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Ostrovityanka [42]

3 years ago

6

What element is stored in the shells and skeletons of living organisms

2 answers:

trasher [3.6K]3 years ago

8 0

Hown work
$2$

LuckyWell [14K]3 years ago

3 0

Carbon is the answer

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_honest person is always at peace<br>

borishaifa [10]

Hmmm I dunno buddy sorry

5 0

3 years ago

How does work affect energy between objects so it can cause a change in the form of energy? Work transfers energy. Work changes

IceJOKER [234]

'Doing work' is a way of transferring energy from one object to another, energy is transferred when a force moves through a distance.

If i explain with formula

Work done (J) = Energy transferred (J)

So more energy, more work done bc u transferred more energy to move the object and doing the work. and if you only use a little of energy, the work done also only a little.

4 0

3 years ago

What does Newton's third law describe?

Dennis_Churaev [7]

A. Equal and opposite forces

Explanation:
History law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exert an equal and opposite force on object A

7 0

3 years ago

A block weighing 3.7 kg is suspended from the ceiling of a truck trailer by a hanging bungee cord. The cord has a cross-sectiona

sweet [91]

Answer:

$Y = 2.27 \times 10^{10} N/m^2$

Explanation:

Natural length of the string is given as

$L_o = 43 cm$

length of the string while block is hanging on it

$L = 53 cm$

extension in length is given as

$\Delta L = 10 cm$

now we have strain in the string is given as

$strain = \frac{\Delta L}{L}$

$strain = \frac{{10 cm}{43 cm}$

$strain = 0.23$

similarly we will have cross-sectional area of the string is given as

$A = 40 \times 10^{-6} m^2$

now the stress in the string is given as

$Stress = \frac{T}{A}$

$Stress = \frac{mg}{A}$

$Stress = \frac{3.7 \times 9.81}{40 \times 10^{-6}}$

$stress = 9.07 \times 10^5 N/m^2$

Now Young's Modulus is given as

$Y = \frac{stress}{strain}$

$Y = \frac{9.07 \times 10^5}{40\times 10^{-6}}$

$Y = 2.27 \times 10^{10} N/m^2$

5 0

3 years ago

A ball is dropped from rest at a point 12 m above the ground into a smooth, frictionless chute. The ball exits the chute 2 m abo

Nonamiya [84]

Answer:

29,7 m

Explanation:

We need to devide the problem in two parts:

A) Energy

B) MRUV

<u>Energy:</u>

Since no friction between pint (1) and (2), then the energy conservatets:

Energy = constant ----> Ek(cinética) + Ep(potencial) = constant

Ek1 + Ep1 = Ek2 + Ep2

Ek1 = 0 ; because V1 is zero (the ball is "dropped")

Ep1 = m*g*H1

Ep2= m*g*H2

Then:

Ek2 = m*g*(H1-H2)

By definition of cinetic energy:

m*(V2)²/2 = m*g*(H1-H2) ---> $V2 = \sqrt{(2*g*(H1-H2)}$

Replaced values: V2 = 14,0 m/s

<u>MRUV:</u>

The decomposition of the velocity (V2), gives a for the horizontal component:

V2x = V2*cos(α)

Then the traveled distance is:

X = V2*cos(α)*t.... but what time?

The time what takes the ball hit the ground.

Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²

In the vertical axis:

Y3 = 0 ; Y2 = H2 = 2 m

Reeplacing:

-2 = 14*t + (1/2)*(-9,81)*t²

solving the ecuation, the only positive solution is:

t = 2,99 sec ≈ 3 sec

Then, for the distance:

X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m

6 0

4 years ago

Read 2 more answers