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ANEK [815]
1 year ago
6

Table 2.4 shows how the dispacement of a runner changed during a sprint race. Draw a dispacement-time graph to show this data, a

nd use it to deduce the runner's speed in the middle of the race​
Physics
1 answer:
GalinKa [24]1 year ago
8 0
4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12
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You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
2 years ago
Kathmandu lies at high altitude than biratnagar from sea level.Where does an object has more weight between two places?Give reas
Alla [95]

Answer:

Kathmandu

Explanation:

As the altitude get higher, the gravitational pull of the earth on the object increases, therefore, the mass is higher up above.

8 0
2 years ago
An electric heater draws a steady current = 20.0 A on a 120-V line. (a) Calculate how much power does it require.
babymother [125]

Answer:

The heater power required is 2400 W. The power in the heater can be calculated as the product of the voltage line and the steady current:

P=V.I

P=120 V * 20 A = 2400 VA = 2400 W

Explanation:

8 0
2 years ago
Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic ener
Flauer [41]
<span>Answer: KE = (11/2)mω²r², particle B must have mass of 2m, while A has mass m. Then the moment of inertia of the system is I = Σ md² = m*(3r)² + 2m*r² = 11mr² and then KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2 So I'll proceed under that assumption. For particle A, translational KEa = ½mv² but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r² For particld B, translational KEb = ½(2m)v² but v = ω*r, so KEb = ½(2m)ω²r² so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2 which is equal to our rotational KE.</span>
5 0
1 year ago
Which of the following statements cannot be supported by Kepler's laws of planetary motion?
gladu [14]

Answer:

The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.

Explanation:

Kepler's three laws are:

1) The orbits of the planets around the Sun are ellipses, with the Sun at one of the focii

2) A line connecting the Sun with each planet sweeps out equal areas in equal time intervals

3) The cube of the semi-major axis of the orbit of one planet is proportional to the square of its orbital period

There 3 laws help explaining the following statements:

- <em>A planet's distance from the sun will not be the same in six months. --> </em>using the 1st law. In fact, since the orbit is an ellipse (and not a circle), and the Sun is at one of the focii, the distance of the planet from the Sun keeps changing during the year.

-<em> A planet's speed as it moves around the sun will not be the same in six months. -</em>-> using the 2nd law. In fact, since the line connecting the Sun to the planet must cover equal areas in the same time interval, it follows that the speed of the planet cannot be constant during the year (it will be faster when closer to the sun and slower when far from the sun).

- <em>The average distance of Saturn can be calculated using the average distance of Neptune and the orbital period of both planets. </em>--> using the 3rd law. In fact, the ratio \frac{a^3}{T^2} (where a is the semi-major axis of the orbit and T the orbital period) is constant and it is the same for every planet orbiting the sun, so by knowing the data of Neptune and the orbital period of Saturn, it is possible to calculate Saturn's average distance.

Instead, the following statement:

<em>The rotational speed of the four smallest planets can be determined using the rotational speeds of the four largest planets and their orbital periods.</em>

Is not supported by any Kepler's law.

8 0
2 years ago
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