is the drill's angular displacement during that time interval?

Answers

Answer:

The  angular displacement  is  [tex]\theta = 29.6 \ rad[/tex]

Explanation:

From the question we are told that

     The initial angular speed is  [tex]w = 5.35 \ rad/s[/tex]

      The angular acceleration is  [tex]\alpha = 0.331 rad /s^2[/tex]

      The time take is  [tex]t = 4.81 \ s[/tex]

     

Generally the angular displacement is mathematically represented as

          [tex]\theta = w * t + \frac{1}{2} \alpha * t^2[/tex]

substituting values

         [tex]\theta = 5.35 * 4.81 + \frac{1}{2} * 0.331 * (4.81)^2[/tex]

         [tex]\theta = 29.6 \ rad[/tex]

A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and
its mass density is 5.0 g/m2, determine the tension of the wire.

Answers

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g[tex]/m^{2}[/tex] = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = [tex]\frac{1}{2L}[/tex][tex]\sqrt{\frac{T}{M} }[/tex]

⇒      T = 4[tex]L^{2}[/tex][tex]f^{2}[/tex]M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × [tex]2^{2}[/tex] ×[tex]65^{2}[/tex] × 0.01

   = 4 × 4 ×4225 × 0.01

   = 676 N

Tension of the wire is 676 N.

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.

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 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

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