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AnnZ [28]
3 years ago
9

Every year, new records in track and field events are recorded. Let's take an historic look back at some exciting races.

Physics
2 answers:
Alexeev081 [22]3 years ago
5 0

Answer:

Average velocity = 10.08 m/s

Explanation:

Given

Length of the track d = 100 m

Duration of the run t = 9.92 s

Solution

Velocity = \frac{Distance}{Time} \\\\Velocity = \frac{100}{9.92} \\\\Velocity = 10.08 m/s

Gala2k [10]3 years ago
4 0
100 meters in 9.92 seconds,
=distance/time
=100m/9.92s 
=10.0806 m/s
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A mass slider m = 0.200 kg rests on a frictionless horizontal air rail connected to a spring with a force constant k = 5.00 N /
Step2247 [10]
Va ser 0.0900 yo creo preo que esta respuesta te ayude
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3 years ago
A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
Which variable mentioned in Table 2-1 is kept constant? a. amount of time spent swimming b. type of swimming stroke c. number of
Vladimir [108]

Answer:

a. amount of time spent swimming

7 0
2 years ago
What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg
scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

F_c = F_g

\frac{mv^2}{r} = \frac{GmM}{r^2}

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

v^2 = \frac{GM}{r}

At the same time we know that the period is equivalent in terms of the linear velocity to,

T = \frac{2\pi}{\frac{v}{r}}

v = \frac{2\pi r}{T}

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

v^2 = \frac{GM}{r}

( \frac{2\pi r}{T})^2 =  \frac{GM}{r}

T = \sqrt{\frac{4\pi^2 r^3}{GM}}

Replacing we have,

T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}

T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})

T= 1.74hour

Therefore the correct answer is C.

7 0
3 years ago
Please don't post anything unless it's the answer
Alja [10]
52800000000000000000000000000000000000000000
7 0
3 years ago
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