Identifying Sources Used to Study Earth's Interie

Which would be the most reliable source for information about the toxicity of an industrial chemical?

kvasek [131]

A scientific journal article that is peer reviewed. This is because it is more likely not have factual information and sources to that information.

3 0

2 years ago

Read 2 more answers

What is the function of the wire in

Naily [24]

Answer:

I think the right answer is option B.

3 0

3 years ago

A body moving with an acceleration 2 m/s?then what is the change in velocity in 4sec.

enot [183]

Answer:

As Per Provided Information

Moving body has 2m/s² acceleration

Time taken by body is 4 second

We are asked to find the 'change in velocity' ( ∆V) by the body.

Formula Used here

$\boxed{\bf{\Delta \: V \: = acceleration \: \times time \:}}$

Substituting the given value

$\sf\longrightarrow\Delta\:V \: = 2 \times 4 \\ \\ \\ \sf\longrightarrow\Delta\:V \: =8m {s}^{ - 1}$

Therefore,

Change in velocity is 8 m/s

7 0

1 year ago

Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has

a_sh-v [17]

Answer:

h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

Em₀ = U = m g h

final point. Lower, just before the crash

Em_f = K = ½ m $v_{b}^2$

energy is conserved

Em₀ = Em_f

m g h = ½ m v²

v_b = $\sqrt{2gh}$

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

p₀ = 2m 0 + m v_b

final instant. Right after the crash

p_f = (2m + m) v

the moment is preserved

p₀ = p_f

m v_b = 3m v

v = v_b / 3

v = ⅓ $\sqrt{2gh}$

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

Em₀ = K = ½ 3m v²

Final point. Higher

Em_f = U = (3m) g h'

Em₀ = Em_f

½ 3m v² = 3m g h’

we substitute

h’= $\frac{v^2}{2g}$

h’ = $\frac{1}{3^2} \ \frac{ 2gh}{2g}$

h’ = 1/9 h

6 0

2 years ago

A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet. If the bullet has a mass of 22.

Bogdan [553]

Answer:

A. 1.172 metres

B. 6.82 Ns

C. 4.796 m/s

Explanation:

The total initial momentum is gotten by multiplying the mass and initial velocity of the both bodies.

The 1.40 kg block is at rest so velocity is zero and has no momentum.

The bullet of mass 22 g = 0.022 kg with velocity of 310 m/s

Momentum = 310*0.022

Momentum = 6.82 Ns.

If the bullet gets embedded they will both have common velocity v

6.82 = (0.022+1.40)v

6.82 = 1.422v

V = 6.82/1.422

V = 4.796 m/s

How high the block will rise after the bullet is embedded is given by

H = (U²Sin²tita)/2g

Where tita is 90°

H = (4.796² * sin²(90))/(2*9.81)

H =( 23.001616*1)/19.62

H = 1.172 metres

6 0

3 years ago