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Delvig [45]
3 years ago
11

A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds

of its travel. What is its average speed during the time from t = 5 seconds to t = 10 seconds?Select one of the options below as your answer:. A. 2 meters/second. B. 3 meters/second. C. 4 meters/second. D. 5 meters/second. E. 6 meters/second . will hand out medals
Physics
2 answers:
igomit [66]3 years ago
8 0
The total distance traveled by the body is the sum of the distance it traveled for the first five seconds which is 10 m and that of the next five seconds which is 30 meters. Thus, the total distance traveled is 40 m. Dividing this by the time, will give the average speed. The average speed is therefore, 4 m/s. The answer is letter C. 
natita [175]3 years ago
7 0

Answer:

c

Explanation:

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A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
ivann1987 [24]

Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

Radius of the ring R = 86.0 m

distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,

r = 31m, g = 9.8m/s² , t = 10800sec

F = \frac{(2.1 \times 10^5 \times86^2)+(2\times3.5v10^4\times31^2) }{2\times86\times10800} \times\sqrt{\frac{9.8}{86} } \\\\F = 294.47N\\\approx294.5N

Each thruster has to applied a force of 294.5N in tangential direction

3 0
3 years ago
Two children ride side-by-side on a carousel. Their paths are shown in the image below.
Sonbull [250]

Answer:

The child represented by a star on the outside path.

Explanation:

5 0
3 years ago
A white dwarf in a binary system suddenly flares, but in a few months it settles back to its original luminosity. This process m
Anvisha [2.4K]

Answer:

A Pulsar

Explanation:

Such white dwarf is a Pulsar (from pulsating star). Three main features we find in this kind of celestial body

1. Incredibly high density, in a pulsar there is not what we can define as free interspace. Everything is mass

2.- Huge rotational speed (more than hundreds of kilometer) spinning at more than hundreds of kilometer per second

3.-Intense magnetic field

Combinations of thse features make pulsar emits important amount of energies with pricese periods of time. And if on earth we look a jet of electromagnetic radiation with a precise rigurosity (at pricese period of time) is because we are looking a pulsar spinning and each time the radiation appears is a turn of the pulsar

4 0
3 years ago
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force
liraira [26]

Answer:

Explanation:

Let the equal charges be Q .

Force between charges = k Q² / R where R is distance between charges , k = 9 x 10⁹.

Putting the charges

Force F = 9 x 10⁹ x Q² /( .72 x 10⁻²)²

= 17.36 x 10¹³ Q² N

This force equals weight so

17.36 x 10¹³ Q² = 15.7 x 10⁻³

Q² = .90 x 10⁻¹⁶

Q = .95 x 10⁻⁸

= 9.5 x 10⁻⁹

= 9.5 nC.

8 0
3 years ago
Calculate change in height of a 2kg ball moving a speed of 10m/s up a frictionless ramp until it stops
Nataliya [291]

The change in the height of the object is 5.1 m.

<h3>Conservation of mechanical energy</h3>

The principle of conservation of mechanical energy states that the total energy of an isolated system is always conserved.

The change in the height of the object is calculated by applying the principle of conservation of mechanical energy as follows;

P.E = K.E

mg \Delta h = \frac{1}{2} mv^2\\\\g \Delta h = \frac{1}{2}v^2\\\\\Delta h = \frac{v^2}{2g} \\\\\Delta h = \frac{(10)^2}{2(9.8)} \\\\\Delta h = 5.1 \ m

Thus, the change in the height of the object is 5.1 m.

Learn more about conservation of mechanical energy here: brainly.com/question/6852965

6 0
2 years ago
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