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lora16 [44]
3 years ago
15

PLZ HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!! What is a front?

Physics
2 answers:
Phoenix [80]3 years ago
7 0

It’s a set of type of one particular face and size

Shkiper50 [21]3 years ago
4 0

Answer:

in computer and writing context it is many styles of writing that you can make the text look like. I.E: <em>helllo  </em>hello

Explanation:

A font is the combination of typeface and other qualities, such as size, pitch, and spacing. For example, Times Roman is a typeface that defines the shape of each character. Within Times Roman, however, there are many fonts to choose from -- different sizes, italic, bold, and so on.

Mark me the brainliest plz

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What is the logical oreder for these words : verse, interlude,verse,refrain, introduction,refrain,coda,refrain,verse
Nikolay [14]

Sorry to say but I know that t(e introduction is first and the coda is last


3 0
3 years ago
When force is applied to a breaker bar the torque can be calculated by multiplying the length of the lever by the?
Nimfa-mama [501]

When a force applied to a breaker bar the torque can be calculated by multiplying the<u> length of the lever</u> by the tangential component of force on the lever.

<h3>What is torque?</h3>

Torque is the <u>rotating equivalent</u> of force in physics and mechanics. Depending on the subject of study, it is also known as the moment, moment of force, rotating force, or turning effect. It illustrates how a force can cause a change in the body's rotational motion.

Torque is given by the formula :

                          α = r x F ( bold letters represent vector quantities)

The S.I. unit for torque is :  N - m ( Newton - meter)

<h3>How do we define 1 N-m of torque?</h3>

The newton-metre is a torque unit (also known as a moment) in the SI system. The torque produced by a one newton force applied <u>perpendicularly to the end of a one metre long</u> moment arm is known as a newton-metre.

To learn more about torque:

brainly.com/question/14970645

#SPJ4

5 0
2 years ago
A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.
SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of x_0 = 5.0 m

- at t = 0, the initial velocity of the motorcyclist is v_0 = 15.0 m east

- The acceleration of the motorcyclist is constant and it is a=4.0 m/s^2

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

x(t)=x_0 + v_0t + \frac{1}{2}at^2

where t is the time.

Substituting t = 2.0 s, we find the position:

x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

v(t)=x'(t)=v_0 + at

where:

v_0=15.0 m/s is the initial velocity

a=4.0 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s

And therefore, the position at t = 2.5 s is:

x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m

Learn more about accelerated motion:

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3 0
3 years ago
Which is an example of potential en
SIZIF [17.4K]

a yoyo in someones hand is an example of potential energy

8 0
3 years ago
Read 2 more answers
A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s. Assume in this problem that air drag i
stiks02 [169]

Answer:

P.E = 0.068 J = 68 mJ

Explanation:

First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = -9.8 m/s²  (negative sign due to upward motion)

h = height attained by the ball toy = ?

Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)

Vi = Initial Velocity = 3 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²

h = (9 m²/s²)/(19.6 m/s²)

h = 0.46 m

Now, the gravitational potential energy of ball at its peak is given by the following formula:

P.E = mgh

P.E = (0.015 kg)(9.8 m/s²)(0.46 m)

<u>P.E = 0.068 J = 68 mJ</u>

4 0
3 years ago
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