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maksim [4K]
3 years ago
6

Four football players are running down the field at the same speed. Player 1 weighs 180 lbs and is running toward the south goal

, player 2 weighs 200 lbs and is running toward the north goal, player 3 weighs 190 lbs and is running toward the north goal, and player 4 weighs 165 lbs and is running toward the south goal. Which player has the most momentum?
Physics
1 answer:
Serggg [28]3 years ago
7 0
Player 2 because moment is mass times acceleration and since they are all going the same speed. Speed doesn't matter so the only thing that is left is mass/ weight and he has the most
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Two wheels with fixed hubs and radii 0.51 m and 1.9 m, each having a mass of 3 kg, start from rest. Forces 5 N and F2 are applie
Katarina [22]

Answer:

18.63 N

Explanation:

Assuming that the sum of torques are equal

Στ = Iα

First wheel

Στ = 5 * 0.51 = 3 * (0.51)² * α

On making α subject of formula, we have

α = 2.55 / 0.7803

α = 3.27

If we make the α of each one equal to each other so that

5 / (3 * 0.51) = F2 / (3 * 1.9)

solve for F2 by making F2 the subject of the formula, we have

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

F2 = 18.63 N

Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.

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3 years ago
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An automobile with an initial speed of 5.13 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed of the car a
Digiron [165]

Answer:

Final velocity v=19.83 m/sec

Explanation:

We have given initial velocity u =5.13 m/sexc

Acceleration of automobile a=3m/sec^2

Time t =4.9 sec

We have to find the final velocity v

According to first law of motion v = u+at ,here v is the final velocity , a is acceleration and t is time

So v=5.13+3\times 4.9=19.83m/sec

So the final velocity is 19.83 m/sec

7 0
3 years ago
If a simple truss member is known to carry a tensile force, then the internal force drawn on a free-body at a section cut when u
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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

-s = -1/2gt²

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

7 0
3 years ago
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