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katrin2010 [14]
3 years ago
11

Suppose you studied the velocity of a cheetah overtime after it starts running to chase an deer. After the linear fit, a spreads

heet program gives you an equation y = 2.2 x + 1.2. You know from physics class that velocity and time can be modeled by the equation v = v0 + at. In this equation, v is velocity after time t, v0 is the velocity at zero time and a is acceleration during the same time interval. According to this model, what does the number 2.2 represents in the equation given by Excel? Group of answer choices Velocity after t. Velocity at t = 0 We cant tell without knowing what was plotted on the horizontal and vertical axes. Acceleration
Physics
1 answer:
EastWind [94]3 years ago
6 0

Explanation:

While studying the velocity of a cheetah over time in a spreadsheet program is given by :

y = 2.2 x + 1.2 ...(1)

We know that,

v=v₀+at ...(2)

v₀ is velocity when t = 0, v is velocity after time t, a is acceleration and t is time.

If we consider time t on x-axis and v on y axis, then only we can draw a plot of equation (2). On comparing equation (1) and (2) we get :

a = 2.2 (but it is not correct as we don't know about axes).

Hence, the correct option is (a) "We cant tell without knowing what was plotted on the horizontal and vertical axes"

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Guys can you text here please i have way to many textes on the other one and also remember was told to paste this to other comme
Nadusha1986 [10]

Answer:

ok

Explanation:

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6 0
3 years ago
Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.
GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately \boxed{\sf{2.9 \times 10^{17} \: N}}

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}

With the following condition :

  • F = gravitational force (N)
  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_1} = mass of the first object (kg)
  • \sf{m_2} = mass of the second object (kg)
  • r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_X} = mass of the planet X = \sf{1.55 \times 10^{22}} kg.
  • \sf{m_Y} = mass of the planet Y = \sf{3.95 \times 10^{28}} kg.
  • r = distance between two objects = \sf{3.75 \times 10^{11}} m.

What was asked :

  • F = gravitational force = ... N

Step by step :

\sf{F = G \times \frac{m_X \times m_Y}{r^2}}

\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}

\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}

\sf{F \approx 2.9 \times 10^{39 - 22}}

\sf{F \approx 2.9 \times 10^{17} \: N}

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

\boxed{\sf{2.9 \times 10^{17} \: N}}

<h3>See More</h3>
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8 0
1 year ago
A wire of resistance R is cut into ten equal parts which are then connected in parallel. The equivalent resistance of the combin
Greeley [361]

Answer:

<em>The equivalent resistance of the combination is R/100</em>

Explanation:

<u>Electric Resistance</u>

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

Parallel connection of resistances: If R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}

If we have 10 wires of resistance R/10 each and connect them in parallel, the equivalent resistance is:

\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}...+\frac{1}{R/10}

This sum is repeated 10 times. Operating each term:

\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}

All the terms have the same denominator, thus:

\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}

Taking the reciprocals:

R_e=R/100

The equivalent resistance of the combination is R/100

6 0
3 years ago
Differentiate between center of mass and center of gravity
ELEN [110]

Answer:

Centre of mass is the point at which the distribution of mass is equal in all directions, and does not depend on gravitational field. Centre of gravity is the point at which the distribution of weight is equal in all directions, and does depend on gravitational field.

Explanation:

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3 years ago
Someone pls help with only if you know correct answer!???
Roman55 [17]

Answer:

A 2.0 kg ball, A, is moving with a velocity of 5.00 m/s due west. It collides with a stationary ball, B, also with a mass of 2.0 kg. After the collision

Explanation:

5 0
2 years ago
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