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2 years ago

Pls help i have test

Physics

2 answers:

Luden [163]2 years ago

8 0

Answer:

A. something pushes or pulls it to stop.

Explanation:

Newtons first law. hope this helps

GenaCL600 [577]2 years ago

4 0

Something pushes or pulls it to stop

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If force remains constant and acceleration decreases what must happen to the mass

blondinia [14]

That can only be happening if the mass mysteriously increased somehow. I'd like to know how in the world THAT happened.

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3 years ago

A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it hits th

Inessa05 [86]

initially coin is at rest and then it drop for total time t = 1.5 s

so here the speed of the coin at which it will hit the floor is to be find

$v_f = v_i + at$

here we know that

$v_i = 0$

a = 9.8 m/s^2

t = 1.5 s

now from above equation

$v_f = 0 + 1.5 (9.8)$

$v_f = 14.7 m/s$

so it will hit the floor with speed 14.7 m/s

3 0

3 years ago

Weightlessness is experienced by an astronaut in space. This means that the astronaut's muscles have to be stronger to move his

just olya [345]

The answer is false. The speed of the astronaut cancels out the force of gravity, causing a 'stationary freefall'. While under these effects, it is not required for an astronaut to 'strengthen' his body.

4 0

3 years ago

Read 2 more answers

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

A wind turbine works by slowing the air that passes its blades and converting much of the extracted kinetic energy to electric e

ddd [48]

Answer:

2649600 Joules

Explanation:

Efficiency = 40%

m = Mass of air = 92000 kg

v = Velocity of wind = 12 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 92000\times 12^2\\\Rightarrow K=6624000\ J$

The kinetic energy of the wind is 6624000 Joules

The wind turbine extracts 40% of the kinetic energy of the wind

$E=0.4\times K\\\Rightarrow E=0.4\times 6624000\\\Rightarrow E=2649600\ J$

The energy extracted by the turbine every second is 2649600 Joules

8 0

3 years ago