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nevsk [136]
2 years ago
14

A boy and a girl are riding a merry-go-round which is turning at a constant rate. the boy is near the outer edge, while the girl

is closer to the center. who has the greater tangential acceleration?
Physics
2 answers:
Artyom0805 [142]2 years ago
8 0
If the ride is turning at a constant rate, then the acceleration is all centripetal, and zero tangential.
scZoUnD [109]2 years ago
7 0
I think that the girl has greater tangential acceleration because she is closer to the center and the acceleration is greater there. 
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The lightest duty and most widely used non flexible metal conduit is
Alenkinab [10]
Aluminium, or heavier version copper.
6 0
1 year ago
A point charge q1 = 4.50 nC is located on the x-axis at x = 1.95 m , and a second point charge q2 = -6.80 nC is on the y-axis at
Vinvika [58]

Answer:

Explanation:

One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m

These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.

Applying Gauss's theorem

Flux through spherical surface = charge inside sphere / ε₀

= 0 / ε₀

= 0 Ans .

3 0
2 years ago
A steel sphere sits on top of an aluminum ring. The steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at
mote1985 [20]

Answer:

C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

8 0
1 year ago
If Katie swims from one end of the pool, to the other side, and then swims back to her original spot, her average velocity is ha
Crank

Answer:

false.

Explanation:

Ok, we define average velocity as the sum of the initial and final velocity divided by two.

Remember that the velocity is a vector, so it has a direction.

Then when she goes from the 1st end to the other, the velocity is positive

When she goes back, the velocity is negative

if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:

AV = (V + (-V))/2  = 0

While the average speed is the quotient between the total distance traveled (twice the length of the pool) and the time it took to travel it.

So we already can see that the average velocity will not be equal to half of the average speed.

The statement is false

4 0
1 year ago
An experiment to measure the speed of light uses an apparatus similar to Fizeau's. The distance between the light source and the
Marina86 [1]

Answer:

2.88*10^{8} m/s

Explanation:

The speed of light is given by

c=\frac {2d}{t} and t=\frac {\theta}{\omega} hence

c=\omega\frac {2d}{\theta}

Speed of light is given by

c=900\times 2\pi(\frac {2\times 10}{\frac {2\pi}{2*800}}})=2.88*10^{8} m/s

4 0
2 years ago
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