An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s). If a particular disk is spun at 646.1 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds, what is the magnitude of the average angular acceleration of the disk?

Answers

Answer 1

Answer:

Angular acceleration will be [tex]1135.5rad/sec^2[/tex]

Explanation:

We have given initial angular speed of a particular disk [tex]\omega _i=646.1rad/sec[/tex]

And it finally comes to rest so final angular speed [tex]\omega _f=0rad/sec[/tex]

Time is given as t = 0.569 sec

From third equation of motion we know that

[tex]\omega _f=\omega _i+\alpha t[/tex]

So angular acceleration [tex]\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-646.1}{0.569}=1135.5rad/sec^2[/tex]


Related Questions

A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick should a 5.00-g tack be placed so that the stick will balance at the 50.0 cm mark?

Answers

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

[tex]80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)[/tex]

[tex]80=5(50-x)[/tex]

[tex]80=250-5x[/tex]

[tex]5x=170[/tex]

[tex]x=\frac{170}{5}[/tex]

[tex]x=34 cm[/tex]

Final answer:

To balance the nonuniform meterstick, the 5.00-g tack should be placed at approximately 50.4 cm from the pivot point.

Explanation:

To find the location on the meterstick where the 5.00-g tack should be placed so that the stick will balance at the 50.0 cm mark, we can use the principle of torque balance.

The torque balance equation is given by:

Torque(counter-clockwise) = Torque(clockwise).

In this case, the torque is equal to the force multiplied by the distance from the pivot point. The force exerted by the 5.00-g tack can be calculated using the formula:

Force = mass * acceleration due to gravity.

By substituting the known values into the equation and solving for the distance, we can determine where the tack should be placed on the meterstick.

In this case, the tack should be placed at approximately 50.4 cm from the pivot point.

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The Explorer VIII satellite, placed into orbit November 3, 1960, to investigate the ionosphere, had the following orbit parameters: perigee, 459 km; apogee, 2 289 km (both distances above the Earth's surface); period, 112.7 min. Find the ratio vp/va of the speed at perigee to that at apogee

Answers

Answer:

[tex]\frac{v_p}{v_a}=1.268[/tex]

Explanation:

1) Basic concepts

Apogee: is the maximum distance for an object orbiting the Earth. For this case the distance for the apogee would be the sum of radius for the earth and 2289 km.

Perigee: Minimum distance of an object orbiting the Earth. For this case the distance for the perigee would be the sum of radius for the earth and 459 km.

A good approximation for these terms are related to the figure atached.

Isolated system: That happens when the system is not subdued to an external torque, on this case the change of angular momentum would be 0.

2) Notation and data

[tex]r_{earth}[/tex] radius for the Earth, looking for this value on a book we got [tex]r_{earth}=6.37x10^{3}km[/tex]

[tex]r_p =459km +r_{earth}=459km +6.37x10^3 km=6.829x10^3 km[/tex], represent the radius for perigee

[tex]r_a =2289km +r_{earth}=2289km +6.37x10^3 km=8.659x10^3 km[/tex], represent the radius for perigee

[tex]T=112.7min[/tex], period

[tex]v_p[/tex] velocity of the perigee

[tex]v_p[/tex] velocity of the apogee

[tex]r=\frac{v_p}{v_a}[/tex] is the variable of interest represented the ratio for the speed of the perigee to the speed of the apogee

3) Formulas to use

We don't have torque from the gravitational force since is a centered force. So the change of momentum is 0

[tex]\Delta L=0[/tex]   (1)

[tex]L_a=L_p[/tex]   (2)

From the definition of angular momentum we have [tex]L_i=mv_i r_i[/tex], replacing this into equation (2) we got:

[tex]mv_a r_a =mv_p r_p[/tex]   (3)

We can cancel the mass on both sides

[tex]v_a r_a =v_p r_p[/tex]   (4)

And solving [tex]\frac{v_p}{v_a}[/tex] we got:

[tex]\frac{v_p}{v_a}=\frac{r_a}{r_p}[/tex]   (5)

And replacing the values obtained into equation (5) we got:

[tex]\frac{v_p}{v_a}=\frac{8.659x10^3 km}{6.829x10^3 km}=1.268[/tex]  

And this would be the final answer [tex]\frac{v_p}{v_a}=1.268[/tex]

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by 0.50 m, the mass that moves downward is 80 kg, and the collision on the floor lasts 0.081 s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

Answers

Answer:

Impulse, J = 250.4 kgm/s,  Avg Force F=3091.4 N

Explanation:

Since we know that impulse is the change in momentum i.e. Δp and   Δp[tex]=mv[/tex], therefore, to calculate the velocity we perform:

As the person has fallen by a 0.50m height, its potential energy changes into kinetic energy, therefore,

K.E.=P.E.

[tex]\frac{1}2}mv^{2}[/tex]=[tex]mgh[/tex]

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2*9.8*0.50}[/tex]

[tex]v=3.13ms^{-1}[/tex]

(a) Impulse [tex]J[/tex] = Δp[tex]=mv[/tex]

[tex]J= 80*3.13[/tex]

[tex]J = 250.4 kgms^{-1}[/tex]

(b)  Avg Force F = Δp/Δt

[tex]F=\frac{250.4}{0.081}[/tex]

[tex]F=3091.4[/tex] N

A paper filled capacitor is charged to a potential difference of 2.1 V and then disconnected from the charging circuit. The dielectric constant of the paper is 3.7. Keeping the plates insulated, the paper filling is withdrawn, allowing air to fill the capacitor instead. Find the resulting potential difference off the capacitor.
(volts)
While continuing to keep the capacitor's plates insulated, an unknown substance is inserted between them. The plates then attain a potential difference that is 0.59 times the original potential difference (when paper filled the capacitor). What is the substances dielectric constant?

Answers

Answer:

Part a)

[tex]V' = 7.77 Volts[/tex]

Part b)

[tex]k' = 6.27 [/tex]

Explanation:

As we know that capacitor plate is connected across 2.1 V and after charging it is disconnected from the battery

So here we can say that charge on the plates will remain conserved

So we will have

[tex]Q = kC(2.1)[/tex]

now dielectric is removed between the plates of capacitor

so new potential difference between the plates

[tex]V' = \frac{Q}{C'}[/tex]

[tex]V' = \frac{kC(2.1)}{C}[/tex]

[tex]V' = 3.7 \times 2.1 [/tex]

[tex]V' = 7.77 Volts[/tex]

Part b)

Now the capacitor plates are again isolated and unknown dielectric is inserted between the plates

So again charge is same so potential difference is given as

[tex]V" = \frac{Q}{k'C}[/tex]

[tex]0.59 V = \frac{kCV}{k'C}[/tex]

[tex]0.59 = \frac{3.7}{k'}[/tex]

[tex]k' = 6.27 [/tex]

Final answer:

The resulting potential difference of the capacitor after the paper is withdrawn is 7.77V. The dielectric constant of the unknown substance, where the potential decreased to 0.59 times the original, is approximately 6.27.

Explanation:

The potential difference across a capacitor is inversely proportional to the dielectric constant of the material between the plates. When a paper dielectric with a constant of 3.7 is removed from a charged capacitor, the potential difference increases because the dielectric constant of air is nearly 1.

Find the resulting potential difference of the capacitor

The initial potential difference (V1) with paper is 2.1V, and the dielectric constant for paper (k1) is 3.7. When the paper is removed, the dielectric constant becomes that of air, which is approximately 1 (k2). For an isolated capacitor, the charge (Q) remains constant, so we can use the relation that the product of the potential difference (V) and the dielectric constant (κ) is a constant (Q=CV and C=κC0). Therefore, V1k1 = V2k2. By substituting the known values: 2.1V * 3.7 = V2 * 1, we find that V2 = 2.1V * 3.7 = 7.77V.

Find the dielectric constant of the inserted unknown substance

If the potential difference with the new substance is 0.59 times the initial potential (when paper filled the capacitor), then V3 = 0.59 * 2.1V. The dielectric constant of the new substance (k3) can be found using the same principle where V1k1 = V3k3. With V3 = 0.59 * 2.1 = 1.239V, we get 2.1V * 3.7 = 1.239V * k3, leading to k3 = (2.1V * 3.7) / 1.239V ≈ 6.27.

A 94.0 N grocery cart is pushed 17.6 m along an aisle by a shopper who exerts a constant horizontal force of 42.6 N. The acceleration of gravity is 9.81 m/s 2 . If all frictional forces are neglected and the cart starts from rest, what is the grocery cart’s final speed

Answers

Answer:

vf = 12.51 m/s

Explanation:

Newton's second law to the grocery cart:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the grocery cart  and the y-axis in the direction perpendicular to it.

Forces acting on the grocery cart

W: Weight of the block : In vertical direction  downward

N : Normal force : In vertical direction  upward

F : horizontal force

Calculated of the mas of the grocery cart (m)

W = m*g

m = W/g

W = 94.0 N  , g = 9.81 m/s²

m = 94/9.81

m = 9.58 Kg

Calculated of the acceleration of the grocery cart (a)

∑F = m*a

F = m*a

42.6 = (9.58)*a

a = (42.6) / (9.58)

a = 4.45 m/s²

Kinematics Equation of the grocery cart

Because the grocery cart moves with uniformly accelerated movement we apply the following formula to calculate its final speed :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

a: acceleration (m/s²)

Data:

v₀ = 0

a = 4.45 m/s²

d = 17.6 m

We replace data in the formula (2) :

vf²=v₀²+2*a*d

vf² = 0+2*(4.45)*(17.6)

vf² = 156.64

[tex]v_{f} = \sqrt{156.64}[/tex]

vf = 12.51 m/s

Which vector below goes from (0,0) to (-2,3)?

A. d
B. c
C. b
D. a

Answers

Answer:

The vector a, the answer is the point D

Explanation:

We need to make the make the subtraction from the x,y points. It will be the points of the head minus the points of the tail.

x=(-2 - 0) and y = (3-0) vector = -2 + 3

The Answer is—————————D. a

A 0.026 kg bullet is fired straight up at a falling wooden block that has a mass of 5.0 kg. The bullet has a speed of 750 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occured. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

Answers

After defining variables and initial conditions, analyzing the block's motion before collision, and applying the conservation of momentum principle, we found that the block had been falling for approximately 0.198 seconds before colliding with the bullet.

1. Defining Variables and Initial Conditions:

m: Mass of the bullet (0.026 kg)

v_b: Velocity of the bullet before collision (750 m/s)

M: Mass of the block (5 kg)

v_f: Final velocity of the block before collision (unknown)

t: Time the block falls before collision (unknown)

V: Final velocity of the combined bullet-block system after collision (unknown)

2. Analyzing Block's Motion before Collision:

The block is initially dropped from rest, so its initial velocity (v_o) is 0 m/s.

The block experiences acceleration due to gravity (g) in the downward direction, represented by a negative sign (-g).

We want to find the final velocity (v_f) of the block just before the collision using the kinematic equation:

v_f = v_o + at

Since v_o = 0 and a = -g, we get:

v_f = -gt

3. Applying Conservation of Momentum:

During the collision, momentum is conserved, meaning the total momentum before the collision equals the total momentum after.

Before the collision, the momentum is the sum of the bullet's momentum (mv_b) and the block's momentum (Mv_f).

After the collision, the combined bullet-block system moves together with a final velocity (V).

We can express the conservation of momentum equation as:

mv_b + Mv_f = (m + M)V

4. Solving for Time (t):

Substitute the expressions for v_f and V from previous steps:

(0.026)(750) + (5)(-gt) = (0.026 + 5)(gt)

Simplify and solve for t:

19.5 - 49t = 49.2548t

98.7548t = 19.5

t = 0.198 seconds

Therefore, the block had been falling for 0.198 seconds before the collision with the bullet.

The bicycle wheel is spinning counterclockwise when viewed from above (so as you view the picture, the right hand side of the wheel is moving away from you). The man and stool are initially at rest. If the man now rotates the wheel’s axis 180˚ so that his right hand is above the wheel, the wheel/man/stool system will:__________

Answers

Answer:

a) rotate counterclockwise as viewed from above.

Explanation:

The text tells us it is spinning counterclockwise.

A shallow groove on the surface of the cortex is called a ________.

Answers

Answer:

sulcus

Explanation:

A sulcus is an indentation or depression in the brain that causes it to look like it  ridges or folds

Cerebral sulci and fissures are grooves between the adjacent gyri on the surface of the cerebral hemispheres.

Sulci can be basically can be divided into three basic function

limiting sulcus: This happens to  develop between areas differing in structure and function, for example central sulcus

axial sulcus: This develops along the axis of a rapidly growing/developing area (e.g. calcarine sulcus)

operculated sulcus: a sulcus may be between two structurally-different areas and a third sulcus may lie in its wall and does not appear on the surface (e.g. lunate sulcus)

A canister is released from a helicopter 500 m above the ground. The canister is designed to withstand an impact speed of up to 100 m/s.
(a) Ignoring air resistance, find an equation of the height at any time t.
(b) Find the impact speed of the canister.
(c) Your answer to (b) should be less than 100 m/s.
Rather than just releasing it, let’s see if we could break it by throwing it down with an initial velocity. Re-do parts (a) and (b) with an initial velocity v0, then find the value of v0 required to break the canister.

Answers

Answer:

(a) h = 500 - 4.9t²

(b) 98.99 m/s

(c) h = 500 - Vot - 4.9t²

    14.14 m/s

Explanation:

(a) You can use the equation of linear motion s = vi.t + 0.5gt²

(b) [tex]v_{f} ^{2} = v_{i} ^{2} +2ax\\v = \sqrt{0^{2} + 2*9.8*500 } = 98.99 m/s[/tex]

(c) [tex]v_{f} ^{2} = v_{i} ^{2} + 2ax\\v_{i} = \sqrt{100^{2} -2*9.8*500}=14.14 m/s[/tex]

The initial velocity mucg be greater than 14.14 m/s to break the cannister

A 5.90-g lead bullet traveling at 410 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in temperature of the bullet?

Answers

Answer:

rise in temperature = 333.53°C

Explanation:

mass of lead bullet, m = 5.9 g

initial velocity, u = 410 m/s

specific heat of lead = 0.126 J/gm - K = 126 J/kg K

Initial kinetic energy of the bullet

K = 1/2 mu^2 = 0.5 x 5.9 x 10^-3 x 410 x 410 = 495.895 J

Half of the kinetic energy is used to raise the temperature of lead bullet

K / 2 = mass of bullet x specific heat of lead x rise in temperature

247.95 = 5.9 x 10^-3 x 126 x rise in temperature

rise in temperature = 333.53°C

Thus, the rise in temperature of lead bullet is 333.53°C.

Two sound waves are moving towards each other. A region of rarefaction overlaps with a region of rarefaction. Which of the following choices correctly describes the behavior of sound waves when they overlap as described?
a. Constructive interference; the amplitude increases but the volume remains the same.
b. Destructive interference; the sound becomes quieter because the amplitude decreases.
c. Constructive interference; the sound becomes louder because the amplitude increases.
d. Destructive interference; the sound becomes louder because the amplitude increases.

Answers

Final answer:

When a region of rarefaction from one sound wave overlaps with a similar region in another wave, they cause constructive interference. The result is an increased amplitude, which makes the sound louder.

Explanation:

When a region of rarefaction of one sound wave overlaps with a region of rarefaction in another sound wave, it results in constructive interference. This is because both the waves are essentially 'in sync', and they amplify each other. The amplitude of the resulting wave increases, and in terms of sound, this means that the sound becomes louder, because amplitude is directly related to the volume or loudness of the sound. Hence, the correct answer is: Constructive interference; the sound becomes louder because the amplitude increases.

For instance, if you're playing two sounds of the same frequency and amplitude together, and their waves align such that their areas of rarefaction meet, the combined sound would be louder than the individual sounds. This is a direct application of the physics concept of wave interference.

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A cubical box 25.0 cm on each side is fully immersed in a fluid. The pressure at the top surface of the box is 109.40 kPa and the pressure on the bottom surface is 112.00 kPa. What is the density of the fluid?

Answers

Answer:

ρ = 1061 kg/m³

Explanation:

The pressure at the top pf the box is clearly different from the pressure at the bottom of of the  box due to the depth. So the problem can be solve the following way:

Difference in pressure =  density * acceleration due to gravity * height

ΔP = ρgΔh

112 000 Pa - 109 400 Pa = ρ * 9.8 m/s² * 0.25 m

                                     ρ = 1061 kg/m³

Final answer:

The density of the fluid can be calculated using the formula P=hρg, where P is the pressure, h is the height or depth, ρ is the density, and g is the acceleration due to gravity. The pressure difference is due to the fluid itself and can be used to find the density when the height and acceleration due to gravity are given.

Explanation:

The student's question is about finding the density of a fluid given the pressure at the top and bottom of the cube. The key to finding this is using the formula that relates pressure, depth, and density in a fluid, which is P=hρg where P is the pressure, h is the height or depth, ρ is the density, and g is the acceleration due to gravity.

Given that the difference in pressure between the bottom and top of the cube is 112.00 kPa - 109.40 kPa = 2.60 kPa, this represents the change in hydrostatic pressure due to the fluid itself. Now, convert this pressure difference from kPa to Pa (since 1 kPa = 1000 Pa), we get 2.60 kPa * 1000 = 2600 Pa.

The height of the cube is given as 25.0 cm which needs to be converted to meters (since 1m = 100 cm), giving us a height of 0.25 m. We can calculate the density of the fluid using the rearranged formula ρ = P/gh. Substituting known values: ρ = 2600 Pa / (9.81 m/s² * 0.25 m). The answer will then give us the density of the fluid.

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In Fig. 10-37, two particles, each with mass m 0.85 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length d ! 5.6 cm and mass M ! 1.2 kg. The combination rotates around the rotation axis with the angular speed v ! 0.30 rad/s. Measured about O, what are the combination’s (a) rotational inertia and (b) kinetic energy

Answers

Answer:

(a). The total moment of inertia of the system is 0.0233 kg-m².

(b). The kinetic energy is 0.0011 J.

Explanation:

Given that,

Mass of each particle m= 0.85 kg

Length = 5.6 cm

Mass of each rod M= 1.2 kg

Angular speed = 0.30 rad/s

The moment of inertia of the rod  between  axis of rotation and mass  is

[tex]I_{1}=\dfrac{Md^2}{3}[/tex]

The moment of inertia of the rod  between masses about center of mass is

[tex]I_{cm}=\dfrac{Md^2}{12}[/tex]

Moment of inertial of the rod between masses about point O is

[tex]I_{2}=M(d+\dfrac{d}{2})^2+\dfrac{Md^2}{12}[/tex]

Moment of inertia of two masses is

[tex]I_{m}=md^2+m(2d)^2[/tex]

(a). We need to calculate the total moment of inertia of the system

Using formula of moment of inertia

[tex]I_{t}=I_{1}+I_{2}+I_{m}[/tex]

Put the value into the formula

[tex]I_{t}=\dfrac{Md^2}{3}+M(d+\dfrac{d}{2})^2+\dfrac{Md^2}{12}+md^2+m(2d)^2[/tex]

[tex]I_{t}=\dfrac{Md^2}{3}+\dfrac{9Md^2}{4}+\dfrac{Md^2}{12}+md^2+4md^2[/tex]

[tex]I_{t}=\dfrac{32Md^2}{12}+5md^2[/tex]

[tex]I_{t}=2.67Md^2+md^2[/tex]

Put the value into the formula

[tex]I_{t}=2.67\times1.2\times(5.6\times10^{-2})^2+5\times0.85\times(5.6\times10^{-2})^2[/tex]

[tex]I_{t}=0.0233\ kg-m^{2}[/tex]

(b). We need to calculate the kinetic energy

Using formula of kinetic energy

[tex]K.E=\dfrac{1}{2}I\omega^2[/tex]

Put the value into the formula

[tex]K.E=\dfrac{1}{2}\times0.0233\times(0.30)^2[/tex]

[tex]K.E=0.0011\ J[/tex]

Hence, (a). The total moment of inertia of the system is 0.0233 kg-m².

(b). The kinetic energy is 0.0011 J.

Final answer:

The rotational inertia of the combination is 0.041 kg·m² and the kinetic energy is 80.93 J.

Explanation:

The rotational inertia (moment of inertia) of the combination can be calculated using the formula I = 2mL², where m is the mass and L is the length of the rod. Substituting the given values, the rotational inertia is 0.041 kg·m².

The kinetic energy of rotation can be calculated using the formula K = ½Iω², where I is the moment of inertia and ω is the angular velocity. Substituting the given values, the kinetic energy is 80.93 J.

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You are asked to construct a mobile with four equal m = 141 kg masses, and three light rods of negligible mass and equal lengths. The rods are of length 55 cm. (a) At what location on the level 1 rod should the free end of rod 2 be attached?

Answers

Answer:

The free end must be attached at a distance of 27.5 cm

Solution:

Mass, m = 141 kg

Length of the rods, L= 55 cm

Now,

As clear from fig. 1:

The free end of the rod 2 must be attached at:

F = 2 W

[tex]WL = W(55 - L)[/tex]

2L = 55

L = 27.5 cm

7. An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where the Earth’s horizontal magnetic field is 5.5×10−5T. What is the maximum possible emf induced in the antenna due to this motion?

Answers

Final answer:

The maximum possible emf induced in the antenna due to the motion of the automobile can be calculated using Faraday's law of electromagnetic induction. Plugging in the given values, the maximum possible emf is calculated as 2.4 x 10^-8 V.

Explanation:

The maximum possible emf induced in the antenna due to the motion of the automobile can be calculated using Faraday's law of electromagnetic induction. The equation for calculating the emf is given by μoθlvB, where μo is the permeability of free space, l is the length of the antenna, v is the velocity of the automobile, and B is the magnetic field strength. Plugging in the given values, we have:

μo = 4π x 10^-7 Tm/A

l = 1.0 m

v = 100.0 km/h = 27.78 m/s

B = 5.5 x 10^-5 T

Using these values, we can calculate the maximum possible emf induced in the antenna as:

emf = (μoθlvB) = (4π x 10^-7 Tm/A)(1.0 m)(27.78 m/s)(5.5 x 10^-5 T) = 2.4 x 10^-8 V

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The maximum possible emf induced in the 1.0 m long antenna moving at 100.0 km/h in a 5.5×10⁻⁵ T magnetic field is approximately 1.53 mV. The calculation involves using the motional emf formula: emf = vBL. Converting the velocity to meters per second and plugging in the values gives the result.

Motional EMF Calculation

The question is about finding the electromotive force (emf) induced in a radio antenna moving through the Earth's magnetic field. This situation involves motional emf, which can be determined using the formula:

emf = vBL

where:

v = velocity of the antenna = 100.0 km/h = 27.78 m/s (converted to meters per second)B = magnetic field strength = 5.5 × 10⁻⁵ TL = length of the antenna = 1.0 m

Now, substituting these values into the formula:

emf = 27.78 m/s × 5.5 × 10⁻⁵ T × 1.0 m

emf ≈ 1.53 × 10⁻³ V

Therefore, the maximum possible induced emf in the antenna due to its motion is approximately 1.53 mV.

HELP ME PLEASE
Question 1 (1 point)

When reporting the speed, we need to include the value, units and direction the object was traveling.

True

False

Question 2 (1 point)

When reporting the velocity of an object, we need to include the value, the units and the direction the object was traveling.

True

False

Question 3 (1 point)

An object travels 10 meters to the right and 7 meters to the left, what distance has it traveled?


17



17 meters



3



3 meters


Question 4 (1 point)

An object travels 5 meters to the north and 7 meters south, what was its total displacement?


2 meters



2 meters south



12 meters



12 meters north


Question 5 (1 point)

An object travels 10 meters to the right before reversing direction and going 13 meters to the left. Jenny reported its displacement as 3 meters, but she was not correct. What did she do wrong?


She reported the distance instead of the displacement.



She forgot to include the units.



She did not give the direction as either left, or negative.



She did not do anything wrong.


Question 6 (1 point)

A car travels 5 miles north and then 2 miles south in 1/4 hour. What was its average speed?


7 mph



3 mph



28 mph



14 mph


Question 7 (1 point)

A car travels 5 miles north and then 2 miles south in 1/4 hour. What was its average velocity?


28 mph



12 mph



12 mph south



12 mph north


Question 8 (1 point)

Mya is running an experiment to determine if a car with large tires will move faster than a car with small tires. Her ramp is 0.75 meters long. She recording the following times on each of her trials: Trial 1: 2.5 seconds, Trial 2: 2.75 seconds, Trial 3: 2.98 seconds. What was the average speed of her car?


2.74 m/s



3.7 m/s



0.27 m/s



2.7


Question 9 (1 point)

Data that is accurate, is data that is really close to the actual value, but data that is precise is data that is repeated over and over again.

True

False

Question 10 (1 point)

Accuracy and precision mean the same thing.

True

False

Answers

1) False

2) True

3) The distance is 17 m

4) The displacement is 2 m south

5) She did not give the direction as either left, or negative.

6) The average speed is 28 mph

7) The average velocity is 12 mph north

8) The average speed is 0.27 m/s

9) False

10) False

Explanation:

1)

Speed is a scalar quantity which tells how fast an object is moving regardless of its direction, and it is calculated as:

[tex]speed=\frac{d}{t}[/tex]

where d is the distance covered by the object and t is the time taken. Being a scalar quantity, speed consists only of a value and its units, so no direction needs to be specified.

2)

Velocity is a vector quantity, defined as

[tex]velocity = \frac{d}{t}[/tex]

where d is the displacement of the object (a vector connecting the initial position to the final position of motion) and t is the time taken. Being a vector, velocity has both a magnitude and a direction (the same direction as the displacement), so direction here should also be specified.

3)

The distance travelled by the object is just the total length of the path taken, regardless of the direction of each part of the motion.

Here the object moves:

10 meters to the right

7 meters to the left

So, the distance travelled is

d = 10 + 7 = 17 m

4)

The displacement is a vector connecting the initial position to the final position of motion, so we have to compare the starting position with the final position.

Taking x = 0 as initial position, and north as positive direction:

- The object moves 5 m north first (+5)

- The object moves 7 m south (-7)

So, the displacement is

d = +5 + (-7) = -2 m

which means 2 meters south.

5)

As we said previously, displacement is a vector connecting the initial position to the final position of motion. Being a vector, it must have:

- A magnitude (the shortest distance between the initial and final position, in a straight line)

- A direction

Here Jenny reported only the magnitude (3 meters), but not the direction, so she forgot to include the direction of the displacement (which is to the left).

6)

The average speed is given by

[tex]speed=\frac{d}{t}[/tex]

where d is the distance and t is the time taken.

The distance is just the total length covered, so:

d = 5 + 2 = 7 miles

The time taken is

t = 1/4 h = 0.25 h

So, the average speed is

[tex]speed=\frac{7}{0.25}=28 mph[/tex]

7)

The average velocity is given by

[tex]velocity=\frac{d}{t}[/tex]

where d is the displacement and t is the time taken.

The displacement is, taking north as positive direction:

d = +5 + (-2) = 3 miles (north)

The time taken is

t = 1/4 h = 0.25 h

So, the average velocity is

[tex]velocity=\frac{3}{0.25}=12 mph[/tex] (north)

8)

We can calculate the average speed by adding the single measurements and dividing by the number of trials done:

[tex]speed_{avg}=\frac{s_1+s_2+s_3}{3}[/tex]

where in this case, N = 3. For this experiment we have:

[tex]s_1 = \frac{0.75 m}{2.5s}=0.30 m/s\\s_2 = \frac{0.75 m}{2.75 s}=0.27 m/s\\s_3=\frac{0.75 m}{2.98 s}=0.25 s[/tex]

So the average is

[tex]speed_{avg}=\frac{0.30+0.27+0.25}{3}=0.27 m/s[/tex]

9)

Data are said to be:

- Accurate, when the average value of the measurements is close to the actual value

- Precise, when the spread of the measurements done in the different trials is small

Therefore, the first part of the sentence "Data that is accurate, is data that is really close to the actual value" is correct, and the second part "data that is precise is data that is repeated over and over again" is not correct, since we may have several measurements but their spread may be large.

10)

As we said in part 9):

- Accuracy refers to how close the measured value is to the actual value

- Precision refers to the spread (or the uncertainty) on the measured value: the smaller it is, the better the precision

For instance, let's assume that the actual value of a certain variable is 3.0. If we get the following set of data:

2.4, 2.5, 2.4, 2.3

it is precise (the spread is small) but not accurate (since the average, 2.4, is far from the actual value)

while the following set:

3.1, 3.6, 2.4, 3.0

is accurate (the average is around 3.0, so close to the actual value), but not precise (the spread is very large).

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Two charges q1 and q2 are separated by a distance d and exert a force F on each other. What is the new force F', if charge 1 is increased to q'1=5q1, charge 2 decreased to q'2=q2/2, and the distance is decreased to d'=d/2?

Answers

Answer:

F'=  10F (N)

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F = K*q₁*q₂ / d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters(m)

Calculating of the new force F'

Data:

q'= 5q

q'₂ = q₂/2

d' = d/2

We apply the Coulomb's law:

F' = K*q'₁*q'₂ / d'²

F'= K*(5q₁)*(q₂/2) / (d/2)²

F'= K*(5q₁*q₂/2) / (d²/4)

F'= K*20q₁*q₂) / (2d²)

F'=  10(K*q₁*q₂) / (d²)

F'=  10(K*q₁*q₂) / (d²)  , F = K*q₁*q₂ / d²

F'=  10F (N)

Final answer:

The force between two charges can be calculated using Coulomb's Law. We can use the given values to calculate the new force F' and determine its value.

Explanation:

The force between two charges is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Using the given information, we can calculate the new force F' using the formula:

F' = (k * q'1 * q'2) / (d'^2)

Substituting the given values into the formula, we get:

F' = (k * 5q1 * (q2/2)) / ((d/2)^2)

After solving the equation, we can determine the value of the new force F'.

Because of the general expansion of space, all distant galaxies appear to be moving away from us, with speeds that increase with distance from our galaxy. An observer in one of these distant galaxies would apparently see ______.

Answers

Answer:all galaxies Moving away from him

Explanation:

When we talk about the expanding universe, it means it has grown with the Big Bang ever since it started.  

Space expansion makes galaxies seem to move apart from each other. Although the galaxies themselves may seem to move through space, in reality, it is the space that is growing between the galaxies.

Thus the observer sees all the galaxies moving away from him, the more distant galaxies faster they move.

An oscillating block-spring system has a mechanical energyof 1.00 J and amplitude of 10.0 cm and a maximum speed of 1.20m/s. Find (a) the spring constant and (b) the mass of theblock, and (c) the frequency of oscillation.

Answers

Answer:

Explanation:

Given

Mechanical Energy of Spring-Block system is 1 J

Maximum Amplitude is [tex]A=10 cm[/tex]

maximum speed [tex]v_{max}=1.2 m/s[/tex]

Suppose [tex]x=A\sin \omega t [/tex]be general equation of motion of spring-mass system

where A=max amplitude

[tex]\omega [/tex]=Natural frequency of oscillation

t=time

[tex]v_{max}=A\omega =1.2[/tex]

[tex]0.1\cdot \omega =1.2[/tex]

[tex]\omega =12 rad/s[/tex]

maximum kinetic Energy must be equal to total Mechanical Energy when spring is un deformed i.e. at starting Position

[tex]\frac{1}{2}mv_{max}^2=1[/tex]

[tex]m=\frac{2}{1.2^2}=\frac{2}{1.44}=1.38 kg[/tex]

Also [tex]\omega [/tex]is also given by

[tex]\omega =\sqrt{\frac{k}{m}}[/tex]

[tex]k=\omega ^2\cdot m[/tex]

where k= spring constant

[tex]k=12^2\cdot 1.38 [/tex]

[tex]k=200 N/m[/tex]

             

The amount of light entering the eye is regulated by the

Answers

Answer:

Iris

Explanation:

The pupil is where the light enters the eye, however, the iris is pupil tissue and is the one who regulates the amount of light that it lets through.

The iris opens or closes to allow a greater or lesser flow of light through the pupil.

In summary, the iris is responsible for regulating the amount of light that enters the eye.

Final answer:

The amount of light entering the eye is mainlly regulated by the pupil, which adjusts its size based on the surrounding light levels. This is achieved by the action of muscles connected to the iris. Furthermore, the light that enters the eye is focused on the retina and processed to the brain via the optic nerve.

Explanation:

The amount of light entering the eye is majorly managed by the pupil. The pupil is the small opening in the center of the eye that adjusts its size based on the light levels in the environment. This adjustment is achieved by the contraction and relaxation of muscles connected to the iris, which is the colored part of the eye.

When the light levels are high, the pupil contracts (or constricts) to limit the amount of light that enters the eye. On the other hand, when the light levels are low, the pupil expands (or dilates) to allow more light to enter the eye. This process is regulated by specific nerves and structures, specifically the optic nerve and the oculomotor nerve in a process called the Pupillary Light Response.

Furthermore, light waves cross the cornea and focus on the retina, a light-sensitive layer lining the back of the eye. The image processing is then carried forward to the brain through the optic nerve.

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You are the Engineering Duty Officer getting your submarine, the USS GREENVILLE, ready to put to sea. When nuclear material in the reactor is fissioning at an increasing rate, this is known as a:__________

Answers

Answer:

Super-critical mass

Explanation:

This term refers to the mass, in which the amount of fission processes per unit of time increases to the point, where some intrinsic feedback mechanism causes the reactor to reach an equilibrium point at a high temperature or power, that is, It becomes critical again, or it is destroyed due to the amount of processes.

"When nuclear material in the reactor is fissioning at an increasing rate, this is known as a ""criticality.""

When nuclear material in the reactor is fissioning at an increasing rate, it is said to be in a state of criticality. This term refers to the condition where the reactor is sustaining a nuclear chain reaction at a steady rate. In a nuclear reactor, criticality is the normal operating condition, where one fission event leads to exactly one subsequent fission event, on average, in the next generation of the reaction. This ensures a controlled and sustained release of energy.

To achieve criticality, the reactor must have enough fissile material (such as uranium-235 or plutonium-239) and the correct geometric arrangement to maintain a chain reaction. Control rods, which are made of materials that absorb neutrons, are used to regulate the rate of the reaction. When the reactor is critical, the rate of neutron production is balanced by the rate of neutron absorption and leakage, resulting in a steady power output.

If the rate of fission increases beyond this balance, the reactor is said to be supercritical, which can lead to an uncontrolled increase in power and is a dangerous condition that must be avoided. Conversely, if the rate of fission decreases, the reactor is said to be subcritical, and the power output will decrease.

In summary, criticality in a nuclear reactor is the state where a self-sustaining chain reaction occurs at a constant rate, providing a stable source of energy for the submarine's propulsion system. It is carefully managed by the reactor's control systems and the vigilance of the Engineering Duty Officer and their team."

Draw a free-body diagram of the rod ab. Assume the contact surface at b is smooth.

Answers

Answer:

See attachment

Explanation:

In a Compton scattering experiment, a beam of x-rays of wavelength 0.07 nm scatters of nearly free electrons. The scattered radiation is investigated as a function of the angle theta between it and the primary beam. The scattered radiation contains__________.
a) a component shifted in wavelength by an amount that decreases as the scattering angle increases.
b) a component shifted in wavelength by an amount that is independent of the scattering angle.
c) a component shifted in wavelength by an amount that is independent of the wavelength of the primary beam.
d) two components with wavelength shifted up and down by an equal amount.
e) no wavelength other than that of the primary beam.

Answers

Answer:

Option (e).

Explanation:

The Compton's equation is:

[tex] \lambda^{'} = \frac{h}{m_{e} c} \cdot (1 - Cos(\theta)) + \lambda = \lambda_{c} (1 - Cos(\theta)) + \lambda [/tex] (1)

where λ': is the wavelength scattered, λ: is the initial wavelength, h: is Planck's  constant, [tex] m_{e} [/tex]: is the electron rest mass, c: speed of light, Θ: scattering angle (between λ and λ'), and [tex] \lambda_{c} [/tex]: is a constant known as the Compton wavelength of the electron = 0.00243 nm.          

From equation (1), the scattered radiation is directly proportional to the scattering angle, having the maximum value when Θ=90°:

[tex] \theta = 90 ^{ \circ} \rightarrow \lambda^{'} = \lambda_{c} (1-Cos(90)) + \lambda = \lambda_{c} + \lambda [/tex]

And the minimum value when Θ=0°:

[tex] \theta = 0 ^{ \circ} \rightarrow \lambda^{'} = \lambda_{c} (1-Cos(0)) + \lambda = \lambda [/tex]

Hence, the options (a) and (b) are incorrect.

Similarly, we can see from equation (1) that the scattered radiation depends also on the wavelength of the primary beam and no wavelength other than that (since [tex] \lambda_{c} [/tex] is a constant), so the correct option is (e).  

Have a nice day!

The neurons that select a particular motor program are the __________.

Answers

Answer:

upper motor neurons in the premotor cortex

Explanation:

Motor neurons are a type of nervous system cells that are located in the brain  and in the spinal cord. They have the function of producing the stimuli that cause the contraction of the different muscle groups of the organism. They are therefore essential for daily activities that require muscle contraction: walking, talking, moving hands and in general all body movements.

Photon A has twice the energy of photon B.
Is the momentum of A less than, equal to, or greater than that of B?
Less Equal Greater Is the wavelength of A less than, equal to, or greater than that of B?
Less
Greater
Equal

Answers

Answer:

Explanation:

Given

Photon A has  twice the Energy of Photon B

i.e. [tex]E_a=2E_b[/tex]

de Broglie wavelength is given by

[tex]\lambda =\frac{h}{P}[/tex]

[tex]\frac{1}{\lmabda }=\frac{P}{h}[/tex]

and Energy of Proton is given by

[tex]E=\frac{hc}{\lambda }[/tex]

[tex]E=hc\times \frac{P}{h}[/tex]

[tex]E=Pc[/tex]

where P=momentum

c=velocity of Light

[tex]P=\frac{E}{c} [/tex]

Momentum of A [tex]P_a=\frac{E_a}{c}=\frac{2E_b}{c}[/tex]

Momentum of B is [tex]P_b=\frac{E_b}{c}[/tex]

Thus [tex]P_a>P_b[/tex]

For wavelength

[tex]\lambda _a=\frac{h}{P_a}[/tex]

[tex]\lambda _b=\frac{h}{P_b}[/tex]

since [tex]P_a>P_b[/tex] therefore

[tex]\lambda _a<\lambda _b[/tex]      

Final answer:

The momentum of photon A is greater than that of photon B, and the wavelength of A is less than that of B.

Explanation:

When comparing the momentum of two photons, we can use the equation p = E/c, where p is momentum, E is energy, and c is the speed of light. Since photon A has twice the energy of photon B, its momentum will also be greater. Therefore, the momentum of A is greater than that of B.

As for the wavelength, we know that the equation c = λf relates the speed of light (c) to the wavelength (λ) and frequency (f). Since the speed of light is constant, if the energy of photon A is greater, its frequency must be higher as well. As a result, the wavelength of A must be shorter than that of B. Therefore, the wavelength of A is less than that of B.

Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz.

Answers

Final answer:

The question is about the angles of maxima and minima in a double-slit interference pattern for two identical loudspeakers placed 1.00 m apart.

Explanation:

The distance between two identical loudspeakers placed 1.00 m apart is given as 'd'.

In the given situation, the listener stands 4.00 m from the wall directly in front of one of the speakers. This distance can be considered as 'L'.

The formula to calculate the distance between adjacent maxima and minima in a double-slit interference pattern is:

d*sin(theta) = m*lambda

- Where 'd' is the distance between the speakers
- 'theta' is the angle at which the maxima or minima occur
- 'm' is the order of the maxima or minima
- 'lambda' is the wavelength of the sound

Applying this formula, we can calculate the angles at which the maxima and minima occur for the given situation.

The phase difference between the waves reaching the observer is approximately 0.66 radians. The frequency closest to 300 Hz for minimal sound due to destructive interference is approximately 286 Hz.

To solve the problem, we need to find the phase difference and the frequency closest to 300 Hz for minimal sound.

a.) The speed of sound, v, is typically about 343 m/s in air. Using the formula for wavelength, λ = v/f, we get:

λ = 343 m/s / 300 Hz λ = 1.143 m

For the listener standing 4.00 m from one speaker and 4.12 m from the second speaker, the difference in the path length, ΔL, is:

ΔL = sqrt((4.00 m)² + (1.00 m)²) - 4.00 m ΔL = 4.12 m - 4.00 m = 0.12 m

The phase difference, φ, is given by:

φ = (2π/λ) * ΔL φ = (2π/1.143 m) * 0.12 m φ ≈ 0.66 radians

b.) For destructive interference, the path difference should be half-integer multiples of the wavelength:

(2n + 1)λ/2 = ΔL

Simplifying, we find the frequency:

0.12 m = (2n + 1) * (343 m/s) / (2f)

Solving for f when n = 0 (closest to 300 Hz):

f = 343 m/s / (2 * 0.12 m) f ≈ 286 Hz

Thus, the frequency closest to 300 Hz for minimal sound is approximately 286 Hz.

complete question:

Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz.

(a) What is the phase difference in radians between the waves from the speakers when they reach the observer? (Your answer should be between 0 and 2?.)

rad

(b) What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?in Hz

A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor at 25◦C. To achieve the desired amount, we cool the tank while we fill it slowly, keeping the tank and its content at 30◦C. Find the final pressure to reach before closing the valve and the heat transfer.

Answers

Answer:

592.92 x 10³ Pa

Explanation:

Mole of ammonia required = 10 g / 17 =0 .588 moles

We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .

From the relation

PV = nRT

P x 2.5 x 10⁻³ =  .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

Why is Saturn almost as big as Jupiter, despite its smaller mass?
a. Saturn's rings make the planet look bigger. Jupiter's strong magnetic field constrains its size.
b. Saturn has a larger proportion of hydrogen and helium than Jupiter, and is therefore less dense.
c. Jupiter's greater mass compresses it more, thus increasing its density.
d. Saturn is further from the Sun, thus cooler, and therefore less compact.

Answers

Answer:

c. Jupiter's greater mass compresses it more, thus increasing its density.

Explanation:

The mass of Jupiter is greater in its interior, this mass compresses Jupiter to some extent. Thus, its density is increased. Now, more mass is compressed in the lesser volume. Hence, its size does not increase enormously. On the other hand the mass of Saturn is lesser  and also density lower. this gives Saturn a reasonably higher volume.

Hence, option C is correct.

Saturn appears almost as large as Jupiter due to its lower mass leading to less gravitational compression of its hydrogen and helium composition, resulting in a larger volume for its mass. The correct answer from the provided choices is c. Jupiter's greater mass compresses it more, thus increasing its density.

The reason why Saturn appears almost as large as Jupiter despite having a smaller mass is largely due to its composition and structure. We see that Jupiter, which is 318 times more massive than Earth, has a significantly higher density because its greater mass compresses the planets' internal hydrogen and helium more than Saturn, which is about 25% as massive as Jupiter. In comparison, Saturn's lower mass results in less compression and hence a larger volume for its mass. So, the correct answer is c. Jupiter's greater mass compresses it more, thus increasing its density.

Jupiter and Saturn are both composed mainly of hydrogen and helium, which become liquid at great depths due to their large size and the resulting high pressure. However, Saturn, with its lower density, is the least dense planet in the solar system, even less dense than water. Contrary to the choices provided, Saturn's rings do not play a significant role in making the planet appear larger in terms of its physical size, and the effect of distance from the Sun on the compactness of the planets is negligible compared to the impact of their massive gravitational compression.

The sides of a composite-cone volcano may be interrupted with which feature?

Answers

Answer:

fumaroles or smokers

Explanation:

Fumaroles: they are gaseous emissions from lavas in craters at more or less elevated temperatures. Its composition varies according to the temperature of the lavas, in such a way that it changes since the fumaroles appear until their extinction.

Water flows in the form of steam continuously through the cracks of a volcano or volcanic surface because its temperature exceeds 100 °C.

Groups of fumaroles

- Dry fumaroles: They are those that come off the melting lava, in the vicinity of the crater. Its temperature is higher than 500oC. They are mainly composed of sodium, potassium and sulfur dioxide and carbon dioxide.

- Acid fumaroles: with temperatures between 300oC and 400oC, they are constituted by a large amount of water vapor, with hydrochloric acid and sulphurous anhydride.

- Alkaline fumaroles: Temperature close to 100oC, contain water vapor with hydrogen sulfide and ammonium chloride.

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