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labwork [276]
2 years ago
13

Difference between text-talk novel and flash fiction?​

Physics
1 answer:
amid [387]2 years ago
3 0

See the flash fictions are small nov.els with para.graphs where as ch*.t story is a storyline that just look like some c

c.h.*.t.s.

You might be interested in
What is the net work done on the 20kg block while it moves the 4 meters?
Vinil7 [7]

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

     Weight = 20 x 9.81

     Weight = 196.2 N

2.- Calculate the work done

     Work = 196.2 x 4

     Work = 784.8 J

5 0
3 years ago
Please help me!! I can't figure the answer out, AND haven't gotten past this (being the block test) for a couple weeks now!!
ololo11 [35]

The density of the nugget is 19 g/cm^3 and is made of gold

Explanation:

The density of an object can be calculated as

d=\frac{m}{V}

where

d is the density

m is the mass

V is the volume of the object

We have to note that density of an object actually depends on the material the object is made of (therefore, two objects made of the same material can have different mass and different volume, but they have same density).

For the nugget in this problem, we have:

mass: m = 38 g

volume: V=2 cm^3

So, its density is

d=\frac{38}{2}=19 g/cm^3

And by looking at the table, we see that this value corresponds approximately to the density of gold, so the nugget is made of gold.

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

5 0
3 years ago
Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and
djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

x_f = Final length of the spring .

= \sqrt{1.25^2+1.8^2}  -0.60

= 1.591 m

x_i= The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

=  1.80 m.

m= The mass of the collar.

=3.55 kg

v_c = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Or, \frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2}  = 3.55\times 9.81\times 1.80

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

Learn more about friction:

brainly.com/question/24338873

#SPJ4

Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

6 0
2 years ago
Energy from the Sun arrives at the top of the Earth’s atmosphere with an intensity of 1.30kW/m21.30⁢kW/m2. How long does it take
vivado [14]

Answer:

T=1.384×10⁶seconds

Explanation:

Given data

p (Intensity)=1.30 kw/m²

E (Energy)=1.8×10⁹ J

A (Area)=1.00 m²

T (Time required)=?

Solution

E=PT   ................eq(i)

where E is energy

P is radiation power

T is time

Radiating Power is given as

P=pA

Where p is intensity

A is Area

Put P=pA in eq(i) we get

E=pAT

T=E/pA

T=\frac{(1.8*10^{9}}{(1.30*10^{3} )*(1.00)}  )\\T=1.38*10^{6} seconds

3 0
3 years ago
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
storchak [24]

Answer:

\large \boxed{42\, \mu \text{C}}$

Explanation:

The formula for the force exerted between two charges is

F=k \dfrac{ q_1q_2}{r^2}

where k is the Coulomb constant.

The charges are identical, so we can write the formula as

F=k\dfrac{q^{2}}{r^2}

\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

7 0
3 years ago
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