What happens to a nucleus undergoing alpha decay?

You drive 4 km at 30 km/h and then another 4 km at 50 km/h. What is your average speed for the whole 8-km trip?

Setler79 [48]

Answer:

option A

Explanation:

given,

drive 4 km at 30 km/h and

another 4 km at 50 km/h

average speed for the whole 8 Km trip = ?

time for the trip of first 4 km

distance = speed x time

$t_1 = \dfrac{d}{s}$

$t_1 = \dfrac{4}{30}$

t₁= 0.1333 hr

time from another 4 km trip

$t_2 = \dfrac{d}{s}$

$t_2= \dfrac{4}{50}$

t₂ = 0.08 hr

average speed

$s= \dfrac{d}{t}$

$s= \dfrac{8}{0.1333 + 0.08}$

$s= \dfrac{8}{0.2133}$

s = 37.50 Km/hr

Less than 40 Km/h

the correct answer is option A

6 0

3 years ago

Iron atoms have been detected in the sun's atmosphere, some with many of their electrons stripped away. What is the net electric

BaLLatris [955]

Answer:

2.88 × 10⁻¹⁸ C

Explanation:

The charge on a proton and electron are the same and it is = 1.6 × 10⁻¹⁹ C

The net charge electric charge of an iron atom = (number of proton × charge on a proton) + ( number of electron × charge on an electron ) = 26 (e) + 8 (-e) = 26 e - 8 e = 18 e = 18 × 1.6 × 10⁻¹⁹ C = 2.88 × 10⁻¹⁸ C

8 0

3 years ago

Read 2 more answers

Suppose that you measure the length of a spaceship, at rest relative to you, to be 400 m. How long will you measure it to be if

Rashid [163]

Answer:

264 m

Explanation:

The complete question is

Suppose that you measure the length of a spaceship, at rest relative to you, to be 400 m. How long will you measure it to be if it flies past you at a speed of v = 0.75c

using the length contraction relationship,

$l = l_{0} \sqrt{1 - \beta ^{2} }$

where $\beta = \frac{v}{c}$

$l$ is the relativistic length

$l_{0}$ is the actual length = 400 m

v is the velocity of the spaceship

c is the speed of light

since v = 0.75c

v/c = 0.75

substituting, we have

$l = 400 * \sqrt{1 - 0.75 ^{2} }$ = 400 x 0.66 = 264 m

5 0

3 years ago

How would a person's mass be affected by traveling to Saturn and why?

IrinaK [193]

A person's mass would not change before, during, or after a trip to Saturn, unless he gained or lost some weight as a result of unrelated eating habits, because mass doesn't depend on an object's location, gravitational environment, color, cost,
texture, temperature, political affiliation or gender orientation.

7 0

3 years ago

Three children are riding on the edge of a merry‑go‑round that has a mass of 105 kg105 kg and a radius of 1.80 m1.80 m . The mer

rewona [7]

Answer:

New angular speed changes from 16 rpm to 19.95 rpm

Explanation:

In the given system the angular momentum of the system is conserved

initial angular momentum of the system

$L_{i}=I_{1}\omega _{1}\\\\\omega _{1}=\frac{8\pi }{15}rad/sec\\\\I_{1}=\frac{1}{2}mr^{2}+(m_{1}+m_{2}+m_{3})r^{2}\\\\I_{1}=\frac{1}{2}\times 105\times 1.8^{2}+(33+28+28) \times 1.8^{2}=458.46kgm^{2}\\\\\therefore L_{i}=458.46\times \frac{8\pi }{15}=768.16$

When 28 kg child moves to the center the moment of inertia changes thus we have

$I_{2}=\frac{1}{2}mr^{2}+(m_{1}+m_{2})r^{2}\\\\I_{1}=\frac{1}{2}\times 105\times 1.8^{2}+(33+28) \times 1.8^{2}=367.74kgm^{2}\\\\\therefore L_{f}=367.74\times \omega _{f}$

Equating initial and final angular momentum we get

$I_{2}=\frac{1}{2}mr^{2}+(m_{1}+m_{2})r^{2}\\\\I_{1}=\frac{1}{2}\times 105\times 1.8^{2}+(33+28) \times 1.8^{2}=367.74kgm^{2}\\\\\therefore L_{f}=367.74\times \omega _{f}\\\\\omega _{f}=\frac{768.16}{367.74}=2.1rad/sec\\\\N_{2}=19.95rpm$

7 0

3 years ago