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Butoxors [25]
3 years ago
10

Astronaut Rob leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. Rob's destination is a star sys

tem 14.4 light-years away (one light-year is the distance light travels in one year). Relative to a frame of reference that is fixed with respect to Earth, how long does it take Rob to complete the trip?
A) 15.0 years
B) 22.7 years
C) 4.20 years
D) 14.4 years
E) 10.8 years
Physics
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

A) 15.0 years

Explanation:

Due to the distance to the star system is in light-year units, we can compute the time by using:

t=\frac{d}{v}=\frac{14.4 l-y}{0.960}=15l-y

then, Rob will take to complete the trip about 15 light-years.

hope this helps!!

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A student is considering doing a complete repeated measures design experiment involving motor skills. The student's advisor has
Lelu [443]

Answer:

<em>c. ABBA counterbalancing </em>

Explanation:

The student should not use the method because it is a progressive error management technique for each subject by introducing all <em>treatment circumstances twice, first in one sequence, then in the other (AB, BA) by subject counterbalancing.</em>

If participants experience conditions more than once, they experience the conditions first in one order, then the opposite order.

3 0
3 years ago
Children are told to avoid standing too close to a rapidly moving train because they might get sucked under it. Is this possible
storchak [24]

Answer:

no its not like the undertow in the ocean

Explanation:

4 0
3 years ago
ANSWER PLS NEED ASAP AS FAST AS HUMANLY POSSIBLE
Soloha48 [4]

Answer:

Should be B

Explanation:

As the light enters the prism at an angle, the glass causes the light to travel slower. The speed is affected by the frequency of the light. So each frequency is slowed by a different amount, causing the colors to separate.

Hope this helps!

6 0
3 years ago
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0
3 years ago
An ordinary flashlight battery has a potential difference of 1.2 V between its positive and negative terminals. How much work mu
Maru [420]

The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.

Given:

Potential difference, V = 1.2 V

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Calculation:

We know that the work done to transport an electron from the positive to the negative terminal is given as:

W.D = (Charge on electron)×(Potential difference)

       = e × V

       = (1.6 × 10⁻¹⁹ C)×(1.2 V)

       = 1.92 × 10⁻¹⁹ J

Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.

Learn more about work done on a charge here:

<u>brainly.com/question/13946889</u>

#SPJ4

4 0
2 years ago
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