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miskamm [114]
3 years ago
8

Two bugs are riding a turntable whose angular speed is constantly increasing with time. Bug A being closer to the edge and Bug B

is farther from the edge of the turntable. Which is true regarding the two bugs' tangential accelerations (in magnitude)?
a.Bug A is experiencing a greater radial acceleration than Bug B.
b.Impossible to answer without knowing the masses of the two bugs.
c.Bug A and Bug B are experiencing the same nonzero radial acceleration.
d.Bug A and Bug B are both experiencing zero radial acceleration.
e.Impossible to answer without knowing the radius of the turntable.
f.Bug B is experiencing a greater radial acceleration than Bug A.
Physics
2 answers:
choli [55]3 years ago
6 0

Answer:

Option A is correct.

Bug A is experiencing a greater radial acceleration than Bug B.

Explanation:

The two bugs have the same angular speed, w, but different radii of the circular motion.

Bug A is closer to the edge of the turntable and bug B is farther from the edge of the turntable.

Hence, Bug A has a bigger radius of circular motion, hence, its radius can be called R and the radius of the circular motion for bug B is r.

v = wr

The radial acceleration of a body in circular motion is given as

α = (v²/r) = rw²

Radial acceleration for bug A = Rw²

Radial acceleration for bug B = rw²

Since we established that R > r and the angular speeds are equal,

Rw² > rw²

Hope this Helps!!!

nika2105 [10]3 years ago
4 0

Answer:

The answer is: f. Bug B is experiencing a greater radial acceleration than Bug A.

Explanation:

The radial acceleration is equal to:

a_{c} =\frac{v^{2} }{r}

If the radial velocity is:

v=rw

Replacing:

a_{c} =\frac{(rw)^{2} }{r} =w^{2} r

According the problem w is the same for both A and B and r is the distance from center, then:

r_{B} >r_{A}

According to this expression, it can be concluded that:

a_{c,B}  >a_{c,A}

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A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

1)From second equation of motion we get

h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

Vf^2=292.04

Vf=17.08 m/s

8 0
4 years ago
f a single circular loop of wire carries a current of 45 A and produces a magnetic field at its center with a magnitude of 1.50
Lelu [443]

Answer:

Radius of the loop is 0.18 m or 18 cm

Explanation:

Given :

Current flowing through the wire, I = 45 A

Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T

Number of turns in circular wire, N = 1

Consider R be the radius of the circular wire.

The magnetic field at the center of the current carrying circular wire is determine by the relation:

B=\frac{N\mu_{0} I}{2R}

Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.

Substitute the suitable values in the above equation.

1.50\times10^{-4} =\frac{4\pi \times10^{-7}\times45 }{2R}

R = 0.18 m

4 0
3 years ago
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