Which type of mirror causes light rays to refract away from each other?

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

6 0

2 years ago

A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int

Inessa05 [86]

Let F be the magnitude of the force applied to the cart, m the mass of the cart, and a the acceleration it undergoes. After time t, the cart accelerates from rest v₀ = 0 to a final velocity v. By Newton's second law, the first push applies an acceleration of

F = m a → a = F / m

so that the cart's final speed is

v = v₀ + a t

v = (F / m) t

If we force is halved, so is the accleration:

a = F / m → a/2 = F / (2m)

So, in order to get the cart up to the same speed v as before, you need to double the time interval t to 2t, since that would give

(F / (2m)) (2t) = (F / m) t = v

3 0

2 years ago

What role does the somatic nervous system play?

kodGreya [7K]

Explanation:

The somatic nervous system coordinates the voluntary control of movement in the body system.

The nervous system is made up of the central nervous system and peripheral nervous system.

The peripheral nervous system is made up of the nerves that branches into the body from the spinal cord.

It consist of the somatic nervous system and the autonomous nervous system.
The somatic nervous system controls the voluntary actions of movement in the body.
It coordinates the reflex arc.

Learn more:

Nervous system brainly.com/question/4662759

#learnwithBrainly

4 0

3 years ago

If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.

Darina [25.2K]

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT = 15 $^0C$ = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

c = $\frac{2050}{150*10^{-3}*15} = 911.11 J/kgK$

So object's specific heat = 911.11 J/kgK

5 0

2 years ago

A child and sled with a combined mass of 48.8 kg slide down a frictionless hill that is 7.05 m high. If the sled starts from res

nexus9112 [7]

Answer:

Speed at bottom of the hill (v) = 11.74 m/s

Explanation:

Given:

Combined mass = 48.8 kg

Height h = 7.05 m

Find:

Speed at bottom of the hill (v)

Computation:

v² = 2gh

v = √2 x 9.8 x 7.05

v = √138.18

v = 11.74 m/s

Speed at bottom of the hill (v) = 11.74 m/s

7 0

3 years ago