A skateboarder has an acceleration of −1.9 m/s2. If her initial speed is 6 m/s, how long does it take her to stop?

1 answer:

6 0

You said that she's losing 1.9 m/s of her speed every second.

So it'll take

(6 m/s) / (1.9 m/s²) = 3.158 seconds (rounded)

to lose all of her initial speed, and stop.

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Why is air considered a fluid

34kurt

Answer:

A fluids is any substance that flows. Air is made of stuff, air particles, that are loosely held together in a gas form. Although liquids are the most commonly recognized fluids, gasses are also fluids. Since air is a gas, it flows and takes the form of its container.

Explanation:

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3 years ago

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1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second.

Alla [95]

Answer:

a.Distance = 150 m

b. Displacement = 50 m

Time lapsed = 5 seconds

Explanation:

a. Distance is the change in the position of an object.

The distance covered by the car = 100 + 50

= 150 m

b. Since displacement is a vector quantity,

Displacement of the car = 100 - 50

= 50 m due east

c. Time elapsed is the time taken for the motion of the car starting from when its starts to when it stops.

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3 years ago

If 2 grams of element X combine with 4 grams of element Y to form compound XY, how many grams of element Y would combine with 11

kykrilka [37]

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3 years ago

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c

Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

Since the block slides across the floor at constant speed, this means that it's not accelerated.
According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

$F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)$

In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

$F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)$