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

Which Degree Would A Person Most Likely Pursue If They Are Interested In Howcomputers Work To Solve Problems?A.

Answers

Answer 1

Answer:

Computer science

Explanation:

You would get everything you need to know about computers if you took computer science. I hope this helps

Answer 2

Answer:

Computer science

Explanation:


Related Questions

Please help me! Some people have proposed a new way to build houses in areas that are likely to experience tsunamis. In this design, a house wouldn’t have solid walls on all four sides. Instead, some of the wall areas would be replaced by substances that water can travel through quickly, as shown in the diagram. How would this design help a house survive a tsunami? What drawbacks might there be to this design?

Answers

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

Define faith in your own words.

Answers

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:

The maximum speed with which an 1000 kg car makes a 180-degree turn is 10 m/s. The radius of the circle through which the car is turning is 24 m. Determine the force of friction and the coefficient of friction acting upon
the car.

Answers

109 is the answer There hope you do well


9. P waves move faster than S waves A. True B. False

Answers

Answer: true
(I need 20 characters to submit the answer so I wrote this not to be confused with the answer)

A guitar string 63.6 cm long vibrates with a standing wave that has five antinodes. Which harmonic is this

Answers

Answer:

fifth harmonic

Explanation:

You are standing 30 m due east of a 50 kg person who is running at a speed of 20 m/s due west. What is the magnitude of that person's angular momentum about you (in units of )

Answers

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

a car moves at a speed of 30m/s to the west of 3hr, what is its displacement of the car in km?​

Answers

Answer:

Explanation:

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

What happens to the gravitational force between two objects if the distance between them triples?

A. The force increases by a factor of 9

B. The force decreases by a factor of 9

C. The force decreases by a factor of 3

D. The force increases by a factor of 3

Answers

So as two objects are separated from each other, the force of gravitational attraction between them also decreases. ... If the separation distance between any two objects is tripled (increased by a factor of 3), then the force of gravitational attraction is decreased by a factor of 9 (3 raised to the second power).

So the answer you are looking for is B.
The answer will be B.

Gravity obeys something called the inverse square law.

This means if distance increases by a factor of x, the force of gravity will decrease by a factor of x^2

For example, if distance increases by 2, the force would be 4 times weaker

If Distance increases by a factor of 3, the force will be 9 times weaker

Someone please help with this question. From my knowledge the answer I believe to be correct is 4Em but I’m still not so sure. Please explain!
Answer choices:
1/2 Em
Em
2Em
4Em

Answers

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!

As a truck rounds a curve, a box in the bed of the truck slides to the side farthest from the center of the curve. This movement of the box is a result of

Answers

Because I don’t have choices I’m not sure what to chose from or what the subject is about but I can tell you that the box moves from the result of Inertia, and is kenetic energy

Answer:

inertia   .

because yes

A block slides on a rough 45 degree incline. The coefficient of friction is µk what is the ratio of acceleration when the block accelerates down the incline to the acceleration when the block is projected up the incline​

Answers

it is the same, force diagram is the same, acceleration depends on forces. the only thing different is the initial velocity.

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

I need help with this answer I believe it's a democracy​

Answers

Answer:

a. democracy

Explanation:

beacouse the government control of their members

How large is the acceleration of a 25 kg mass with a net force of 75 N applied horizontally to it?

Answers

Answer:

Explanation:

F = ma

a = F/m

a = 75/25

a = 3 m/s²

A woman skis from rest down a hill 20 m high. If friction is negligible, what is her speed at the bottom of the slope? Select one: O a. 20 m/s O b. 12 m/s O c. 400 m/s O d. 6 m/s​

Answers

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

I need help with us history

Answers

Answer:

English is the language

Answer: im most likely wrong but i think its A

Explanation:

Determine the rTo understand the concept of nodes of a standing wave.
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 (i.e., the point of attachment does not move).
Consider a standing wave, where y represents the transverse displacement of a string that extends along the x direction. Here is a common mathematical form for such a wave:

y(x,t)=Acos(kx)sin(ωt),

where A is the maximum transverse displacement of the string (the amplitude of the wave), which is assumed to be nonzero, k is the wavenumber, ω is the angular frequency of the wave, and t is time.
Part A
Which one of the following statements about wave y(x,t) is correct?

adius of the 236U nucleus.

Answers

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

Three people are trying to move a box. Which set of forces will result in a net
force on the box of 20 N to the left?

Answers

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

4. Protons and neutrons are held together to form this _________

Answers

Answer:

strong nuclear force.

Explanation:

hope this helps you!!

A 4.0 kilogram projectile is fired horizontally from a 500 kilogram cannon initially at rest. The momentum of the projectile after being fired is 600 kilogram-meters per second to the north
(neglecting friction).
What is the speed of the cannon after firing?
0.83 m/s
1.2 m/s
o
3.3 m/s
150 m/s

Answers

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.

MV = P................ Equation 1

⇒ Where :

M = mass of the cannon,V = velocity of the cannonP = momentum of the projectile.

⇒ make V the subject of the equation

V = P/M.................. Equation 2

From the question,

⇒ Given:

P = 600 kgm/sM = 500 kg

⇒ Substitute these values into equation 2

V = 600/500V = 1.2 m/s

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

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

What are the characteristics of high energy wave?

A. Low frequencies and short wavelengths.

B. High frequencies and long wavelengths.

C. Low frequencies and long wavelengths.

D. High frequencies and short wavelengths

Answers

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

Need help ASAP, 1 MC

Answers

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

Which statement is true for a series circuit

Answers

Answer: they have one path to flow

Explanation: share the same current

In a series circuit, the current is the same at each resistor. If the light bulbs are identical, then the resistance is the same for each resistor. The voltage drop (I•R) will be the same for each resistor since the current at and the resistance of each resistor is the same.

A student connects a 21.0 V battery to a capacitor of unknown capacitance. The result is that 52.8 µC of charge is stored on the capacitor. How much energy (in J) is stored in the capacitor?

Answers

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

a rocket ship is moving through space at 1000 m/s. It accelerates in the same direction at 4m/s/s. What is its speed after 100 seconds

Answers

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.

What is velocity?

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

#SPJ5

As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constants. The angular speed changes from 3.70 rad/s at t = 0 to 2.00 rad/s at t = 8.60 s.

a. Use this information to determine σ and ω0.
σ = _______s−1
ωo = ______rad/s

b. Determine the magnitude of the angular acceleration at t = 3.00 s.
______rad/s2
c. Determine the number of revolutions the wheel makes in the first 2.50 s
_______rev

d. Determine the number of revolutions it makes before coming to rest.
_______rev

Answers

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]

explain the different conditions that can result in hot and cold lahars, and explain how lahars change the earth's surface?

Answers

Lahars can also be formed when high-volume or long-duration rainfall occurs during or after an eruption. It can change the shape of a mountain by blowing parts of it away, but volcanic eruptions can also build up the land around a volcano when lava flows out and hardens on the surface. The surface of the Earth can crack and shift during an earthquake above the point where the crust moves.

Một chất điểm khối lượng m=200g chuyển
động chậm dần với vận tốc biến đổi theo qui luật
v=30 – 0,4t2 (SI). Lực hãm tác dụng vào chất điểm
lúc t = 5 giây là
A. 8 N B. 0,8 N C. 4 N D. 0,4 N

Answers

Can you please translate to English?


Two Blocks are connected by a massless rope over a massless,
frictionless pulley, as shown in the figure. The mass of block 2
is m2 = 10.1 kg, and the coefficient of kinetic friction
between block 2 and the incline is Mk = 0.200. The angle 0 of
the incline is 27.5º. If block 2 is moving up the incline at
constant speed, what is the mass mi of block 1?

Answers

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.

What is tension force?

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

#SPJ2

What is the weight of a 5kg object at the surface of the earth?

A. 49N
B. 49kg
C. 5.0kg
D. 25N

Answers

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

Two steel guitar strings have the same length. String A has a diameter of 0.489 mm and is under 410 N of tension. String B has a diameter of 1.27 mm and is under a tension of 809 N. Calculate the ratio of the wave speeds, vA/vB, in these two strings.

Answers

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

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