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3 years ago

The atomic mass is important because?

Physics

2 answers:

Lunna [17]3 years ago

6 0

Im not quite sure but it might be a or c... ill double check

ratelena [41]3 years ago

4 0

Hello, my name is Charlie.

On this is about science.

The atomic mass is most important because it tells you the charge of the atom. <span>Because from atomic mass number it came from periodic table.</span>

The answer is B.

Hope it helped you.

-Charlie

Have a great day!

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Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

$mv_{1i}+Mv_{2i}=(m+M)v$ (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

$v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}$

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

3 0

3 years ago

In these chemical equations, each of the letters J through M represents a different element. Which equation is balanced?

Kobotan [32]

Answer:

D. 2JK3 + 3L2M --> 6LK + J2M3

Explanation:

5 0

3 years ago

Mass of 25 kg and a KE of 450 J.

anyanavicka [17]

Answer:

6m/s

Explanation:

8 0

3 years ago

An airplane is traveling at a fixed altitude with an outside wind factor. The airplane is headed N 40° W at a speed of 600 miles

bija089 [108]

Answer: $\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]$

Explanation:

Given

Plane is initially flying with velocity of magnitude $v=600\ mph$

at angle of $40^{\circ}$ with North towards west

Velocity of plane airplane can be written as

$v_a=600(-\sin 40\hat{i}+\cos 40\hat{j})$

Now wind is encountered with speed of $v=80\ mph$ at angle of $N45^{\circ}E$

$v_w=80(\cos 45\hat{i}+\sin 45\hat{j})$

resultant velocity

$\vec{v_R}=\vec{v_a} +\vec{v_w}$

$\vec{v_R}=600(-\sin 40\hat{i}+\cos 40\hat{j})+ 80(\cos 45\hat{i}+\sin 45\hat{j})$

$\vec{v_R}=\hat{i}[-385.67+56.56]+\hat{j}[459.62+56.56]$

$\vec{v_R}=\hat{i}[-329.11]+\hat{j}[516.18]$

for direction $\tan \theta =\frac{516.18}{329.11}$

$\tan \theta =1.568$

$\theta =57.47^{\circ}$ west of North

3 0

4 years ago

A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s

____ [38]

Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}$

v = 2 m/s

Hence, their speed after collision is 2 m/s.

7 0

4 years ago