What scientific breakthrough did Charles Darwin conclusion lead to? ANSSER ASAP PLZ

002 10.0 points note: report your answer in joules here. a proton and an antiproton, both at rest with respect to one another, m

Ulleksa [173]

The mass of the proton is:
$m_p = 1.67 \cdot 10^{-27} kg$
and the mass of the antiproton is exactly the same, so the total mass of the two particles is $2m_p$ .

In the annihilation, all the mass of the two particles is converted into energy, and the amount of this energy is given by Einstein's equivalence between mass and energy:
$E=Mc^2$
where M is the mass converted into energy and c is the speed of light. In this example, $M=2m_p$ , therefore the energy released is
$E=2m_p c^2 = 2 (1.67 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2=3 \cdot 10^{-10}J$

3 0

3 years ago

A student launches a small 0.5 kg rocket with an initial speed of 30 m/s at an angle of 60°. Approximately how much time will th

Elena L [17]

Answer:

$t=5.30s$

Explanation:

The high reached by a proyectile in an uniformly accelerated motion is given by:

$y=v_{0y}t-\frac{gt^2}{2}$

The time that the rocket spends in the air is obtained for y = 0, since this is the time that the rocket travels before touching the ground. Recall that $v_{0y}=v_0sin\theta$ . Solving for t:

$0=(v_0sin\theta) t-\frac{gt^2}{2}\\\frac{gt}{2}=v_0sin\theta\\t=\frac{2v_0sin\theta}{g}\\t=\frac{2(30\frac{m}{s})sin(60^\circ)}{9.8\frac{m}{s^2}}\\t=5.30s$

5 0

4 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0

4 years ago

Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to

jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

7 0

2 years ago

Define distance and displacement with illustration

Deffense [45]

Answer:

<h2>Distance</h2>

The length of the actual path travelled by a body is called distance travelled by a body.It is a scalar Quantity.It is measured in meter(m) in SI system.

<h2>Displacement</h2>

The shortest distance from initial position to the final position of a body is called displacement of the body.It is a vector Quantity.It is measured in meter(m) in SI system..

Please see the attached picture...

It is the example of distance and displacement....

Hope this helps...

Good luck on your assignment...

3 0

3 years ago