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oee [108]
2 years ago
7

(a) Draw a Velocity - Time graph for an Object with Constant Acceleration.

Physics
1 answer:
den301095 [7]2 years ago
7 0

Explanation:

answers to 1 and 2 above/below respectively but can't do 3 because I don't know what graph you're talking about

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A football player runs from his own goal line to the opposing team's goal line, returning to his forty-yard line, all in 22.4 s.
Bumek [7]
I assume L=120 yards as the length of the football field.

1) The average speed is given by the total distance covered by the player divided by the time taken.
The total distance covered to go from one goal line to the other and then back to the 40-yards line is
S=120y+(120-40)y=120y+80y=200 y
And the time taken is t=22.4 s, so the average speed of the player is
v= \frac{S}{t}= \frac{200 y}{22.4 s}=8.93 y/s

2) The find the average velocity, we should also consider the direction (and the sign) of the velocity.
In the the first part of the motion, the player goes from one goal line to the other one, so he covers 120 y. However, in the second part of the motion he goes back by 80 y. Therefore, the net displacement of the player is
S=120 y-80 y=40 y
and so, the average velocity is
v= \frac{S}{t}= \frac{40 y}{22.4 s}= 1.79 y/s
6 0
3 years ago
Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

T^2 \propto R^3

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

T^2 \propto R^3

T^2 = kR^3

Substituting the values, we get the value of constant k for mars.

687^2 = k\times (2.279 \times 10^{11})^3

k = 3.92 \times 10^{-29}

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

365^3 = 3.92\times 10^{-29} R^3

R^3 = 3.39 \times 10^{33}

R= 1.50 \times 10^{11}\;\rm m

Hence we can conclude that the distance of the earth from the sun is 1.50 \times 10^{11}\;\rm m.

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0
3 years ago
¿Qué es una onda electromagnética?
Leya [2.2K]

Answer:

Son aquellas ondas que no necesitan un medio material para propagarse. Incluyen, entre otras, la luz visible y las ondas de radio, televisión y telefonía. Todas se propagan en el vacío a una velocidad constante, muy alta (300 0000 km/s) pero no infinita.

3 0
3 years ago
Who needs to know physics when they are making a satellite
vladimir2022 [97]
First satellite was Sputnik and it was a race then to send satellites!! Everyone starts experimenting how to launch.

first try was on a dog , Laika !! It was send to space as it is long after sputnik is placed in orbit !!

Now you know everyone knows rocket science and the most active are ISRO (India) and NASA (USA) !!
4 0
4 years ago
A plane flies from alphaville to betaville and then back to alphaville. when there is no wind, the round trip takes 4 hours and
kykrilka [37]
Refer to the diagram shown below.

d =  distance (miles) from Alphaville to Betaville.
v = speed (mph) of the plane with no wind.

With no wind:
The time taken to travel a distance of 2d is 4 hrs, 48 min = 4.8 hrs.
Therefore
2d/v = 4.8
v = 2d/4.8 = 0.4167d mph             (1)

With the wind:
The velocity from Alphaville to Betaville is (v + 100) mph.
The time of travel is
t₁ = d/(v+100) h
The velocity from Betaville to Alphaville is (v - 100) mph.
The time of travel is
t₂ = d/)v-100) h
Because the return trip takes 5 hours, therefore
t₁ + t₂ = 5
\frac{d}{v+100} + \frac{d}{v-100} =5 \\ \frac{2vd}{v^{2}-100^{2}} =5 \\ 2vd = 5(v^{2}-10^{4})            (2)

From (1), obtain
2(0.4167)d² = 5[(0.4167d)² - 10⁴]
0.8334d² = 0.8682d² - 5 x 10⁴
0.0348d² = 5 x 10⁴
d = 1198.7 mi

Answer: 1199 miles (nearest integer)

8 0
4 years ago
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