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3 years ago

It is the obligation of researchers to review and comment on the research of other researchers. True or False.

2 answers:

8 0

It is false that it is the obligation of researchers to review and comment on the research of other researchers. It is not their obligations - they don't have to do it, although they can if they want to and if they are allowed by the author him or herself. However, they are not bound by law or something like that to do this, it's just due to their kindness or genuine interest that they do this.

false

romanna [79]3 years ago

4 0

<span>It is false that it is the obligation of researchers to review and comment on the research of other researchers. It is not their obligations - they don't have to do it, although they can if they want to and if they are allowed by the author him or herself. However, they are not bound by law or something like that to do this, it's just due to their kindness or genuine interest that they do this.</span>

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Answer:

Explanation:19,2 or 0/4 or 5 or 40,4

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3 years ago

During an experiment, a scientist places a heat lamp above a bowl of water and uses the lamp to heat up the water. How does heat

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3 years ago

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jolli1 [7]

Answer:

The second answer from the top, no the energy in the wave pushed the water particles from above the earthquake in the opposite direction.

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3 years ago

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A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t

HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0

4 years ago

Two people stand facing each other at roller skating rink then push off each other

9966 [12]

a) 0 kg m/s

b) 0 kg m/s

c) +3 m/s

d) 60 N

Explanation:

The momentum of an object is a vector quantity given by:

$p=mv$

where

m is the mass of the object

v is the velocity of the object

In this problem, we have a system of two people, so the total momentum will be the sum of the individual momenta of the two people:

$p=p_1 + p_2$

Which can be rewritten as

$p=m_1 u_1 + m_2 u_2$

where $m_1,m_2$ are the masses of the two people and $u_1,u_2$ their initial velocities.

However, the two people are initially at rest, so

$u_1 = 0\\u_2 = 0$

Therefore the total momentum is

$p=0+0=0$

The principle of conservation of momentum states that when there are no external forces acting on a system, the total momentum of the system is conserved, so we can write:

$p_i = p_f$

where

$p_i$ is the total momentum of the system before

$p_f$ is the total momentum of the system after

In this problem,

$p_i = 0$

As we calculated in part a: this is because the total momentum of the two people before they push off each other is zero.

Therefore, according to the law of conservation of momentum,

$p_f = p_i = 0$

So the total momentum is zero also after they push off each other.

The total momentum of the girl and the boy after they push off each other can be written as:

$p_f = m_1 v_1 + m_2 v_2$ (1)

where:

$m_1 = 30 kg$ is the mass of the girl

$v_1 = -5 m/s$ is her velocity (she moves backward, so the negative sign)

$m_2 = 50 kg$ is the mass of the boy

$v_2$ is the velocity of the boy

As calculated in part b), we also know that the total momentum of the girl and the boy is

$p_f = 0$ (2)

By combining eq(1) and eq(2) we get

$0=m_1 v_1 + m_2 v_2$

And solving for v2 we find the velocity of the boy:

$v_2=-\frac{m_1 v_1}{m_2}=-\frac{(30)(-5)}{50}=+3 m/s$

and the positive sign means he is moving forward.

We can solve this part by applying the impulse theorem, which states that the change in momentum of an object is equal to the product between the force applied on it and the duration of the collision:

$\Delta p = F\Delta t$