Which degree would a person most likely pursue if they are interested in how
computers work to solve problems?
A. Business management
B. Mechanical engineering
C. Computer science
D. Education technology
SUBMIT

Answer:

[tex]\frac{a_{d}}{a_{i}} = \frac{(1 -mu)}{mu}[/tex]

= (1 - μ)/μ

Explanation:

Always draw a diagram!

Up the incline:

[tex]Fr_{max}[/tex] = maximum friction

[tex]Fr_{max}[/tex] = μk

k = R = mg.cos(45) = mg.sin(45)

Resolution of forces parallel to the slope:

F (Fp in the diagram) = force of propulsion

g = gravity

[tex]F - Fr_{max} = ma_{i}[/tex]

[tex]F -[/tex] μ.mg.cos(45) [tex]= ma_{i}[/tex]

Down the decline:

Resolution of forces:

[tex]mg.sin(45) - Fr_{max} = ma_{d}[/tex]

[tex]mg.sin(45) -[/tex] μ.mg.cos(45) [tex]= ma_{d}[/tex]

Then, find the ratio:

[tex]\frac{ma_{d}}{ma_{i}} = \frac{mg.sin(45) - mu.mg.cos(45)}{-F + mu.mg.cos(45)} \\\\ \frac{a_{d}}{a_{i}} = \frac{k - k.mu}{-F + k.mu} \\\\ = \frac{k(1 -mu)}{-F + k.mu}[/tex]

Potentially, there is no need to consider F in this situation, in which case:

[tex]\frac{a_{d}}{a_{i}} = \frac{k(1 -mu)}{k.mu} \\\\ = \frac{(1 -mu)}{mu}[/tex]

= (1 - μ)/μ

Answer: The answer is A) 49 N(Newtons).

Answer:

English is the language

Answer: im most likely wrong but i think its A

Explanation:

Answer:

Explanation:

30 m/s • 3 hr •3600 s/hr / 1000 m/km = 324 km west

The mass of block 1 will be 1.99 kg.The tension force is applied along the whole length of the wire, pulling energy equally on both ends.

The tension force is described as the force transferred through a rope, string, or wire as it is pulled by opposing forces.

Given that,

Mass of block 1=? kg

The coefficient of the kinetic friction,μ=0.200

Now consider the weight component in the uphill direction.The weight is found as;

[tex]\rm W=m_1gsin \theta[/tex]

The force is balanced in the vertical direction as;

[tex]\rm T=F_f-W[/tex]

When the force of friction is;

[tex]\rm F_F=\mu_k N[/tex]

[tex]\rm F_f=(m_1 gcos \theta)[/tex]

Substitute the value in the vertical balanced equation;

[tex]\rm T=m_1gsin(27.5)^0-\mu_kmgcos27.5^0[/tex]

[tex]\rm T-m_2g=0\\\\T=m_2g[/tex]

[tex]\rm (10.1) g=m_1g(0.699-0.2 \times(-0.714) ) \\\\ (10.1) g=m_1g (0.699+0.1428) \\\\\ (10.1) g= m_1 \times 0.8418 \\\\ m_1 =11.99 \ kg[/tex]

Hence the mass of block 1 will be 1.99 kg.

To learn more about the tension force refer to the link;

https://brainly.com/question/2287912

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Answer:

Acceleration = (final velocity - starting velocity) / time

4 = (x-1000) / 100

<br/>x = 1400 m/s

Explanation:

The final velocity of the rocket ship which is moving with an initial velocity of 1000 m/s and acceleration of 4 m/s² after 100 seconds is 1400 m/s.

Velocity of a moving body is the rate of its speed. Mathematically velocity is the ratio of distance travelled to the time taken with a unit of m/s. Acceleration is the rate of change in velocity of  moving body. The unit of acceleration is 4 m/s² .

Thus acceleration can be determined from the change in velocity with respect to the change in time. Now, the relation between initial velocity, acceleration,  a and time,  t with the final velocity is written in the equation below:

v = u + at.

Where, v is the final velocity and u be the initial velocity.

Given here the initial velocity is 1000 m/s. Acceleration of the rocket is 4 m/s² . Thus the velocity after 100 seconds is calculated as follows:

v = 1000 m/s +  ( 4 m/s² × 100 s )

  = 1400 m/s.

Hence, the speed of the rocket after 100 seconds will be 1400 m/s.

To find more about velocity, refer the link:

https://brainly.com/question/28738284

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Hi there!

We can use the following equation for angular momentum:

[tex]\large\boxed{L = mrv}[/tex]

m = mass (kg)

r = distance from reference point (m)

v = velocity (m/s)

We can simply plug-and-chug the values provided in the question.

L = (50)(30)(20) = 30000 kgm²/s

Answer:

strong nuclear force.

Explanation:

hope this helps you!!

Answer:

Explanation:

F = ma

a = F/m

a = 75/25

a = 3 m/s²

Answer: push force gravity tension and reverse force

Explanation: sense they are pushing the box there is a push force gravity because this is likely on earth tension because it is the reverse of gravity and the reverse force because you have to have the reverse of push

Hi there!

We can use the work-energy theorem to solve.

Recall:

[tex]\large\boxed{E_i = E_f}}[/tex]

Initial energy = final energy

The initial energy is purely potential (she starts from rest), and, if we assign the bottom of the slope as the 0 line, her energy at the bottom is purely kinetic.

PE = mgh

KE = 1/2mv²

We can begin by setting the two equal:

mgh = 1/2mv²

Cancel out the mass and rearrange to solve for velocity:

2gh = v²

v = √2gh

Plug in given values and use g ≈ 10 m/s²:

v = √2(10)(20) = 20 m/s

Answer:

fifth harmonic

Explanation:

Answer:

inertia   .

because yes

Answer:

D. High frequency and short wavelengths.

Explanation:

If a wave is high in energy it will have a higher frequency.

High frequency = short wavelengths

Answer:

1.108 ×  [tex]10^{-3}[/tex]J

Explanation:

v=21.0v

Q=52.8×  [tex]10^{-6}[/tex]

E=?

V=E/Q

E=v ×Q

 =21 ×52.8 ×[tex]10^{-6}[/tex]

 =1108.8   ×[tex]10^{-6}[/tex]

E= 1.108 ×  [tex]10^{-3}[/tex]J

Answer:

Explanation:

For an ideal spring over a frictionless horizontal surface, stored energy is only a function of the spring constant k and the distance of compression. The mass of the block doing the compressing is irrelevant

Energy stored in the first example is

Em = ½kd²

Energy stored in the second example is

E₂m = ½k(2d)² = 4(½kd²) = 4Em

So the second situation has four times as much stored spring potential energy as the first situation

4 Em is correct

Good job!

The speed of the cannon after firing is 1.2 m/s

This can be solved using the law of conservation of momentum.

From the law of conservation of momentum,

⇒ The momentum of the projectile is equal to the momentum of the cannon.

⇒ Where :

⇒ make V the subject of the equation

From the question,

⇒ Given:

⇒ Substitute these values into equation 2

Hence, The speed of the cannon after firing is 1.2 m/s

Learn more about momentum here: https://brainly.com/question/14015797

Answer:

Explanation:

vA / vB = √(TA/(m/L)) / √(TB/(m/L))

The lengths are the same, so the L divides out to 1

The material is identical so the mass will be directly proportional to the cross sectional area of the string

vA / vB = √(TA/(πdA²/4)) / √(TB/(πdB²/4))

π/4 is common so divides out to 1

vA / vB = √(TA/dA²) / √(TB/dB²)

vA / vB = √(410/0.489²) / √(809/1.27²)

vA / vB = 41.407 / 23.396

vA / vB = 1.8488961...

vA / vB = 1.85

Answer:

I think some drawbacks are that since there are no solid walls meaning it is weak and if murphy's law is in place, the water will destroy the substance. Tsunami waves also happen very quickly so even if the water can travel thru the substance quickly, it probably won't be quick enough. This design could help if the wave is smaller because less destruction would occur.

Explanation:

yeah

Answer: they have one path to flow

Explanation: share the same current

Answer:

a. democracy

Explanation:

beacouse the government control of their members

Answer:

The nodes of a standing wave are points where the displacement of the wave is zero at all times nodes are important for matching boundary conditions for example that the point at which a string is tied to a support has zero displacement at all times ie the point of attachment does not move consider a standing

Answer:

Faith, to me is almost like hope but for faith you need to believe. Some people associate the word faith with religion. But that's not it. Faith is really just a stronger term than believing someone. Like you can believe in someone but not have faith in that same person. You can have faith in someone but to do that you have to believe and trust them.

Explanation:

Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]

We are given that at t = 0, ω =  3.7 rad/sec. We can plug this into the equation:

[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]

Now, we can solve for sigma using the other given condition:

[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]

c.

The angular displacement is the INTEGRAL of the angular velocity function.

[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]

[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]

[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]

[tex]\theta = 8.471 rad[/tex]

Convert this to rev:

[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]

Evaluate the improper integral:

[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]

[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]

[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]

Convert to rev:

[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]

Answer:

The first one is the only one that is true all the time

Explanation:

The second one may be true if friction is high enough.

The other three are false all the time