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ch4aika [34]
3 years ago
7

Heat likes to remain ONMFRIU ← Whats that unscrambled?

Physics
1 answer:
OlgaM077 [116]3 years ago
5 0
<span>So the question is how does heat prefer to remain and to unscramble the letters ONMFRIU. The unscrambled letters mean: UNIFORM. The heat likes to remain uniform because thermodynamic systems always tend to reach thermal equilybrium after some period of time that is specific for each system. </span>
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List the types of electromagnetic radiation in order from lowest energy photons to highest energy photons.
frutty [35]

radio waves,X-rays,

Explanation:

In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves. Microwaves (like the ones used in microwave ovens) are a subsection of the radio wave segment of the EM spectrum.

8 0
2 years ago
In what type of change is matter not destroyed
Margaret [11]

There are two types of change in matter: physical change and chemical change. ... This is called the Law of Conservation of Matter. It states that matter can never be created or destroyed, only changed and rearranged.

6 0
3 years ago
Read 2 more answers
The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared
velikii [3]

Answer:

\displaystyle a(5)=-32

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

\displaystyle v(t)=\frac{df}{dt}

And the acceleration is

\displaystyle a(t)=\frac{dv}{dt}

Or equivalently

\displaystyle a(t)=\frac{d^2f}{d^2t}

The given height of a projectile is

f(t)=-16t^2 +238t+3

Let's compute the speed

\displaystyle v(t)=-32t+238

And the acceleration

\displaystyle a(t)=-32

It's a constant value regardless of the time t, thus

\boxed{\displaystyle a(5)=-32}

3 0
3 years ago
What is the equation that defines work
mina [271]
Work=force*displacment
3 0
3 years ago
A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
1 year ago
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