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White raven [17]

3 years ago

10

Astronaut Rob leaves Earth in a spaceship at a speed of 0.960crelative to an observer onEarth. Rob's destination is a star syste

m 14.4 light-years away (one light-year is the distance lighttravels in one year). Relative to a frame of reference that is fixed with respect to Earth, how longdoes it take Rob to complete the trip?

1 answer:

Slav-nsk [51]3 years ago

5 0

To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to

$1Ly =9.4605284*10^{15}m \rightarrow 'Ly'$ means Light Year

Then

$14.4Ly = 1.36231609*10^{17} m$

If we have that

$v= \frac{x}{t} \rightarrow t = \frac{x}{t}$

Where,

v = Velocity

x = Displacement

t = Time

We have that

$t = \frac{1.36231609*10^{17}}{0.96c} \rightarrow c$ = Speed of light

$t = \frac{1.36231609*10^{17}}{0.96(3*10^8)}$

$t= 454105363 s (\frac{1hour}{3600s})$

$t= 126140 hours(\frac{1day}{24hours})$

$t= 5255.85 days(\frac{1 year}{365days})$

$t = 14.399 years$

Therefore will take 14.399 years

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Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the fre

liq [111]

Answer: D. the wave has traveled 97.2 cm in 1 second.

Explanation:

Got it right on edge, hope this helps :)

7 0

3 years ago

A neutral atom of tin (Z = 50) has 69 neutrons. (a) How many protons does it have? (b) What is its mass number?

Olenka [21]

Answer:

Protons = 50

Mass number = 119

Explanation:

A neutral of tin, Sn, has a Z value of 50 and neutrons are 69.

Z is the atomic number. The atomic number is equal to number of protons or number of electrons in an atom that is neutral i.e:

Z = PROTONS = ELECTRONS

So the number of protons = 50 since Z is 50

>>>>>>>>>>>>>>>¦

The bulk of the mass of an atom is concentrated in the nucleus. To calculate mass number, we simply sum all the elementary particles in the nucleus i.e protons and neutrons. These particles are called the nucleons.

Mass number (A) = P + N

= 50 + 69

= 119

3 0

3 years ago

In the compound MgS, the sulfide ion has 1. lost one electrons. 2. lost two electrons. 3. gained one electron. 4. gained two ele

never [62]

Answer:

4.has gained two electrons

Explanation:

There exist electrovalent bonding the compound MgS . In electrovalent bonding, there is a transfer of electrons from the metal to non-metal.

Magnesium atom has an atomic number 12 and its electron configuration is 2,8,2

Sulfur atom , a non-metal has atomic number of 16 and its electron configuration = 2,8,6

This means that magnesium as a metal needs to loose two electrons from its valence shell to attain its stable structure.Also sulfur requires two more electron to achieve its octet structure.

Hence a transfer of electrons will take place from magnesium atom to sulfur atom, sulfur gaining two electrons.

6 0

3 years ago

Read 2 more answers

A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t

Shalnov [3]

Answer:

a) yield strength

$\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa$

b) modulus of elasticity

strain calculation

$\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046$

strain for offset yield point

$\varepsilon_{new} = \varepsilon_0 -0.002$

=0.0046-0.002 = 0.0026

now, modulus of elasticity

$E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}$

= 184615.28 MPa = 184.615 GPa

c) tensile strength

$\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa$

d) percentage elongation

$\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%$

e) percentage of area reduction

$\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%$

7 0

3 years ago

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

$\theta=35^{\circ}$

a.Let $v_0$ be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

$v_x=v_0cos\theta=v_0cos35$

$v_y=v_0sin\theta=v_0sin35$

$x=v_0cos\theta\times t=v_0cos35\times t$

$t=\frac{30}{v_0cos35}$

$h=v_yt-\frac{1}{2}gt^2$

Substitute the values

$1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2$

$1.8=30tan35-\frac{6574.6}{v^2_0}$

$\frac{6574.6}{v^2_0}=21-1.8=19.2$

$v^2_0=\frac{6574.6}{19.2}$

$v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s$

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

$t=\frac{30}{18.5cos35}$

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0

3 years ago