How can you use biome in a sentence?

A 2,000 kg car travels with a tangential

Gekata [30.6K]

A= v²/R
a = 12²/30 =4.8 m/s²

3 0

3 years ago

A uniform plank of mass 10kg and length 10m rests on two supports, A and B as shown. A boy of weight 500N stands at a distance o

kifflom [539]

Answer:

U² = 142.86 N

U¹ = 357.14 N

Explanation:

Taking summation of the moment about point A, we get the following equilibrium equation: (taking clockwise direction as positive)

$W(2\ m) - U^2(7\ m) = 0$

where,

W = weight of boy = 500 N

U² = reaction ay B = ?

Therefore,

$(500\ N)(2\ m)-(U^2)(7\ m)=0\\U^2=\frac{1000\ Nm}{7\ m}\\$

Now, taking summation of forces on the plank. Taking upward direction as positive, for equilibrium position:

$W-U^1-U^2=0\\500\ N - 142.86\ N = U^1\\$

3 0

3 years ago

What is one way to describe elements

Talja [164]

Answer:

They are the smallest pancise of any matere

4 0

3 years ago

The older, geocentric model of the solar system placed the__________ at the center. Today's heliocentric model places the_______

DedPeter [7]

the answer is earth for the first one and sun for the second one

3 0

3 years ago

Read 2 more answers

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

3 years ago