A truck moves 60 km West, and then 80 km North, and then
travels in a straight line back to its starting point. The distance
travelled by the truck is ____km and its displacement is _____km

Hi there!

We can begin by using the work-energy theorem in regards to an oscillating spring system.

Total Mechanical Energy = Kinetic Energy + Potential Energy

For a spring:

[tex]\text{Total ME} = \frac{1}{2}kA^2\\\\\text{KE} = \frac{1}{2}mv^2\\\\PE = \frac{1}{2}kx^2[/tex]

A = amplitude (m)

k = Spring constant (N/m)

x = displacement from equilibrium (m)

m = mass (kg)

We aren't given the mass, so we can solve for kinetic energy by rearranging the equation:

ME = KE + PE

ME - PE = KE

Thus:

[tex]KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2\\\\[/tex]

Plug in the given values:

[tex]KE = \frac{1}{2}(20)(0.3^2) - \frac{1}{2}(20)(0.3^2) = \boxed{0 \text{ J}}[/tex]

We can also justify this because when the mass is at the amplitude, the acceleration is at its maximum, but its instantaneous velocity is 0 m/s.

Thus, the object would have no kinetic energy since KE = 1/2mv².

Answer:

4600 Kg/m³

Explanation:

Volume of block=5m³

Mass of block= 920 kg/m³

Density=mass × volume

=920 × 5³

=4600 /m³

The density of ice is 4600 Kg/m³

___________________________________

(Hope this helps can I pls have brainlist (crown)☺️)

Explanation:

Solution:

Here

volume=5

Density=920

Density =Mass/Volume

or,Mass=Density*Volume

or,M=920*5

so,M=4600kg

Answer:

A

Explanation:

L=1m

=2mm²=(2/1000²)=2(10^-6)m

y=4x10¹¹N/m²

∆l=2mm=2/10.00=0.002m

=(4x10¹¹x2x10^-6x2x10^-3)÷1m

=16x10¹¹-⁶-³

=16x10¹¹-⁹

=16x10²

=1600N

where a=cross sectional area=2x10^-6m

C=2mm= 2x10^-3m

L=1m

hope it helps

Answer:

No

Explanation:

Changing momentum of any kind requires work. Work is a force acting over a distance. While holding the weights at arms length and spinning will create a force (centripetal), there is no radial distance change incurred. Releasing the weights will reduce the force to zero, still no work done and no change in angular momentum.

If he was holding the weights at arms length while spinning and he pull his hands to his chest, there now exists both the centripetal force and a distance in the direction of that force (inward radial) this work will result in an increase in angular velocity as moment of inertia has decreased with the work done.

No, the skater doesn't succeed in changing his angular velocity.

The final angular velocity of the skater is determined by applying the principle of conservation of angular momentum as shown below;

Li = Lf

[tex]Ii\omega _i = I_f \omega _f[/tex]

where;

When the skater holds the weight, the momnet of inertia of both arms is the same. Also when the skater drops the weight, the moment of inertia of both arms is still the same. Thus, at any instant, the moment of inertia of the two arms is the same.

To change the angular speed, the initial and final moment of inertia of the two arms must be different. Thus, the skater doesn't succeed in changing his angular velocity.

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Answer:

a battery, wires, and a switch.

Explanation:

All circuits include?

The magnitude of the load L on the crane before the collapse is 3211.81 N

To determine the magnitude of the load on the crane (L), we will need to make use of the equilibrium conditions of the torque.

It is always an ideal process to list out all the parameters given as this will let you understand how you can determine the answer to the question from the given parameters.

From the given information;

From the question;

the angle at which the crane is positioned can be determined by taking the tangent of the angle θ. i.e.

[tex]\mathbf{tan \ \theta = \dfrac{h}{d-s}}[/tex]

[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{h}{d-s} \Big )}[/tex]

[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{2.070 }{5.580 - 0.522} \Big )}[/tex]

[tex]\mathbf{\theta =22.26^0}[/tex]

Consider the equilibrium conditions of the torques with respect to the magnitude of the load at point P.

∴

[tex]\mathbf{Tsin \theta (d-s) - W_L (d-x) -(Mg) (\dfrac{d}{2}) = 0}[/tex]

By making the magnitude of the load [tex]\mathbf{W_L}[/tex] the subject of the formula, we have:

[tex]\mathbf{W_L = \dfrac{Tsin \theta (d-x) -(Mg) (\dfrac{d}{2})}{ (d-s) } }[/tex]

[tex]\mathbf{W_L = \dfrac{(11650 )sin (22.26) (5.580-1.350) -(88.50\times 9.81) (\dfrac{5.580}{2})}{ (5.580-0.522) } }[/tex]

[tex]\mathbf{W_L = 3211.81 \ N }[/tex]

Therefore, we can conclude that the magnitude of the load is 3211.81 N

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Explanation:

F = Icurrent×length×Bfieldstrength×sin(angle field to wire)

in our case

Icurrent = 10 A

length = 0.02km = 20 meters

B = 10^-6 T

angle = 30 degrees.

F = (20 A)(20m)(10^-6 T)×sin(30) = 400× 10^‐6 ×0.5 N =

= 200 × 10^-6 = 2 × 10^‐4 N

Answer:

I don't know he he.

just joking

Answer:

Force = mass × acceleration.

Hi there!

We can use the following equation to find the frequency of each harmonic:

[tex]f_n = \frac{n}{2L} \sqrt{\frac{T}{\lambda}}[/tex]

n = nth harmonic

L = length of string (m)

T = Tension of string (N)

λ = linear density (kg/m)

Begin by converting the linear mass density to kg:

2.00g /m · 1 kg / 1000g = 0.002 kg/m

Now, we can use the equation to find the first three harmonics.

First harmonic:

[tex]f_1 = \frac{1}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{131.76 Hz}[/tex]

Second harmonic:

[tex]f_2 = \frac{2}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{263.52Hz}[/tex]

Third harmonic:

[tex]f_3 = \frac{3}{2(0.6)} \sqrt{\frac{50}{0.002}} = \boxed{395.28Hz}[/tex]

Answer:

[tex]k.e. = \frac{1}{2}mv {}^{2} \\ 800= \frac{1}{2} \times 40 \times v {}^{2} \\ 800 = 20v {}^{2} \\ v {}^{2} = 400 \\ v {}^{2} = 20ms {}^{ - 1} [/tex]

k.e. = kinetic energy

hope helpful <3

Answer:

Explanation:

F = ma

a = F/m

a = mBg / (mB + mA)

a = 3.5(9.8)/(3.5 + 3.4)

a = 4.971014...

a = 5.0 m/s²

If you want to use individual Free Body Diagrams

mass A will have downward weight and upward normal forces equal at mAg

and a horizontal force of string tension T

F = ma

T = mAa

mass B will have a downward force of mBg and an upward force of T

mBg - T = mBa

substitute for T

mBg - mAa = mBa

mBg = a(mB + mA)

a = mBg / (mB + mA)   which is identical to the above answer.

Answer:

The 7 states of matter are solid, loquid, gas, fermionoc condensate, quark gluton plasm, bose einetein condensate amd ionised plasm but its usually only 3 they teach you

Answer:

7

Explanation:

solid, liquid,gas,fermionoc condensate,quark glutton plasm,bose einetein condensate amd ionised plasm.

The rolling disk's initial angular momentum is mR√[2(gR + v²)]/2

Using the law of conservation of energy, the initial mechanical energy E of the disk equals its final mechanical energy E' as it climbs the step.

So, E = E'

1/2Iω + 1/2mv² + mgh = 1/2Iω' + 1/2mv'² + mgh'

where I = rotational inertia of disk = 1/2mR² where m = mass of disk and R = radius of disk, ω = initial angular speed of disk, v = initial velocity of disk, h = initial height of disk = 0 m, ω' = final angular speed of disk = 0 rad/s (assumung it stops at the top of the step), v' = final velocity of disk = 0 m/s (assumung it stops at the top of the step), and h' = final height of disk = R/2.

Substituting the values of the variables into the equation, we have

1/2Iω² + 1/2mv² + mgh = 1/2Iω'² + 1/2mv'² + mgh'

1/2(1/2mR² )ω² + 1/2mv² + mg(0) = 1/2I(0)² + 1/2m(0)² + mgR/2

mR²ω²/4 + 1/2mv² + 0 = 0 + 0 + mgR/2

mR²ω²/4 + 1/2mv² = mgR/2

R²ω²/4 = gR/2 + 1/2v²

R²ω²/4 = (gR + v²)/2

ω² = 2(gR + v²)/R²

ω² = √[2(gR + v²)/R²]

ω = √[2(gR + v²)]/R

Since angular momentum L = Iω, the rolling disk's initial angular momentum is

L = 1/2mR² ×√[2(gR + v²)]/R

L = mR√[2(gR + v²)]/2

the rolling disk's initial angular momentum is mR√[2(gR + v²)]/2

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Answer:

340

Explanation:

Sorry I don't know how to do this one yet, I just found the answer in a textbook.

The angle that vector a makes with the +x axis is closest to 23.

The direction of a vector is represented tangent of angle equal to the ratio of the y component and the x component of the vector quantity.

tangent of angle = y/x

angle = tan⁻¹ (-2.3/5.3)

angle = 23.46°

Thus, the angle that vector makes with +x is 23.

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Answer:

200 kilocalories

Explanation:

Answer:

78

Explanation:

x=xi+vi(t)+1/2a(t)^2

x=10+16(4)+1/2(0.50)(4)^2

x=74+4

x=78 m

Answer:

626m

Explanation:

Answer:

In 1924 Louis de Broglie introduced the idea that particles, such as electrons, could be described not only as particles but also as waves. This was substantiated by the way streams of electrons were reflected against crystals and spread through thin metal foils.

Explanation:

I know I probably didn't answer your question, I just used all of my knowledge that I learned about Louie De Broglie. Hope it helps!

Answer:

B. . by consuming plants or other animals

The units of the constant of proportionality between pressure and volume in alphabetical order are

1. Celsius (°C)

2. Fahrenheit (°F)

3. Kelvin (K)

The units for R, that is, the ideal gas constant are

1. J K⁻¹ mol⁻¹

2. L atm K⁻¹ mol⁻¹

We will start by completing the Boyle's Law stated

Boyle's Law states the volume and pressure of a gas are inversely proportional, provided that the temperature remains constant.

This means temperature is the constant of proportionality.

Now, we will name the three units of the constant of proportionality, that is, temperature. The units are

1. Degree Celsius (°C)

2. Degree Fahrenheit (°F)

3. Kelvin (K)

2. In the ideal gas equation (PV/nT=R), R represents the ideal gas constant.

The units for R, that is, the ideal gas constant are

1. J K⁻¹ mol⁻¹

2. L atm K⁻¹ mol⁻¹

Hence,

The units of the constant of proportionality between pressure and volume in alphabetical order are

1. Celsius (°C)

2. Fahrenheit (°F)

3. Kelvin (K)

The units for R, that is, the ideal gas constant are

1. J K⁻¹ mol⁻¹

2. L atm K⁻¹ mol⁻¹

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Answer:

heating prosedure takes place the opposite of condensation

Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant, which has the same value throughout our universe.

D is the distance between both objects.

so, now some numbers change

Fgravitynew = G*((1/3)*mass1*2*mass2)/(2D)² =

= G*((2/3)*mass1*mass2)/(4D²) =

= (2/3)* (G*(mass1*mass2)/D²) / 4 =

= ((2/3)/4) * G*(mass1*mass2)/D² =

= (2/12) * Fgravity = Fgravity/6

the new gravitational force will be 16/6 = 8/3 units.

Hello!

To solve, we can begin by using the kinematic equation:

[tex]d = v_it + \frac{1}{2}at^2[/tex]

Where:

vi = initial velocity (m/s)

t = time (s)

a = acceleration (in this case, due to gravity. g = 9.8 m/s²)

Since the object falls from rest, the initial velocity is 0 m/s.

[tex]d = \frac{1}{2}at^2[/tex]

Plug in the given values:

[tex]d = \frac{1}{2}(9.8)(3.5^2) = \boxed{60.025 m}[/tex]

Answer:

batteries in parallel connection and bulbs in serial connection

To get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.

When two-terminal components and electrical networks that can be connected in series or parallel. This will result in two terminals in the electrical network, and may themselves participate in a series or parallel topology. When a two-terminal "object" is an electrical component or electrical network is a matter of perspective.

A circuit is said to be in series when the same current flows through all the components in the circuit where the current has only one path. A circuit is said to be parallel when there are multiple paths for the electric current to flow through it where the components which are part of the parallel circuit will have a constant voltage across all their ends.

Thus, to get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.

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Answer:

C number is write i think

Explanation:

The buoyant force [tex]F_B[/tex] is defined as

[tex]F_B = \rho_wgV[/tex]

where [tex]\rho_w[/tex] is the density of the displaced fluid (freshwater), g is the acceleration due to gravity and V is the volume of the submerged object. In the case of freshwater, its density is [tex]997\:\text{kg/m}^3.[/tex] Since the buoyant force is 20 N, we can solve for the volume of the displaced fluid:

[tex]F_B = \rho_wgV \Rightarrow V = \dfrac{F_B}{\rho_wg}[/tex]

Plugging in the values, we get

[tex]V = \dfrac{20\:\text{N}}{(997\:\text{kg/m}^3)(9.8\:\text{m/s}^2)}[/tex]

[tex]\:\:\:\:\:= 2.05×10^{-3}\:\text{m}^3[/tex]

Recall that the weight of an object in terms of its density and volume is given by

[tex]W = \rho gV[/tex]

Using the value for the volume above, we can solve for the density of the object as follows:

[tex]\rho = \dfrac{W}{gV} = \dfrac{900\:\text{N}}{(9.8\:\text{m/s}^2)(2.05×10^{-3}\:\text{m}^3)}[/tex]

[tex]\:\:\:\:\:= 44,798\:\text{kg/m}^3[/tex]