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3 years ago

Why is it important to practice a presentation?

Physics

2 answers:

egoroff_w [7]3 years ago

7 0

Answer: D. To ensure the smooth flow of the presentation.

Explanation:

A presentation is a narratative explanation of a topic. A practice of presentation is required before presenting it in the front of the audience. It will help in maintaining the desired flow and sequence of the content which an author want to present in a systematic way. It will help in reducing faults and breaks in the conversation.

-Dominant- [34]3 years ago

6 0

Your answer should be D
let me know if I got it wrong
Hope this helped!

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If the mass of a materials is 44grams and the volume of the material is 11cm^3 what would the density of the material be

jok3333 [9.3K]

Density =mass/volume
density= (0.044kg/(11 x10^-6)
4000kg/m^3

8 0

3 years ago

You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength

Ray Of Light [21]

Answer:

wavelength $\lambda$ = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = $\frac{D\lambda}{d}$ ....................1

$\lambda$ is Wavelength of light and D is Distance between screen and slit and d is slit width

so put here value and we get

$\lambda$ = $\frac{2.96*0.325*10^{-6}}{2.20}$

$\lambda$ = 437.27 × $10^{-9}$ m

wavelength $\lambda$ = 437.27 nm

4 0

3 years ago

A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming

PtichkaEL [24]

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

$u=v-at$

a=-g

$u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s$

The velocity of the shell at launch and 5.4 s after the launch is given by :

$v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s$

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

6 0

3 years ago

A force Fx=cx-3.00x2 acts on a virus as the virus moves along an x axis, with F measured in Newtons, x in meters, and c a consta

andrey2020 [161]

Answer:

Explanation:

We are given that

$F_x=cx-3x^2$

K.E at x=0 m=20 J

K.E at x=3 m=11 J

We have to find the value of c.

By work energy theorem

Work done=Change in kinetic energy

W= $\int_{0}^{3} Fdx=\int_{0}^{3}(cx-3x^2)dx$

$W=[\frac{cx^2}{2}-x^3]^{3}_{0}$

$W=\frac{9c}{2}-27$

$\Delta K=11-20=-9 J$

$-9=\frac{9c}{2}-27$

$-9+27=\frac{9c}{2}$

$18=\frac{9c}{2}$

$c=\frac{2}{9}\times 18=4$

4 0

3 years ago

block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th

Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0

3 years ago