given two vector
p= 2i + 2j + 4k
q = i - 4j + 4k
find p+ q​

Answer:

in general context yes it is closest to 1.04

Explanation:

theres no right or wrong way to scientifically prove this though.

Overall in scale its closest to 1.04 hope that helped

Hi there!

[tex]\large\boxed{E_{total} = 8.0 \text{ J}}[/tex]

For a mass undergoing SHM, the total energy of the system is given as:

[tex]ME = \frac{1}{2}kA^2[/tex]

Where:

k = Spring constant (N/m)

A = amplitude (m)

There is no quantity of mass in the equation, so the total mechanical energy of the system is NOT impacted by the object's mass.

Thus, the energy of the system will still be 8.0 J.

Answer:

Explanation:

As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.

Ff = Fcosθ

if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.

μN = Fcosθ

μ(mg - Fsinθ) = Fcosθ

μmg = Fcosθ + μFsinθ

100μ = F(cos30 + μsin30)

F = 100μ / (cos30 + ½μ)

Ff = 100μcos30 / (cos30 + ½μ)

A 100 N create is being pulled at a constant velocity by a rope a 30 degrees to the horizontal as depicted in the diagram given in question the force of friction Ff = 100μcos30 / (cos30 + ½μ).

A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.

Ff = Fcosθ

if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.

μN = Fcosθ

μ(mg - Fsinθ) = Fcosθ

μmg = Fcosθ + μFsinθ

100μ = F(cos30 + μsin30)

F = 100μ / (cos30 + ½μ)

Ff = 100μcos30 / (cos30 + ½μ)

the force of friction Ff, is 100μcos30 / (cos30 + ½μ).

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Answer:

The acceleration of the car is 8.10 m/s²

Explanation:

d = 110 m

t = 5.21 s

vi = 0 m/s

a = ?

d = Vᵢ × t + 0.5 × a × t²

110 m = (0 m/s) × (5.21 s) + 0.5 × (a) × (5.21 s)²

110 m = (13.57 s²) × a

a = (110 m)/(13.57 s²)

a = 8.10 m/s²

Thus, The acceleration of the car is 8.10 m/s²

-TheUnknownScientist 72

Answer:

D1 FG 12 15×AG+5T×G7+3F

Answer:

physics that deals with the mechanical action or relations of heat.

a.

The components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

The force on a current carrying conductor in a magnetic field is given by F = iL × B where i = current = 9.00 A, L = 25.0 cmk = 0.25 mk (since the conductor is along the z-direction). B = magnetic field. Since B has component Bx = -0.242T, By= -0.985, and Bz = -0.336, B = -0.242i + (-0.985j) + (-0.336)k = -0.242i - 0.985j - 0.336)k.

So, F = iL × B

F = 9.00 A{(0.25 m)k × [-0.242Ti + (-0.985Tj) + (-0.336T)k]T}

F = 9.00 A{(0.25 m)k × (-0.242T)i + (0.25 m)k × (-0.985Tj) + (0.25 m)k × (-0.336T)k]}

F = 9.00 A{-0.0605mT)k × i + (-0.24625 mT)k × j + (-0.084 m)k × k]}

F = 9.00 A{-0.0605mT)j + (-0.24625 mT) × -i + (-0.084 mT) × 0]}

F = 9.00 A{-0.0605mT)j + (0.24625 mT)i + 0 mT]}

F = -0.5445 AmT)j + (2.21625 AmT)i + 0 AmT]}

F = -0.5445j + 2.21625i + 0 k

F = (2.2163i - 0.5445j + 0 k) N

So, the components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

b.

The magnitude of the net force on the wire is 2.282 N

The net force F = √(Fx² + Fy² + Fz²)

F = √[(2.2163 N)² + (-0.5445 N)² + (0 N)²)

F = √[(4.912 N)² + 0.2964 N)² + (0 N)²)

F = √[5.2084 N)²

F = 2.2822 N

F ≅ 2.282 N

So, the magnitude of the net force on the wire is 2.282 N

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[tex]\omega = 21.8\:\text{rad/s}[/tex]

Explanation:

We know that there are [tex]2\pi[/tex] radians in one revolution and 60 seconds in one minute so we can easily convert the rev/min unit to rad/s using the following conversion factors:

[tex]208\:\dfrac{\text{rev}}{\text{min}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}}×\dfrac{1\:\text{min}}{60\:\text{s}}[/tex]

[tex]\;\;\;\;\;=21.8\:\text{rad/s}[/tex]

Answer:

Explanation:

An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.

Answer:

Quiet, is the adjective in this sentence

Explanation:

She had a quiet voice. This statement uses the adjective quiet. Her voice can be characterized as quiet.

What is adjectives?

A word used to characterize a person, place, or thing is called an adjective. Adjectives are used to modify nouns or pronouns.

There are many adjectives that describe quiet, including muted, peaceful, reticent, silent, soft, gentle, mild, placid, private, secluded, sedate, serene, timid, smooth, calm, tranquil, modest, restrained, sober, and subdued.

Hence, the significance of the adjectives is aforementioned.

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Answer:

Explanation:

a) The spring force will equal the weight.

b) If up is positive

kx - mg = 0

mg = kx              kx = 25 N

c) m = kx/g = 25/10 = 2.5 kg

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

Answer:

Light can travel in three ways from a source to another location: (1) directly from the source through empty space; (2) through various media; (3) after being reflected from a mirror.

Explanation:

Hi there!

Part A:

The only quantities explicitly given to us are:

3. mass (m)

4. Maximum velocity (vmax)

5. Amplitude (A)

8. Position x from equilibrium

Part B:

To solve, we must begin by calculating the force constant, 'k'.

We can use the following relationship:

[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]

We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:

[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]

[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]

Now, we can find the velocity at 4.7cm (0.047m) using the equation:

[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]

Plug this value into the equation for kinetic energy:

[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]

Part C:

The potential energy of a spring is given as:

[tex]U = \frac{1}{2}kx^2[/tex]

Find 'k' using the derived quantity above:

[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]

Now, calculate potential energy:

[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]

Answer:

circuit

Explanation:

Answer:

The objects outside the reference frame aren't moving. It appears this way since the vehicle you are inside is moving, but unless the objects are people, animals, or other vehicles, the objects aren't moving.

Answer:

jsbdhdndmlsusgsbkaksudgnslsosufhbf ffb

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.

The gravitational force acting between the two bodies is given by:

F=G

r

2

m

1

m

2

G=

m

1

m

2

Fr

2

The dimension of the force is [MLT

−2

]

=

[M][M]

[MLT

−2

][L

2

]

=M

−1

L

3

T

−2

Answer:

how long does it take? we need it to answer ure question

Explanation:

cause we don't know how many feet until we know how long it was falling

If a  stone is dropped from the edge of a roof and hits the ground with a velocity of -150 feet per second, then the height of the roof would have been 1148 feet.

There are three equations of motion given by  Newton

v = u + at

S = ut + 1/2 × a × t²

v² - u² = 2 × a × s

As given in the problem, a penny is dropped from a building that is 95 m tall, the initial velocity of the penny is zero, and the acceleration acting is due to the acceleration due to gravity,

By using the second  equation of the motion for the vertical motion ,

v²  = ( 2 × g ×h )

150²  = 2 × 9.8 × h

h = 22500 / 19.6

  = 1148 feet

Thus, the height of the roof would have been 1148 feet.

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The expression for the speed of waves in materials allows us to find the result for the speed of sound in aluminum is:

The speed of a wave in a material is determined by the relationship between its volumetric modulus and its density, it is given by the expression.

               v = [tex]\sqrt{ \frac{B}{\rho} }[/tex]  

where v is the speed of the wave in the material (sound), B is the volume modulus and ρ the density.

They indicate that the volumetric or elastic modulus of aluminum is;

            B = 6.89 10⁴ Mpa ( [tex]\frac{10^6 \ Pa}{1 \ MPa}[/tex] ) = 6.89 10¹⁰ Pa

The density of aluminum is tabulated  ρ = 2.7 10³ kg / m³

We calculate.

              v = [tex]\sqrt{ \frac{6.89 \ 10^{10} }{2.7 \ 10^3 }}[/tex]  

              v = 5.05 10³ m / s

In conclusion using the expression for the speed of waves in materials we can find the result for the speed of sound in aluminum is:

     v = 5050 m / s

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Answer:

heat

Explanation:

Answer:

200 kilocalories

Explanation:

Answer:

a

Explanation:

sana po makatulong <3♡♡

Answer:

Explanation:

conservation of momentum during the collision

0.035(214) + 0.15(0) = 0.185v

v = 40.486 m/s

The kinetic energy after impact will convert to gravity potential energy

(ignoring air resistance)

mgh = ½mv²

     h = v²/2g

     h = 40.486² / (2(9.8))

     h = 83.6303...

     h = 84 m

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:

Explanation:

Without friction, a roller coaster continuously converts potential energy to kinetic energy and back again. Total energy will be constant.

Let m be the mass of the car and ground level is the origin.

on the 5.5 m hill, total energy is

E = PE + KE

E = mgh + ½mv²  

E = m(9.8)(5.5) + ½m(9.3)² = 97m J

a) The maximum height will occur when the total energy is all potential energy.

E = mgh

h = E/mg

h = 97m/m(9.8) = 9.9 m  

As this value is greater than the height of the third hill at 5.5 + 4.0 = 9.5 m The car will cross the last hill with some remaining velocity in kinetic energy.

b) As 9.5 m is greater than 9.3 m, the 9.5 m hill will have more of the total energy of the system as potential energy, This mean there is less kinetic energy and therefore less velocity (and speed) on top of the 9.5 m hill.

c) KE = E - PE

KE = 97m - m(9.8)(9.5 - 1.0)

KE = 97m = 83.3m

KE = 13.7m = ½mv²

v² = √(2(13.7)

v = 5.2345...

v = 5.2 m/s

Answer:

15[m].

Explanation:

1) the required distance is AD, for more info see the attached picture.

2) [tex]AD=\sqrt{AA_1^2+A_1D^2} =\sqrt{81+144}=15[m].[/tex]