The planet whose distance from the sun equals 1 <br> a.u. is: venus earth mars jupiter

A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg

motikmotik

Answer:

$\omega_{f}=1.634\ rad/s$

Explanation:

given,

diameter of merry - go - round = 2.40 m

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg

initial angular momentum of the system

$L_i = I\omega_i$

$L_i =356\times 1.80$

$L_i =640.8\ kg.m^2/s$

final angular momentum of the system

$L_f = (I_{disk}+mR^2)\omega_{f}$

$L_f = (356 + 25\times 1.2^2)\omega_{f}$

$L_f= (392)\omega_{f}$

from conservation of angular momentum

$L_i = L_f$

$640.8= (392)\omega_{f}$

$\omega_{f}=1.634\ rad/s$

8 0

2 years ago

Read 2 more answers

Si la aceleración de gravedad en la superficie del planeta Mercurio es de 3,7 m/s2, entonces ¿cuál sería el peso de una persona

erica [24]

Answer:

W = 222 N.

Explanation:

The qiestion says" If the acceleration of gravity on the surface of the planet Mercury is 3.7 m / s2, then what would be the weight of a person with mass 60 kg on its surface? "

Mass of the person, m = 60 kg

The acceleration due to gravity on the surface of gravity is 3.7 m/s²

We need to find the weight of a person on the surface of Mercury.

Weight of an object is given by :

W = mg

So,

W = 60 kg × 3.7 m/s²

W = 222 N

Hence, the person will weigh 2222 N on the surface of Moon.

4 0

2 years ago

A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge

Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

$\tau=RC$

where

$R=50,000 \Omega$ is the resistance

$C=2.0\mu F=2.0\cdot 10^{-6}F$ is the capacitance

Substituting,

$\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s$

The charge on a charging capacitor is given by

$Q(t)=Q_0 (1-e^{-t/\tau} )$ (1)

where

$Q_0$ is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

$Q(t)=0.90 Q_0$

Substituting this into eq.(1) we find

$0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s$

4 0

2 years ago

What affect would using a 12V car battery have on the operation of your circuit? (Do not try this.) What would happen to the cur

k0ka [10]

Answer:

Incomplete question

This is the completed question

If the resistor in the circuit had a larger resistance then the current would be then have to be proportionally smaller. Because the batteries each give off 1.5 volts then the current would have to be the variable that would change. What affect would using a 12V car battery have on the operation of your circuit? (Do not try this.) What would happen to the current? What would happen to the resistor?

Explanation:

Using ohms law as our basis

Ohms law state that, the voltage in an ohmic conductor is directly proportional to the current

V∝I

Resistance is the constant of proportionality

Then

V=iR

Since we want a relationship between current and resistance.

then, I=V/R

So, current is inversely proportional to Resistance

as the current increase the resistance reduce and as the current reduces the resistance increases.

a. So, increasing the voltage from 1.5V to 12V increases the current In the circuit because voltage Is directly proportional to I.

From ohms law

V=iR

When v =1.5V

I=1.5/R

When V increase to 12V

I=12/R

I.e, it increases by a factor of 8. Eight times it's initial value

b. Now, the resistance in the circuit is the constant of proportionality and it doesn't change in a given circuit expect when using a variable resistoa r like rheostat.

6 0

2 years ago

What is the force on a 1,000 kilogram-elevator that is falling freely under the acceleration of gravity only?

likoan [24]

   Force = (mass) · (acceleration)

   = (1,000 kg) · (9.8 m/s²)

   = 9,800 newtons

Why are you still having a problem with F = M · a ?

8 0

2 years ago

The planet whose distance from the sun equals 1 a.u. is: venus earth mars jupiter