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iogann1982 [59]
2 years ago
13

Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?.

Physics
1 answer:
natita [175]2 years ago
3 0

For a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s   is mathematically given as

F= 618.9 N

<h3>What is the centripetal force?</h3>

Generally, the equation for the angular speed  is mathematically given as

w = v/R

Therefore

w= 4.7/1.8

w= 2.611 rad/s

Where total momentum

Tm= 642.96 + 272.32

Tm= 915.28

and total inertia

Ti= 184 + 246.24

Ti= 430.24

In conclusion, centripetal force

F= mrw^2

F = m*R*w2^2

F = 76*1.8*2.127^2

F= 618.9 N

Read more about mass

brainly.com/question/15959704

CQ

Flag

a merry go round with a radius of R=1.8 m and moment of inertia I=184 kg-m^2 is spinning with an initial angular speed of w=1.48 rad/s in the counter clockwise direction when viewed from above a person with mass m=76 kg and velocity v=4.7 m/s runs on a path tangent to the merry go round once at the merry go round the person jumps on and holds on to the rim of the merry go round angular speed of the merry go round after the person jumps on 2.127 rad/s Once the merry go round travels at this new angular speed with what force does the person need to hold on?

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Answer:

electric flux through the three side = 2.35 N m²/C

Explanation:

given,

equilateral triangle of base = 25 cm

electric field strength = 260 N/C

Area of triangle = \dfrac{\sqrt{3}}{4}a^2

                          =  \dfrac{\sqrt{3}}{4} 0.25^2

                          = 0.0271 m³

electric flux = E. A

                   = 260 × 0.0271

                   = 7.046 N m²/C

since, tetrahedron does not enclose any charge so, net flux through tetrahedron is zero.

electric flux through the three side = (electric flux through base)/3

                               = \dfrac{7.046}{3}

electric flux through the three side = 2.35 N m²/C

4 0
3 years ago
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
Sergeeva-Olga [200]

Answer:

The wavelength of the radiation absorbed by ozone is 319.83 nm

Explanation:

Given;

frequency of absorbed ultraviolet (UV) radiation, f = 9.38×10¹⁴ Hz

speed of the absorbed ultraviolet (UV) radiation, equals speed of light, v = 3 x 10⁸ m/s

wavelength of the absorbed ultraviolet (UV) radiation, λ = ?

Apply wave equation for speed, frequency and wavelength;

v = fλ

λ = v / f

λ = (3 x 10⁸) / (9.38×10¹⁴)

λ = 3.1983 x 10⁻⁷ m

λ = 319.83 x  10⁻⁹ m

λ = 319.83 nm

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5 0
3 years ago
A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re
marusya05 [52]

Answer: a) 6.67cm/s b) 1/2

Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

100 + 0 = 15v

v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

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= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

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= 222.44Joules

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3 0
3 years ago
If a reaction starts with 30 grams how many should it end with?
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Sphinxa [80]

Answer: \frac{V^{2}}{r}

Explanation:

According to Newton's 2nd Law of motion the force F is proportional to the mass Fm and acceleration a:

F=m.a (1)

On the other hand, the equation for the Centripetal force is:

F=\frac{mV^{2}}{r} (2)

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Making (1) and (2) equal:

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It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).

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3 years ago
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