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MaRussiya [10]
3 years ago
13

What orientation of velocity and acceleration will cause something to initially speed up?

Physics
1 answer:
Temka [501]3 years ago
7 0
-- If acceleration and velocity are in the same direction,
then the object is speeding up.

-- If acceleration and velocity are in opposite directions,
then the object is slowing down.

-- If acceleration is perpendicular to velocity, then the object
is moving on a circular curve at constant speed.
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You walk 6 m North and then 4 m south what distance did you travel? What is your displacement?
mestny [16]
<span>Distance is the actual path covered and displacement is the shortest distance from the object to the point of origin.

Distance is a total of 10m because you walked 6m initially and then another 4m. 6m + 4m = 10m

Displacement includes the starting point. You walked 6m North. Then you turned around and walked 4m South. Your total displacement is 2m because that is your current distance from the starting point.

Please mark as brainliest if satisfied with answer
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6 0
3 years ago
How long will it take a person walking at 2.1 m/s to travel 13 m?
MrRissso [65]

Answer:

I gonna give you the number so but you need to round 6.19047619048

Explanation:

  • This is a speed formula so you would use the formula speed=distance/time
  • You need to rearrange it to time=distance/speed
  • So you need to divide 13m by 2.1 m/s

7 0
3 years ago
What is the natural pH of the ocean?
Katyanochek1 [597]

Answer: Basic

Explanation: water is "neutral" with a ph of 7, the ocean has like 8 so it is more basic on the scale

7 0
3 years ago
Read 2 more answers
As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in
vlada-n [284]

Answer:

a=-0.33\ m/s^2

Explanation:

<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

\displaystyle a=\frac{\Delta v}{t}

Or, equivalently

\displaystyle a=\frac{v_f-v_o}{t}

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2

The negative sing of a indicates there is deceleration or decreasing speed

7 0
3 years ago
A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed
Dahasolnce [82]
<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>
6 0
3 years ago
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