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SIZIF [17.4K]
3 years ago
13

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular

momentum?
Physics
1 answer:
laiz [17]3 years ago
8 0

Answer:

L=11.3\ kg-m^2/s

Explanation:

Given that,

Angular speed of a skater, \omega=3\ rot/s=18.84\ rad/s

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

L=I\omega

Substitute all the values,

L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s

So, its angular momentum is equal to 11.3\ kg-m^2/s.

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Una pelota de tenis de 100 g de masa lleva una rapidez de 20 m/s. Al ser golpeada por una raqueta, se mueve en sentido contrario
REY [17]

Answer:

Explanation:

El impulso aplicado a la pelota produce una variación en su momento lineal.

J = m (V -Vo)

Conviene elegir positivo el sentido de la velocidad final.

J = 0,100 kg [40 - (- 20)] m/s = 6 kg m/s

Saludos Herminio

7 0
3 years ago
Can someone please help​
coldgirl [10]

Answer:

Making a quick cut left to intercept a pass

Explanation:

It takes more energe to do than running

7 0
3 years ago
A brick sliding across a smooth floor has a coefficient of static friction of s=0.23 and a coefficient of kinetic friction of k=
pantera1 [17]

Answer:

0.029325

Explanation:

0.23x0.15=0.0345

0.0345x0.85=0.029325

4 0
1 year ago
Read 2 more answers
Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?
noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

k = 9 * 10^9

q_1 = 2pC

q_2 = 4pC

r = 6cm = 0.06 m

now by using the above equation we have

F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}

F = 2 * 10^{-11} N

so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

3 0
3 years ago
A cup of coffee is sitting on a table in a train that is moving with a constant velocity. The coefficient of static friction bet
Vikki [24]

Answer:

a = 2.94 m/s²

Explanation:

In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:

Unbalanced Force = Frictional Force

ma = μR = μW

ma = μmg

a = μg

where,

a = maximum acceleration for the cup not to slip = ?

μ = coefficient of static friction = 0.3

g = acceleration due to gravity = 9.8 m/s²

Therefore,

a = (0.3)(9.8 m/s²)

<u>a = 2.94 m/s²</u>

3 0
3 years ago
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