Some students conduct an experiment to prove conservation of momentum. They use two objects that collide Measurements
are taken before and after the collision.

Which two quantities will the students multiply together before and after the collision?


A. mass and velocity

B. distance and time

C. mass and acceleration

D. velocity and time

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:

a battery, wires, and a switch.

Explanation:

All circuits include?

Answer:

heating prosedure takes place the opposite of condensation

Answer:

20 m/s at -35°

Explanation:

Ignoring air resistance, the initial vertical velocity will be reversed and the initial horizontal velocity will remain constant.

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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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

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

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.

Answer:

200 kilocalories

Explanation:

Answer:

Explanation:

1 Object C has an acceleration that is greater than the acceleration for D.

2 B

3 17M

4 The velocity is zero.

5 a straight line with negative slope

just took it

20 seconds

Explanation:

Let [tex]x_a[/tex] be the distance traveled by the accelerating car and [tex]x_c[/tex] be the distance traveled by the car moving with a constant velocity. When they cover the same distance, we can write

[tex]x_a = x_c \Rightarrow v_{0a}t + \frac{1}{2}at^2 = v_ct[/tex]

where [tex]v_c[/tex] is the velocity of car moving at a constant rate and a is the acceleration of the accelerating car. Since the accelerating car started from rest, then [tex]v_{0a}[/tex] is zero so our equation above simplifies to

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

Note that the variable t cancels out so solving for t, we get

[tex]\frac{1}{2}at = v_c \Rightarrow t = \dfrac{2v_c}{a}[/tex]

Plugging in the given values,

[tex]t = \dfrac{2(20\:\text{m/s})}{2\:\text{m/s}^2} = 20\:\text{s}[/tex]

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:

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.

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:

I don't know he he.

just joking

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]

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:

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

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

A.

The energy of the hot water is 482630400 J

Using Q = mcΔT where Q = energy of hot water, m = mass of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 189 L = 0.189 m³,

c = specific heat capacity of water = 4200 J/kg-°C and ΔT = temperature change of water = T₂ - T₁ where T₂ = final temperature of water = 608 °C. If we assume the water was initially at 0°C, T₁ = 0 °C. So, the temperature change ΔT = 608 °C - 0 °C = 608 °C

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

Q = mcΔT

Q = ρVcΔT

Q = 1000 kg/m³ × 0.189 m³ × 4200 J/kg-°C × 608 °C

Q = 482630400 J

So, the energy of the hot water is 482630400 J

B.

The elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.

Using the equation for gravitational potential energy ΔU = mgΔh where m = mass of object = 1000 kg, g = acceleration due to gravity = 9.8 m/s² and Δh = h - h' where h = required elevation and h' = zero level elevation = 0 m

Since the energy of the mass equal the energy of the hot water, ΔU = 482630400 J

So, ΔU = mgΔh

ΔU = mg(h - h')

making h subject of the formula, we have

h = h' + ΔU/mg

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

h = h' + ΔU/mg

h = 0 m + 482630400 J/(1000 kg × 9.8 m/s²)

h = 0 m + 482630400 J/(9800 kgm/s²)

h = 0 m + 49248 m

h = 49248 m

So, the elevation the mass would have to be raised from zero elevation relative to the reference environment for its exergy to equal that of the hot water is 49248 m.

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Answer:

None of these elements.

The magnitude of the total initial velocity of the rocket is determined as 9.82 m/s.

The magnitude of the total initial velocity of the rocket is calculated as follows;

V = D/T

where;

V = (54)/(5.5)

V = 9.82 m/s

Thus, the magnitude of the total initial velocity of the rocket is determined as 9.82 m/s.

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Answer:

Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.

Explanation:

Answer:

C number is write i think

Explanation:

These prefixes are very commonly used in naming chemical compounds.

Di- means two.

For example, carbon dioxide's formula is be written as [tex]\text{CO}_2,[/tex] and it has 2 oxygen atoms, hence "di-oxide."

Tetra - means four.

For example, carbon tetrachloride's chemical formula is written as [tex]\text{CCl}_4[/tex], and there are four chlorine atoms

Deca- means ten

For example, lanthanum decahydride's chemical formula is written as [tex]\text{LaH}_{10}.[/tex] In this case there are 10 hydrogen atoms for every lanthanum atom.

Hepta - means seven

For example, iodine heptafluoride is written as [tex]\text{IF}_7[/tex]. Note the seven fluorine atoms attached to the iodine atom, hence the name "hepta-fluoride."

Answer:

626m

Explanation:

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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