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Nezavi [6.7K]
3 years ago
5

A skateboarder with a mass of 45 kilograms is riding on a skateboard with a mass of 2.5 kilograms. What should be the velocity o

f the skateboarder if she has to attain a momentum of 264 kilogram meters/second?. . Blaaah Physics.-.
Physics
2 answers:
DanielleElmas [232]3 years ago
6 0
Given momentum (P) and mass (m), the velocity of the object is dividing the momentum by the mass. The total mass of the skateboarder and the skateboard is 47.5 kg. Performing the operation,
                              v = P / m
                             v = (264 kg m/s  / 47.5 kg) = 5.56 m/s
Therefore, the velocity would be 5.56 m/s. 

klasskru [66]3 years ago
5 0
So momentum is just velocity times mass, this means Momentum = Velocity x Mass.
We can rearrange this to be Velocity = Momentum/Mass.

Since we know momentum and mass we can now solve.

Velocity = 264/(45+2.5)
              = 5.56 m/s
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Answer:

D

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3 years ago
A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance.
Nookie1986 [14]

Answer:

a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

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s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

v² = u² - 2gs

= 0 - 2 × 9.8 m/s² × (-0.6 m)

= 11.76 m²/s²

v = √11.76 m/s

= 3.43 m/s

So, the magnitude of the resultant velocity is V = √(v² + v'²)

= √((3.43 m/s)² + (1.10 m/s)'²)

= √(11.76 m²/s² + 1.21 m²/s²)

= √12.97 m²/s²

= 3.6 m/s

Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction

= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

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Answer:

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First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

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where u = initial velocity = 0 m/s

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Therefore:

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Then, for the next 30 ms (0.03 secs), we use the formula:

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This speed is the same as the final velocity of the tongue after the first 20 ms.

This can be obtained by using the formula:

v = u + at\\\\=> v = 0 + (250 * 0.02)\\\\\\v = 5 m/s

Therefore:

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