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Mice21 [21]
3 years ago
13

A solid 0.7150-kg ball rolls without slipping down a track toward a loop-the-loop of radius r = 0.9150 m. what minimum translati

onal speed vmin must the ball have when it is a height h = 1.433 m above the bottom of the loop, in order to complete the loop without falling off the track?
Physics
1 answer:
sveticcg [70]3 years ago
7 0
<span>Answer: kinetic energy of a sphere is worth 0.7mV^2 at the top of the loop the sphere must have a tangential speed such that V^2/r = g in order to prevent it from falling dawn...so : V = âšg*r = âš9.8*0.915 = 3.00 m/sec in addition a further potential energy U = m*g*2r must be given to hold the sphere....this means that the total energy E the sphere must posses at the bottom of the loop is equal to : E = 0.7mV^2+m*g*2r = 0.475*(0.7*2.455^2+9.8*1.23) = 7.73 joule @ height H = 0.9133 m : E = 7.73 = m*g*H+0.7mVo^2 7.73 = 0.475*9.8*0.9133+0.475*0.7*Vo^2 Vo = âš(7.73 -(0.475*9.8*0.9133))/(0.475*0.7) = 3.23 m/sec</span>
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Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless.  I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
                                    Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
                                                                 </em>
That's about  <em>2.634 m/s²</em>  <em>

</em>
(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

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7 0
2 years ago
A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu
Bumek [7]

Answer:

a) v1 = 5.52m/s

b) v2 = -1.52m/s

c) v3 = 4.62m/s

d) vt = 3.85m/s

Explanation:

The velocity of the football wide receiver is his displacement per unit time.

Velocity v = (displacement d)/time t

v = d/t .....1

For each of the cases, equation 1 would be used to calculate the velocity.

a) v1 = d1/t1

d1= 16m

t1 = 2.9s

v1 = 16m/2.9s

v1 = 5.52m/s

b) v2 = d2/t2

d2 = -2.5m

t2 = 1.65s

v2 = -2.5/1.65

v2 = -1.52m/s

c) v3 = d3/t3

d3 = 24m

t3 = 5.2s

v3 = 24/5.2

v3 = 4.62m/s

d) vt = dt/tt

dt = 16m - 2.5m + 24m = 37.5m

tt = 2.9 + 1.65 + 5.2 = 9.75s

vt = 37.5/9.75

vt = 3.85m/s

5 0
3 years ago
What is another name for the introitus? psych 230?
Vera_Pavlovna [14]
Vaginal opening. areola is the part of the breast.
7 0
2 years ago
Newton first law of motion ?​
maw [93]

Answer:

The law of inertia

Explanation:

A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force

8 0
2 years ago
Read 2 more answers
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

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The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
2 years ago
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