According to the conservation of mechanical energy, the kinetic energy before plus the kinetic energy after equals the
potential energy before plus the potential energy after.
True
False

Explanation:

The box is accelerating along the y-axis at a rate of [tex]+2.5\:\text{m/s}^2[/tex] as well as along the x-axis at a rate of [tex]+5.1\:\text{m/s}^2.[/tex] So the magnitude of the box's total acceleration is given by

[tex]a_T = \sqrt{a_x^2 + a_y^2}[/tex]

[tex]\:\:\:\:= \sqrt{(5.1\:\text{m/s}^2)^2 + (2.5\:\text{m/s}^2)^2}[/tex]

[tex]\:\:\:\:=5.7\:\text{m/s}^2[/tex]

The direction of the acceleration [tex]\theta[/tex] with respect to the horizontal direction is given by

[tex]\theta = \tan^{-1}\!\left(\dfrac{a_y}{a_x}\right) = \tan^{-1}\!\left(\dfrac{2.5\:\text{m/s}^2}{5.1\:\text{m/s}^2}\right)[/tex]

[tex]\:\:\:\:= 26.1°[/tex]

Answer:

Explanation:

The work done by brake friction changes the kinetic energy,

ASSUMING the road is level and other friction is ignored.

W = KEf - KEi = ½mvf² - ½mvi² = ½m(vi² - vf²)

W = ½(1800)(30² + 10²) = 900(900 - 100) = 720000 J = 720 KJ

A bio psychologist would address questions on stress, abuse and conflict when studying depression.

A bio psychologist is a person that is involved in the study of the brain and peoples behavior. This person tries to understand why people do the things or act the way they do using a biological and psychological approach.

While trying to understand the cause of this depression, the bio psychologist would

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Answer:

Answer:

Change in state from liquid to solid.

Explanation:

Decrease in thermal energy will decrease energy and particles will slow down. Change in state from liquid to solid.

Answer:

The longest wavelength for closed at one end and open at the other is

y / 4      where y is the wavelength - that is node - antinode

The next possible wavelength is 3 y / 4 -    node - antinode - node -antinode

y / 4 = 3 m     y = 12 meters    the longest wavelength

3 y / 4 = 3 m      y = 4 meters   1 / 3 times as long

Answer:

Stress = F / A       force per unit area

A = 3.00 cm^2 = 3 E-4  m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N    max force applied

F/3 = 2.4E4 N  if force not to exceed limit   (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2      about 2 g

dharna is said to be concentration.. it is true

Complete Question

(c) It takes you 5 hours to to bring the turkey from 10.0°C to 75.0 °C. During that time, the electrical grid transfers a constant 2500.0 Watts of power into the the oven. Take the turkey and the air in the oven to be your system. What was the thermal transfer of energy between the system and the surroundings?

Answer:

[tex]Q=4.50 *10^7J[/tex]

Explanation:

From the question we are told that:

Time [tex]t=5hours[/tex]

Temperature rise [tex]dT= 65\textdegree[/tex]

Power [tex]P=2500.0 Watts[/tex]

Generally, the equation for Power is mathematically given by

[tex]P=\frac{Q}{t}[/tex]

Therefore

[tex]Q=2500*5*360[/tex]

[tex]Q=4.50 *10^7J[/tex]

Answer:

wegkwe fhkrbhefdb

Explanation:B

Answer:

6v

Explanation:

V=IR

V= 2* 3

V= 6 volts

Answer:

» The image is laterally inverted.

» The image is upright.

» The image geometry is same as object geometry.

» Image distance is same as object distance.

» Image is not real, it's virtual ( not formed on screen ).

[tex].[/tex]

Comparison of For-profit and Non-profit Fitness Centers

This is how they compare with each other: For-profit centers offer more luxury services than nonprofit centers.

For-profit fitness centers are fitness centers that offer their services to cover their costs and make some profits.  They are business entities that  generate income for their shareholders.  Their fees are based on cost plus profit.  This approach enables them to remain sustainable as business ventures.

They are not like non-profit fitness centers, which may offer their services for free or at cost.  Sometimes, the nonprofit centers may not even cover their fixed and operational costs.  They are usually located in low-income communities, unlike for-profit fitness centers, which are mainly located in rich-income communities.

Considering their clientele, for-profit centers provide modern and better facilities to satisfy their rich-income clients.

Thus, these for-profit centers offer more luxury services than their non-profit counterparts.

For more information about how for-profit centers base their charges, see https://brainly.com/question/21302488.

Answer:

well its about our thinking but i do believe in ghost a little

La longitud del faldón de la rampa es de 5.4 m.

La pendiente expresada en porcentaje sigue la siguiente ecuación:

[tex]m=\frac{y}{x}*100[/tex] (1)

Donde:

Sabemos que la pendiente es de 27%. Por lo tanto, usando la ecuación 1, despejamos y.

[tex]27=\frac{y}{20}*100[/tex]

[tex]y=\frac{27*20}{100}[/tex]

[tex]y=5.4\: m[/tex]        

La longitud del faldón es 5.4 m

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Complete Question

The Complete Question is Attached below

Answer:

[tex]E_t=2.14*10^{-13}J[/tex]

Explanation:

From the question we are told that:

Kinetic energy [tex]K.E=5*10^{-14}[/tex]

Generally, the equation for Total Energy is mathematically given by

[tex]E_t=K.E+2me[/tex]

Where

Mass of an electron in MeV

[tex]m=0.510 998 950 00 MeV[/tex]

And

Electron Charge

[tex]1.6 * 10^{-19}[/tex]

Therefore

[tex]E_t=K.E+2me[/tex]

[tex]E_t=5*10^{-14}*0.510 998 950 00*1.6 * 10^{-19}[/tex]

[tex]E_t=2.14*10^{-13}J[/tex]

Answer:

true

Explanation:

forces require an agent

you should always be able to identify what (the agent) is producing the force

Answer:

c = 3.00E108 m/s = 3.00E5 km/s

t = S / v = 3.84E5 / 3.00E5 = 1.28 sec

Answer:

Answer:30 cm

Answer:30 cmExplanation:

Answer:30 cmExplanation:Given=ROC= 60cm

Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.

The ratio of foort dustance to load distance in a simple machine is called mechanical advantage or MA.

MA= Effort Distance / Load Distance

The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. Mass is the measure of the amount of matter in a body. Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more-approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater than 100 kilograms. In common usage, the mass of an object is often referred to as its weight, though these are in fact different concepts and quantities. In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity. In other words, an object with a mass of 1.0 kilogram weighs approximately 9.81 newtons. Weight and mass are considered to be the same quantities. But many people tend to misuse these terms in their daily conversations. The main difference between weight and mass is that weight is the force of gravity by which the earth attracts towards it whereas mass is the amount of matter in an object.

The answer is 40 kg. m/s.

Formula for momentum:

p=mv

p=(10 kg.)(4 m/s)

So, therefore, the final answer is p=40 kg. m/s.

I hope this helped answer your question. Enjoy your day, and take care!

Answer: its 400 n/s

Explanation:

cus my thing said it was right

F = m • a

F = (15 kg) • (4 m/s^2)

F = 60 Newtons

NOTE:

That's NOT a "resultant force".

That's the force you have to EXERT on 15 kg to MAKE it accelerate at 4 m/s^2 .

Using Newton second law

[tex]\\ \sf \longmapsto Force=Mass\times Acceleration[/tex]

[tex]\\ \sf \longmapsto Force=15(4)[/tex]

[tex]\\ \sf \longmapsto Force=60N[/tex]

Answer:

Horizontal distance, R = vo2 sin(2x45)/g

v = 285 km/h = 79.17 m/s

R = 79.172 x 1/1.6

R = 3917 m

R = 3.9 km

Tiger Woods' golf ball will travel approximately 3925.22 meters horizontally before landing back on the surface of the moon.

To find the horizontal distance traveled by the golf ball on the moon, we can use the following kinematic equation:

d =[tex](v^2 * sin(2\theta)) / g[/tex]

where:

d = horizontal distance traveled

v = initial velocity of the golf ball (converted to m/s)

θ = launch angle

g = acceleration due to gravity on the moon (1.60 m/s²)

First, we need to convert the initial velocity from km/h to m/s:

Initial velocity (v) = 285 km/h = 285 * (1000 m/km) / (3600 s/h) ≈ 79.17 m/s

Now, plug the values into the equation:

d = ([tex]79.17^2[/tex]* sin(2 * 42°)) / 1.60

Calculate the value within the parentheses first:

sin(2 * 42°) = sin(84°) ≈ 0.999

Now, calculate the horizontal distance (d):

d = [tex](79.17^2 * 0.999) / 1.60[/tex]

d ≈ 3925.22 meters

So, Tiger Woods' golf ball will travel approximately 3925.22 meters horizontally before landing back on the surface of the moon.

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#SPJ2

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

Answer:

explained

Explanation:

If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.

A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction.

Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.

An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north.

Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).

In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.

How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at  5  m/s

straight toward the goal and drives the ball in the same direction with a velocity of  30 m/s

relative to her body, then the velocity of the ball is  35  m/s

relative to the stationary, profusely sweating goalkeeper standing in front of the goal.

In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (

The figure shows components of velocity v in horizontal  vx and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta.

Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors  and  

These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.