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3 years ago

Which of the following elements of writing is defined as the choice of words sulted to the type of writing?

Physics

2 answers:

pantera1 [17]3 years ago

5 0

Answer:

Diction

Explanation:

is defined as the choice of words suited to the type of writing. hope this helps you :)

OlgaM077 [116]3 years ago

3 0

Answer:

Diction

Explanation:

Is an element of writing is defined as the choice of words sulted to the type of writing

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3. Three blocks of masses m, 2m and 3m are suspended from the ceiling using ropes as shown in diagram. Which of the following co

Alja [10]

Answer:

The correct option is d: T₁ > T₂ > T₃.

Explanation:

Let's evaluate each tension.

Case T₃.

$T_{3} - W_{3} = 0$

For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.

$T_{3} = W_{3}$

Since W₃ = mg, where m is for mass and g is for the acceleration due to gravity, we have:

$T_{3} = W_{3} = mg$ (1) Case T₂.

$T_{2} - (T_{3} + W_{2}) = 0$

$T_{2} = T_{3} + W_{2}$ (2)

By entering W₂ = 2mg and equation (1) into eq (2) we have:

$T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg$

Case T₁.

$T_{1} - (T_{2} + W_{1}) = 0$

$T_{1} = T_{2} + W_{1}$ (3)

Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:

$T_{1} = 3mg + 3mg = 6mg$

Therefore, the correct option is d: T₁ > T₂ > T₃.

I hope it helps you!

6 0

3 years ago

Read 2 more answers

If a wave traveling through air decreases its wavelength by half, describe what happens to the wave speed and frequency.

natka813 [3]

The frequency, the speed and the wavelength of a wave are related by the following equation:
$f= \frac{v}{\lambda}$ (1)
where
f is the frequency
$\lambda$ is the wavelength
v is the wave speed

The speed of the wave does depend only on the properties of the medium, so since the wave is still traveling in air, the medium has not changed and therefore the speed remains the same. We see instead from eq.(1) that the frequency is inversely proportional to the wavelength, so if the wavelength is decreased by half, we see that the frequency will double.

5 0

3 years ago

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches

yaroslaw [1]

Answer:

(a) The parachutist spent 24.84 secs in air

(b) The height the fall begins is 472 m

Explanation:

Here is the complete question:

A parachutist bails out and freely falls 63 m. Then the parachute opens, and thereafter she decelerates at 1.5 m/s2. She reaches the ground with a speed of 3.3 m/s. (a) How long is the parachutist in the air (b) At what height does the fall begin?

Explanation:

From one of the equations of kinematics for free fall

$H = ut - \frac{1}{2}gt^{2}$

Where H is the height

u is the initial velocity

t is the time

and g is the acceleration due to gravity (Take g = 9.8 m/s2)

Now, we can find the time spent before the parachute opens.

u = 0 m/s (we assume the parachutist starts from rest)

H = - 63 m

∴ $-63 = 0(t) - \frac{1}{2}(0.98) t^{2} \\-63 = -4.9 t\\t^{2} = \frac{63}{4.9} \\t^{2} = 12.86\\t = \sqrt{12.86} \\t = 3.59 s$

This the time spent before the parachute opens

Also from one of the equations of kinematics for free fall

$v = u - gt$

where v is the final velocity

We can determine the final velocity before the parachute opens and she starts to decelerate

∴ $v = - 9.8(3.59)\\v = - 35.18m/s$

Now, we will calculate the time spent after the parachute opens

From one of the equation of kinematics for linear motion,

$v = u + at$

Here, the initial velocity will be the final velocity just before the parachute opens, that is

$u = - 35.18 m/s$

From the question,

$v = - 3.3 m/s$

$a = 1.5 m/s^{2}$

We then get

$-3.3 = - 35.18 + (1.5)t$

$-3.3 + 35.18 = 1.5t\\31.88 = 1.5t$

$t = \frac{31.88}{1.5}$

$t = 21.25 secs$

(a) To determine how long the parachutist is in the air,

That is sum of the time used when falling freely and the time used after the parachute opens

= 3.59 secs + 21.25 secs

24.84 secs

Hence, the parachutist spent 24.84 secs in air

(b) To determine what height the fall begins

First, we will calculate the height from which the parachute opens

From one of the equation of kinematics for linear motion,

$x = ut + \frac{1}{2}at^{2} \\$

$x = -35.18(21.25) + \frac{1}{2}(1.5)(21.25)^{2} \\x = -747.58 + 338.67\\x = -408.91m\\$

$x$ ≅ $- 409m$

Hence, the height the fall begins is 63m + 409m

= 472 m

3 0

3 years ago

Momentum ap physics 1

11111nata11111 [884]

Answer:

The correct option is;

B. Object X travels at -2 m/s and object Y travels at 4 m/s after the spring is no longer compressed

Explanation:

The given parameters are;

The mass of object Y = M

The mass of object X = 2·M

The initial velocity of object X and object Y = 0 m/s

Let A represent the velocity of object X after the spring is released and B represent the velocity of object Y after the spring is released, therefore, by the principle of the conservation of linear momentum, we have;

(M + 2·M) × 0 = M × B + 2·M × A

∴ (M + 2·M) × 0 = 0 = M × B + 2·M × A

M × B = -2·M × A

∴ B = -2·A

Therefore, the velocity of the object Y = -2 × The velocity of the object X

Whereby the velocity of the object X = -2, The velocity of the object Y = -2 × -2 = 4

Which gives, object X travels at -2 m/s and object Y travels at 4 m/s after the spring is no longer compressed.

8 0

3 years ago

What is the acceleration of a car that goes from 0 MS to 60 MS and six seconds

Len [333]

A = Δv/Δt
a = v2-v1/t2-t1
a = 60MS - 0MS / 6 seconds
a = 60 MS/ 6 seconds
a = 10 m/s^2

8 0

3 years ago