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abruzzese [7]
3 years ago
6

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour

s, the velocity of the car is 62 km/h.
Part A: Write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used.

Part B: How can you graph the equation obtained in Part A for the first seven hours?
Physics
1 answer:
Serhud [2]3 years ago
4 0
For the answer to the question above, first find out the gradient. 

<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>

<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>

<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>

<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>

<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
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2 years ago
Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in
charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

  • t = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
  • s = distance to be moved by the rock long the horizontal = 98 yards
  • y = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

  • u = magnitude of initial velocity of the rock
  • \theta = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\

On dividing equation (1) by (2), we have

\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ

Now, putting this value in equation (2), we have

u\cos 51.34^\circ\times  5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

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3 years ago
The innermost satellite of jupiter orbits the planet with a radius of 422 × 103 km and a period of 1.77 days. what is the mass o
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The mass of Jupitar is obtained from the calculations as 5.8 * 10^-14 Kg.

<h3>What is the mass of Jupitar?</h3>

There are nine planets in the solar system and the sun lies at the enter of our solar system. This is the heliocentric model of the solar system.

Given that;

T^2 = GMr^3/4π

T = period

G = gravitational constant

r = radius

M = mass of Jupitar

Now;

1 day = 86400 seconds

1.77 days = 1.77 days *  86400 seconds/1 day

= 152928 seconds

Making M the subject of the formula;

M =4πT^2/Gr^3

M = 4 * 3.142 * (152928)^2/6.67 × 10^-11 * (422 × 10^9)^3

M = 2.9 * 10^11/5.0 * 10^24

M = 5.8 * 10^-14 Kg

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An object has a fixed volume and a variable shape before it changes state.
Wittaler [7]

Answer:

solid to liquid :)

Explanation:

APEX i just took it ;)

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3 years ago
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