What are the two fundamental ways energy and momentum are moved from one place to another?​

Answer:

Explanation:

ASSUMING that the dart is fired horizontally so that gravity potential energy considerations are not needed. Also ignoring friction work.

The spring potential will convert to kinetic.

KE = PS

½mv² = ½kx²

     v = [tex]\sqrt{kx^2/m}[/tex]

     v = [tex]\sqrt{1115(0.05^2)/0.025}[/tex]

     v = 10.55935...

     v = 11 m/s

Answer:

Speed of waves on the rope is 21 m/s

Explanation:

Length of the rope (l) = 5.0 m

Mass of the rope (m) = 0.52 kg

Tension in the rope (T) = 46 N

Formula of speed of waves on the rope:

[tex] \bold{v = \sqrt{\dfrac{T}{\mu}}} [/tex]

[tex] \mu [/tex] = Mass per unit length of the rope (m/l)

By substituting the values in the formula we get:

[tex] \implies \rm v = \sqrt{\dfrac{T}{ \dfrac{m}{l} }} \\ \\ \implies \rm v = \sqrt{\dfrac{Tl}{m}} \\ \\ \implies \rm v = \sqrt{ \dfrac{46 \times 5}{0.52} } \\ \\ \implies \rm v = \sqrt{ \dfrac{230}{0.52} } \\ \\ \implies \rm v = \sqrt{442.3} \\ \\ \implies \rm v = 21 \: m {s}^{ - 1} [/tex]

Speed of waves on the rope (v) = 21 m/s

Answer:

geothermal energy

Explanation:

the energy is obtained from the heat within the surface of earth

Answer:

heat energy

Explanation:

Answer:

Let’s substitute in our givens.

(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)

I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.

Note that I have considered the bullet’s velocity to be in the positive direction,

The answer is vf = 0.280 m/s

Answer:

Explanation:

momentum is mass times velocity

p = mv

so take the momentum of the truck in question 17 and divide by the mass of this car

v = p/m = p / 1400

Answer:

7.5m/s

Explanation:

Force= mass × velocity

Energy is conserved, the car and van should have the same overall force.

1500kg × 30m/s= 6000kg × final velocity

Final velocity = 7.5m/s

Answer:

1)The position change of almost any manually operated room light switch.

2) Sunlight striking a point on the ground on a partly cloudy and windy day

Explanation:

answer = 6n to the right

Explanation:

2n plus 4n equals 6n

since 6n is more than 5n it goes 6n to the right

Explanation:

We apply the definition of work by a constant force in the first three parts, but then in the fourth part we add up the answers. The total (net) work is the sum of the amounts of work done by the individual forces, and is the work done by the total (net) force. This identification is not represented by an equation in the chapter text, but is something you know by thinking about it, without relying on an equation in a list.

The definition of work by a constant force is W=FΔrcosθ.

(a) The applied force does work given by

W=FΔrcosθ=(16.0N)(2.20m)cos25.00=31.9J

(b), (c) The normal force and the weight are both at 900 to the displacement in any time interval. Both do 0 work.

(d) ∑W=31.9J+0+0=31.9J

Answer:

Explanation:

How long does the football need to rise?

4.70/3 = 2.35 s

What height will the football reach?

h = ½(9.81)2.35² = 27.1 m

With what speed does the punter need to kick the football?

vy = g•t = 9.81(2.35) = 23.1 m/s

vx = d/t = 56.0/4.70 = 11.9 m/s

v = √(vx²+vy²) = 26.0 m/s

At what angle (θ), with the horizontal, does the punter need to kick the football?

θ = arctan(vy/vx) = 62.7°

Answer:

Diffusion in gases is quick because the particles in a gas move quickly. It happens even faster in hot gases because the particles of gas move faster.

Answer:

It could be related with the lesson from which this question belongs as far we did not read the lesson

Sorry

[tex]\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time, t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0 +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}[/tex]

solution:

the formula is T = F * r * sin(theta) so just input the numbers and solve it.

Explanation:

Torque is the twisting force that tends to cause rotation. The point where the object rotates is known as the axis of rotation. Mathematically, torque can be written as T = F * r * sin(theta), and it has units of Newton-meters

Answer:

pressure = force / area

then pressure = 25 / 5 = "5" N/m^2

Answer:

magnetic force.

Explanation:bc it makes sense, and can i please get brainliest answer i never asked. its ok if you say no. have a great day <3.

Answer:

i don't know but my father i think he can't answer this

Answer:

Explanation:

I will ASSUME this means torque about the dot.

3) 20(3) + 10(6) - 30(4) = 0 N•m

4) 10(0.5) - 6sin45(1) = -0.7573593... or about 0.76 N•m CCW

5) 25(3) - 40sin30(4) = -5 N•m or 5 N•m CCW

6) 15(3) - 12(2) - 10sin45(6) = -21.4264068... or about 21 N•m CCW

Answer:

Essentially all of it

Explanation:

The potential energy was

PE = mgh = 6.36(9.81)(2.05) = 127.90278 = 128 J

ignoring air resistance, this PE converts to KE

With no rebound final velocity is zero, so Kinetic energy lost = 128 J

Hi there!

Impulse = Change in momentum

I = Δp = mΔv = m(vf - vi)

Where:

m = mass of object (kg)

vf = final velocity (m/s)

vi = initial velocity (m/s)

Begin by converting grams to kilograms:

1 kg = 1000g ⇒ 145g = .145kg

Now, plug in the given values. Remember to assign directions since velocity is a vector. Let the initial direction be positive and the opposite be negative.

I = (.145)(-20 - 17) = -5.365 Ns

The magnitude is the absolute value, so:

|-5.365| = 5.365 Ns

Explanation:

a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:

[tex]x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)[/tex]

[tex]y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)[/tex]

From Eqn(2), we see that

[tex]N = mg\cos 25°\;\;\;\;\;\;\;(3)[/tex]

so using Eqn(3) on Eqn(1), we get

[tex]mg\sin 25° - \mu_kmg\cos 25° = ma[/tex]

Solving for the acceleration, we see that

[tex]a = g(\sin 25° - \mu_k\cos 25°)[/tex]

[tex]\;\;\;\;= 2.45\:\text{m/s}^2[/tex]

b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

[tex]v^2 = v_0^2 + 2ax[/tex]

Since the crate started from rest, [tex]v_0 = 0.[/tex] Thus our equation reduces to

[tex]v^2 = 2ax \Rightarrow v = \sqrt{2ax}[/tex]

[tex]v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}[/tex]

[tex]\;\;\;\;= 6.32\:\text{m/s}[/tex]

Hi there!

In this instance, the object's centripetal force is provided by the horizontal component of the tension, so:

Tsinθ = mv²/r

**We use sine because in this situation, the angle is with the vertical**

We can plug in the known values for tension and theta:

60sin(60) = mv²/r

51.96 = mv²/r

The radius is equivalent to the sine of the string in respect to theta:

sin(60) = O/H = r/L

2sin(60) = 1.732 m

Now, solve for the velocity:

51.96 = mv²/r

51.96r / m = v²

51.96(1.732)/.400 = v²

v² = 225

v = 15 m/s

Answer:

Explanation:

The balloon would require a time of

t = d/v = 13.5/ (23.6cos38) = 0.7259...s

to travel the horizontal distance.

the vertical position relative to the throw point at that time is

h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)

h = 7.9652...

so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.

If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time

Answer:

[tex]\huge\boxed{\sf a = 1200\ m/s\²}[/tex]

Explanation:

Given Data:

Initial Velocity = Vi = 40 m/s

Final Velocity = Vf = 80 m/s

Distance = S = 200 m

Required:

Acceleration = a = ?

Formula:

2aS = Vf² - Vi² (THIRD EQUATION OF MOTION)

Solution:

2a (200) = (80)² - (40)²

400a = 6400 - 1600

400a = 4800

Divide 400 to both sides

a = 4800 / 400

a = 1200 m/s²

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

Answer:

This is the midline or the medium which is the exact middle of the graphs minimum and maximum points(which are the amplitude)

Answer:

An orbit of a spacecraft from which the spacecraft or another vehicle may be launched on a new trajectory.