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Nadusha1986 [10]
3 years ago
10

What subject is he......... now? Vietnamese.A. to learnB. learnC. learningD. learned

Engineering
1 answer:
otez555 [7]3 years ago
5 0

Answer:

<h3>What subject is he Learning now? (≧∇≦)</h3><h3 />
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Where can you go to find the system requirements?
SSSSS [86.1K]

Answer:c

Explanation:

yea, jus go to the systems help center then youll be confronted with a talks. person n they'll walk u threw it.

8 0
1 year ago
Can anyone answer this question
Solnce55 [7]

Answer:

a) 2.18 m/s^2

b) 9.83 m/s

Explanation:

The flywheel has a moment of inertia

J = m * k^2

Where

J: moment of inertia

k: radius of gyration

In this case:

J = 144 * 0.45^2 = 29.2 kg*m^2

The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.

T = p * r

r is the radius of the wheel.

T = m1 * g * r

T = 18 * 9.81 * 0.6 = 106 N*m

The torque will cause an acceleration on the flywheel:

T = J * γ

γ = T/J

γ = 106/29.2 = 3.63 rad/s^2

SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:

at = γ * r

at = 3.63 * 0.6 = 2.81 m/s^2

Now that we know the acceleration of the block we can forget about the flywheel.

The equation for uniformly accelerated movement is:

X(t) = X0 + V0*t + 1/2*a*t^2

We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:

X(t) = 1/2*a*t^2

Rearranging

t^2 = 2*X(t)/a

t = \sqrt{\frac{2*X(t)}{a}}

t = \sqrt{\frac{2*18}{2.81}} = 3.6 s

It will reach the 1.8 m in 3.6 s.

Now we use the equation for speed under constant acceleration:

V(t) = V0 + a*t

V(3.6) = 2.81 * 3.6 = 9.83 m/s

7 0
4 years ago
End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaf
sergiy2304 [10]

Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα +  ∑Mad    ...............1

put here value and we get

so here I = ( \frac{1}{12} ) × m × L²    ................2

I = ( \frac{1}{12} ) × 5 × 0.8²

I = 0.267 kg-m²  

and

a  is =  r × α    

a  = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x)    ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 -  5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y)    .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81  = 5 × 0.4 × 2²

a(y) = 57.1 N

so

total force at A will be

Fa = \sqrt{a(x)^2+a(y)^2}    ...............5

Fa = \sqrt{(57.1)^2+(5)^2}  

Fa = 57.32 N

3 0
4 years ago
What is referred to as "Pyroelectric" materials?
aliina [53]

Answer and Explanation:

Pyroelectric material

Pyroelectric materials have special property of generating potential difference (although it is very less ) when these material are treated with heat or when celled down.

The potential difference generated is for very less time

The generation of potential difference is due to change in position of atoms after heating or cooling.

5 0
3 years ago
Question 8 of 10
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Gtfggyfbkuvjhkbghvfycuyivty
6 0
3 years ago
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