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anzhelika [568]
3 years ago
5

An amusement park ride spins you around in a circle of radius 2.5 m with a speed of 8.5 m/s. If your mass is 75 kg, what is the

centripetal force acting on you? A. 1530 N B. 2168 N C. 255 N D. 2359 N
Physics
2 answers:
nataly862011 [7]3 years ago
8 0

B is the correct answer - APEX

zhannawk [14.2K]3 years ago
3 0
B is the answer to this truly did the math
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7. Un bloque de 700 N se encuentra sobre una viga uniforme de 200 N y 6 m de longitud. El bloque está a una distancia de 1 m del
GrogVix [38]

Answer:

 x =  0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

      sin 60 = T_{y} / T

      T_{y} = T sin 60

       cos 60 = Tₓ / T

      Tₓ = T cos 60

we apply the equation

       ∑ τ = 0

       -W L / 2 - w x + T_{y} L = 0

 

the length of the bar is L = 6m

           -Mg 6/2 - m g x + T sin 60 6 = 0

             x = (6 T sin 60 - 3 M g) / mg

let's calculate

let's use the maximum tension that resists the cable T = 900 N

             x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

             x = (4676 - 5880) / 6860

             x = - 0.176 m

Therefore the block can be up to 0.176m to keep the system in balance.

5 0
3 years ago
~choices below~
mart [117]
A is the correct answer
5 0
3 years ago
Read 2 more answers
Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20
Genrish500 [490]

Answer:

The time it can operate between chargins in minutes is

t=102.8 minutes

Explanation:

Given: m=500kg, r=1.0m, w=200\pi rad/s

a). The rotational kinetic energy

K_R=\frac{1}{2}*I*w^2

I=\frac{1}{2}*m*r^2

I=\frac{1}{2}*500kg*(1.0m)^2

I=250 kg*m^2

K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2

K_R=49.348x10^6J

b). The power average 0.8kW un range time can be find

P=\frac{K_R}{t'}

Solve to t'

t=\frac{K_R}{P}

t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s

t=6168.5s\frac{1minute}{60s}=102.8 minutes

3 0
3 years ago
A girl throws a rock horizontally at 10 m/s from the top of a building, 22 m above street level. Assuming free fall conditions a
MAXImum [283]

Answer:21.18 m

Explanation:

Given

initial speed u=10 m/s

height of building h=22 m

time taken to complete 22 m

h=ut+\frac{1}{2}at^2

initial vertical velocity =0

22=\frac{1}{2}gt^2

t=\sqrt{\frac{22\times 2}{g}}

t=2.11 s

Horizontal Distance moved

R=u_x\times t

R=10\times 2.11

R=21.18 m

6 0
3 years ago
A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl
frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ =  34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

W = 292.9686 J

W ≅ 292.97 J

Hence,  the work done by tension before the block goes up the incline = 292.97 J

8 0
3 years ago
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