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Sati [7]
3 years ago
13

Deena made a chart to summarize features of a velocity vs. time graph for objects with constant acceleration and objects with co

nstant velocity. Which best describes Deena’s error?
The line for constant velocity is not horizontal; it is diagonal and slopes downward.

Include the area of the rectangle under the line to find displacement for constant acceleration.

Find acceleration by finding the area of the rectangle above the line.

Find velocity by calculating the slope of the line.
Physics
2 answers:
Irina-Kira [14]3 years ago
5 0
The error is:
"<span>Find acceleration by finding the area of the rectangle above the line"

The acceleration of any body is the rate of change of velocity of that body. If the motion of the object being considered is plotted on a graph, with the velocity on the y-axis and the time on the x-axis, this ratio will be equivalent to the slope of the graph, not the area above the line.</span>
goldenfox [79]3 years ago
4 0

Answer:

Include the area of the rectangle under the line to find displacement for constant acceleration.

Explanation:

correct on edge

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What is the gravitational force between a 45 kg person, and the Earth at 5.98 x 1024 kg, with a distance of
Assoli18 [71]

Answer:

C. 441 N

Explanation:

Gravitational force between two objects can by calculated by the formula

= G m₁m₂ / r² , m₁ and m₂ are masses at distance r

= ( 6.67 x 10⁻¹¹ x 45 x 5.98 x 10²⁴) / ( 6.38 x 10⁶ )²

= 44.09 x 10

= 440.9 N

= 441 N .  

7 0
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You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use
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P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
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Plss help
Genrish500 [490]

Answer:

b

Explanation:

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3 years ago
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Nadya [2.5K]
<span>Fusion produces large amounts of energy, and the fuel is found on Earth.</span>
4 0
3 years ago
Read 2 more answers
A car that is traveling in a straight line at 40 km/h can brake to a stop within 20 m. If the same car is traveling at 120 km/h
stiks02 [169]

Answer:

180 m

Explanation:

Case 1.

U = 40 km/h = 11.1 m/s, V = 0, s = 20 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0 = 11.1 × 11.1 - 2 × a × 20

a = 3.08 m/s^2

Case 2.

U = 220 km/h = 33.3 m/s, V = 0

a = 3.08 m/s^2

Let the stopping distance be x.

Again use third equation of motion

0 = 33.3 × 33.3 - 2 × 3.08 × x

X = 180 m

8 0
3 years ago
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