How can the momentum of an object be changed?

An atom of sodium has a larger mass than an atom of neon. How is this possible when neon has so many more valence electrons?

jeka94

The electrons contribute just about zero to the mass of an atom.
It takes more than 1,800 electrons to make the mass of one
proton or neutron.

The naturally occurring element with the most complex atom is
Uranium. That's element #92 , so a neutral uranium atom has
92 electrons. It would take almost exactly 20 times that many
electrons to add the mass of one proton or neutron to the atom!
(And no other element has that many electrons in an atom of it.)

5 0

4 years ago

Use the information from the graph to answer the question. What is the acceleration of the object

Art [367]

Answer:

-2.5 m/s²

Explanation:

The acceleration of a body is the change in it's velocity with time.

The change in velocity with time can be obtained as the slope of a velocity time graph ;

Acceleration = (change in velocity / change in time)

Taking the slope :

Change in Velocity = △y = y2 - y1

Change in time = △x = x2 - x1

(10, 15) ; (0, 40)

△y / △x = y2 - y1 / x2 - x1 = (40 - 15) / (0 - 10)

△y / △x = 25 / - 10 = - 2.5 m/s²

4 0

3 years ago

The y component of the electric field of an electromagnetic wave travelling in the +x direction through vacuum obeys the equatio

notsponge [240]

Answer:

λ =8.57 μ m

Explanation:

Given that

Ey = 375 cos [kx − (2.20 × 10¹⁴ rad/s)t] N/C

Standard form

Ey=Eo cos[k x-ωt] N/C

By comparing the given equation with the standard wave equation

Eo = 375 N/C

ω = 2.20 × 10¹⁴ rad/s

We know that ω = 2 π f

$f=\dfrac{\omega}{2\pi }$

$f=\dfrac{2.2\times 10^{14}}{2\pi }\ Hz$

f=3.50×10¹³ Hz

We know that the velocity given as

V = f λ

λ =Wavelength

V=Speed = 3 x 10⁸ m/s

$\lambda =\dfrac{V}{f }$

$\lambda =\dfrac{3\times 10^8}{3.5\times 10^{13}}\ m$

λ =0.00000857 m ( 1 μ m = 10⁶ m)

λ =8.57 μ m

8 0

3 years ago

A 77.8 kg basketball player is running in the positive direction at 8.1 m/s. She is met head-on by a 99.8 kg player traveling at

lilavasa [31]

Answer:

-5.24 m/s

** The minus sign indicates that the velocity vector points in the opposite direction with respect to the initial direction of the 77.8 kg player **

Explanation:

Hi!

We can solve this problem considering each player as a point particle and taking into account the conservation of linear momentum.

Since the 99.8 kg player is moving towards the 77.8kg, the initial total momentum is:

m1*v1_i + m2*v2_i = (77.8kg)(8.1 m/s) - (99.8kg)(6.9 m/s)

** The minus sign indicates that the velocity vector points in the opposite direction with respect to the initial direction of the 77.8 kg player **

The final total momentum is equal to:

m1*v1_f + m2*v2_f = (77.8 kg)v1_f + (99.8 kg)(3.5 m/s)

The conservation of momentu tell us that:

m1v1_i + m2v2_i = m1v1_f + m2v2_f

Therefore:

v1_f =v1_i + (m2/m1)*(v2_i-v2_f)

v1_f = 8.1 m/s + (99.8 / 77.8) * (-6.9 - 3.5 m/s)

6 0

3 years ago

A bullet with mass m = 0.1 kg grams hits a ballistic pendulum with length L = 3 meters and mass M = 2 kg and lodges in it. When

Mnenie [13.5K]

Answer:

e) v₁ = 29.7 m / s

Explanation:

Let's propose the solution of the problem, let's start at the moment

Initial

p₀ = m v₁

Final

$p_{f}$ = (m + M) v

The moment is preserved

p₀ = $p_{f}$

m v₁ = (m + M) v

v = m / (m + M) v₁ (1)

a) energy is conserved

Let's look for kinetic energy

Initial

K₀ = ½ m v₁²

Final

$K_{f}$ = ½ (m + M) v²

Let's replace v

$K_{f}$ = ½ (m + M) [m / (m + M) v₁]²

$K_{f}$ = ½ m² / (m + M) v₁²

Let's look for the relationship of these energies

Ko / $K_{f}$ = ½ m v₁² / (½ m² / (m + M) v₁²)

Ko / $K_{f}$ = (m + M) / m = (1 + M / m)²

The kinetic energy changes therefore it is not conserved in the process, the missing energy is converted into potential heat energy during the crash

b) The impulse is conserved because the system is defined as formed by the two bodies and the externals are of action and reaction, so for the complete system the sum is zero and the moment does not change in value

c) in this case the system is already formed by the two bodies and since there is no rubbing the mechanical energy is conserved, transforming from kinetics to potential

d) when the pendulum oscillates the speed changes from v to zero, so the moment is not conserved, this is because there is an external force acting on the system, the force of gravity

e) For this part let's start at the end of the movement

It is system (bullet + block) moves, energy is conserved

Final. Highest point

$Em_{f}$ = U = (m + M) g h

Initial. Lowest point

Em₀ = K = ½ (m + M) v2

Em₀ = $Em_{f}$

½ (m + M) v² = (m + M) g h

v = √ 2gh

Let's look for the height (h) by trigonometry

Cos 15 = x / L

h = L-x

h = L - L cos 15

h = L (1- cos 15)

We replace

v = √ (2gL (1- cos 15))

Now we use equation (1) of momentum conservation

v = m / (m + M) v1

v₁ = (m + M) / m v

v1 = (0.1 +2.0) /0.1 RA (2 9.8 3 (1- cos 15))

v₁ = 21 √ (2.00)

v₁ = 29.7 m / s

6 0

3 years ago