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8_murik_8 [283]
3 years ago
10

3. What is the mass of a paratrooper who experiences an air resistance of 400 N and an acceleration of 4.5 m/s2

Physics
1 answer:
goblinko [34]3 years ago
4 0

Answer:

88.89kg

Explanation:

The formula for mass is m=F/a. If we plug in the values, we get m=400N/4.5m/s^2. The mass is 88.89kg. We know that the unit is in kg because one newton (N) is 1kg*m/s^2. The m/s^2 is cancelled out by the acceleration, and we are left with kg.

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6 0
3 years ago
Please help, I need help with this test question
Arturiano [62]

I believe this is right-

for the first question, it should be fluffy

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4 0
2 years ago
a wheel 0.35m in diameter rotates at 2200rpm. calculate its angular velocity in rad/s and its linear speed and acceleration of a
bazaltina [42]

Answer:

( Angular Velocity = ( About ) 230 rad / s,

( Linear Speed = ( About ) 40.25 m / s,

( Acceleration = ( About ) 9290 m / s^2

Explanation:

Here we want the angular velocity in radians per second, the linear velocity and acceleration.

The diameter = .35 meters, and thus we can conclude that the radius be half of that, or .175 meters. For part ( a ), or the calculation of the angular velocity, it is given that the diameter rotates at 2200 revolutions per minute - but we need to convert this into radians per second.

We can say that there are 2π radians for every minute, and for every minute there are 60 seconds. Therefore -

( a ) w = 2,200 rpm( 2π rads / rev )( 1 min / 60 sec )...

Hence, w = ( About ) 230 rad / s

_____

For this second part we can calculate the the linear velocity by multiplying the angular velocity ( omega ) by the radius r -

( b ) v = w( r ) - Substitute,

v = ( 230 rad / sec )( .175 m )...

v = ( About ) 40.25 m / s

_____

And for this last bit here, to find the acceleration we can simply take the angular velocity ( omega ) squared, by the radius r -

( c ) a_{rad} = w^2( r ),

a_{rad} = ( ( 230 rad / sec )^2 )( .175 m )...

a_{rad} = ( About ) 9290 m / s^2

7 0
3 years ago
Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to
pishuonlain [190]

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
  • r = √( 0.4² + 0.3²)
  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

6 0
3 years ago
A glass of root beer with a scoop of ice cream floating on top and a straw sticking out.
vampirchik [111]

Answer:

These forces are all equal and cancel each other out. Gravity pushes downward on the ice cream. This can also be called the weight of the ice cream. Buoyant force pushes the ice cream upward

6 0
2 years ago
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