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2 years ago

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]2 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]2 years ago

4 0

100 meters in 9.92 seconds,
=distance/time
=100m/9.92s
=10.0806 m/s

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Based on the information presented in the graph, what is the velocity of the object?

Sladkaya [172]

Answer:

3 m/s

Explanation:

<u>The velocity of a position-time graph is the slope of the line.</u> Slope is rise over run, or rise divided by run. The rise (how many units the line goes up) is 3 units and the run is 1 unit. 3/1 is 3, so the velocity is 3 m/s.

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3 years ago

Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy

erma4kov [3.2K]

Answer:

Explanation:

⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆

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3 years ago

A piece of metal has a mass of 10g and a mass of 2cm

qwelly [4]

Answer:

so whats the questain?!

Explanation:

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2 years ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

3 years ago

A bag of sugar weighs 5.00 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixt

gizmo_the_mogwai [7]

Answer:

Earth: 22.246 N

Moon: 3.71 N

Jupiter: 58.72 N

Explanation:

The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.

To calculate anything, we need to convert to standard measurements.

5.00 lbs = 2.27 kg

On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N

Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N

Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N

3 0

3 years ago