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3 years ago

Nanotechnology is technology that is about 1 nm to ______ nm in size.

Physics

2 answers:

hammer [34]3 years ago

8 0

Answer:

100

Explanation:

AnnyKZ [126]3 years ago

6 0

Explanation:

Nanotechnology is science, engineering, and technology conducted at the nanoscale, which is about 1 to 100 nanometers.

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Q2) The position of an artillery, with a speed of projectile 650, which can fire in any direction above the horizontal plane and

Wittaler [7]

Answer:

The minimum time to reach the target is 2156s

Explanation:

Check attachment

5 0

2 years ago

Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of

kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

$m=\dfrac{F}{g}$

$m=\dfrac{160\ lb}{32\ ft/s^2}$

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

$m=\dfrac{F}{g}$

$m=\dfrac{1.9\ lb}{32\ ft/s^2}$

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

$W=2300\ kg\times 32\ ft/s^2=73600\ lb$

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

$W=0.022\ kg\times 32\ ft/s^2=0.704\ lb$

Hence, this is the required solution.

5 0

3 years ago

A slug has a speed of .00242 miles per hour

Delvig [45]

I like fortnite too:):):)

4 0

3 years ago

A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto

Fed [463]

Answer:

a) a = 34.375 m / s², b) v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

v_f = $\frac{x-x_1}{t}$

we substitute the values

v_f = $\frac{ 6600 -x_1}{4}$

The initial part of the movement is carried out with acceleration

v_f = v₀ + a t

x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

x₁ = ½ a t²

v_f = a t

we substitute the values

x₁ = 1/2 a 16²

x₁ = 128 a

v_f = 16 a

let's write our system of equations

v_f = $\frac{6600 - x_1}{4}$

x₁ = 128 a

v_f = 16 a

we substitute in the first equation

16 a = $\frac{6600 -128 a}{4}$

16 4 a = 6600 - 128 a

a (64 + 128) = 6600

a = 6600/192

a = 34.375 m / s²

b) let's find the time to reach this height

x = ½ to t²

t² = 2y / a

t² = 2 5100 / 34.375

t² = 296.72

t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

v_f = 16 a

v_f = 16 34.375

v_f = 550 m / s

8 0

3 years ago

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small

gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

$T_{simple$ = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / $I$ ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and $I$ is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

$I$ = $\frac{1}{3}$ mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/ $I$ ) OR T = 2π√( $I$ /mgL)

so we can use $I$ = $\frac{1}{3}$ mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L = $\frac{1}{2}$ D.

now, substituting these equations, the period becomes;

T = 2π/√( $I$ /mgL) OR T = $2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }$ OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω $_{simple$ = 2π/ $T_{simple$ OR ω $_{simple$ = √(g/D) OR ω $_{simple$ = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × $T_{simple$

we substitute in value of $T_{simple$

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

8 0

2 years ago