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3 years ago

The data table shows some data related to the Sun and the planets in our solar system.

Physics

2 answers:

AlexFokin [52]3 years ago

8 0

You have selected the correct answer and blobbed over it with your pencil.

I assume you must have looked at Saturn's average distance, found 1427,
divided that number by 6, got 237 and change, then looked at the others,
and found that 228 was the only one that's anywhere close.

DerKrebs [107]3 years ago

7 0

Answer:

Mars

Explanation:

The distance between Sun and Saturn = 1427 Million km

one sixth of the distance between Sun and Saturn will be

$= \frac{1427\times 10^{6} }{6} km$

= 237.83 Million km

From the given data it is evident that the right answer will be Mars.

As the distance between Mars and Sun is 228 million km which is the closest value to the one sixth of the distance between Sun and Saturn.

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the car starts from a stop to travel 1100 meters in 14 seconds. it is clocked at 65 m/s at point k. find its average speed and a

inysia [295]

Answer:

The average velocity of the car is, V = 74.04 m/s

Explanation:

Given data,

The initial velocity of the car, u = 0 m/s

The displacement of the ca, S = 1100 m

The time period of travel, t = 14 s

The velocity of the car at point k, v = 65 m/s

Using the II equation of motion,

S = ut + ½ at²

Substituting the given values,

1100 = 0 + ½ x a x 14²

a = 11.22 m/s²

Using the III equation of motion

v² = u² + 2 as

v = √(2as) (∵ u = 0)

Substituting,

v = √(2 x 11.22 x 1100)

= 157.11 m/s

The average speed of the car,

$V=\frac{0+65+157.11}{3}$

V = 74.04 m/s

Hence, the average velocity of the car is, V = 74.04 m/s

4 0

3 years ago

A ____(concave, convex) mirror is used to collect light in a reflecting telescope

kati45 [8]

Im pretty sure its Concave.

4 0

3 years ago

Read 2 more answers

A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in se

Rudik [331]

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

$x = 32t - 38\frac{t^3}{3} \\\\$

The velocity of the particle is calculated as the change in the position of the particle with time;

$v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s$

Acceleration is the change in velocity with time;

$a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2$

4 0

3 years ago

Who was albert einstien describe his contributions???

Nimfa-mama [501]

Answer:

Albert Einstein was from Germany he was born theoritical physicist, from childhood only he loved mechanical toys, he was highly gifted in Mathematics, He was a world citizen and a scientific genius too. His contribution were:

1) he developed the theory of relativity

2) he also discovered the process of nuclear fission

3)he developed the quantum theory of specific heat

4)theory of stimulated emission, on which laser device technology is based

5) law of photoelectric effect

hope it helped you :)

8 0

3 years ago

A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe

joja [24]

Answer:

The Doppler Effect is given by the following relation;

$f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f$

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

$v_o$ = The velocity of the observer

$v_s$ = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(ii) When the source is receding from the observer, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

$f' = \left (\dfrac{v }{v + v_s} \right) \times f$

(1) Given that (v + $v_s$ ) > v, therefore, v < (v + $v_s$ ), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

$f' = \left (\dfrac{v }{v - v_s} \right) \times f$

(1) Given that (v - $v_s$ ) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

Explanation:

7 0

3 years ago