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xz_007 [3.2K]
3 years ago
8

In 1969, Apollo 11 landed on the Moon. It took four days for the spacecraft to travel from Earth to the Moon. In 2006, New Horiz

ons left Earth and reached Pluto in 2015. Based on this information, what is the most reasonable estimate for a spacecraft to travel from Earth to Mars?
Physics
1 answer:
Bess [88]3 years ago
7 0

Answer:

An estimate for the time it will take for a spacecraft to travel from Earth to Mars is approximately 138.8 days

Explanation:

The distance between Earth and the Moon = 684,400 km

The distance between Earth and Mars = 220.58 × 10⁶ km

The distance between Earth and Pluto = 5.2241 × 10⁹ km

The ratio of the distance between Earth and Pluto and the distance between Earth and Mars = (5.2241 × 10⁹ km)/(220.58 × 10⁶ km) ≈ 23.683

It took 2006 to 2015 (9 years) to travel from Earth to Pluto, therefore, it can take approximately (9 years)/(23.683) ≈ 0.38 of a year which is ((9 years)/(23.683)) × 365.2422 ≈ 138.8 days for a spacecraft to travel from Earth to Mars

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A person jogs for 4.0 km in 32 mins,then 2.0km in 22mins,and finallu 1.0 km in 16 mins.whqt is the joggers average speed in km p
blondinia [14]

Average speed = (total distance covered) / (time to cover the distance)

total distance covered = (4km + 2km + 1km) = 7 km

time to cover the distance = (32min + 22min + 16min) = 70 min

Average speed = (7 km) / (70 min)

Average speed = 0.1 km/minute

5 0
3 years ago
3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0
3 years ago
You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee
Archy [21]

Answer:

V = 4.81 V

Explanation:

  • As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
  • We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

        V_{rint} = I* r_{int}

  • The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

       V = V_{b} - V_{rint}  = 5.03 V = 6.0 V - 0.383 A* r_{int}

  • We can solve for rint, as follows:

         r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega

  • When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

       V = V_{b} - V_{rint}  = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V

  • As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.
5 0
3 years ago
When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is
lukranit [14]

Answer:

The angle of incidence when the reflected ray is perpendicular to the incident ray = 45°

Explanation:

According to Snell's Law,

n₁ sin θ₁ = n₂ sin θ₂

When the angle between the incident ray and reflected ray is 90°, the angle of incidence is θ₁ and the angle of reflection, θ₂ = 90° - θ₁ and the index of refraction in the Snell's Law for both media would be the same, n₁ = n₂ = n

n sin θ₁ = n sin (90° - θ₁)

Note that from trigonometric relations,

Sin (90° - θ₁) = cos θ₁

n sin θ₁ = n cos θ₁

(sin θ₁)/(cos θ₁) = 1

tan θ₁ = 1

θ₁ = arctan 1 = 45°

Hope this Helps!!!

7 0
3 years ago
Read 2 more answers
GETTING TIMED PLS HELP! Do the runners in the picture above represent kinetic or potential energy? you need to explain why.
prohojiy [21]

Answer:

<em>They represent kinetic energy</em>

Explanation:

<u>Kinetic Energy </u>

A body can do work due to some of its attributes or states. For example, its mass can do work if used to provide energy, if the object is at a certain height respect to some reference level, it can do work when going downwards (potential energy), if the object moves at a certain speed, it can do work when transferring part of its speed to other objects. It's called kinetic energy and is given by

\displaystyle K=\frac{mv^2}{2}

Both runners are moving in a horizontal path, thus they have kinetic energy, given by the above equation. If they could jump below ground level, then they will also have potential energy

8 0
3 years ago
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