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3 years ago

Which of the following is a health problem associated with obesity in children?

Physics

1 answer:

mihalych1998 [28]3 years ago

5 0

Risk of not being able to reduce their weight

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(PLEASE HELP THANK YOU) A car is going 8 meters per second on an access road into a highway and then accelerates at 1.8 meters p

jonny [76]

Answer:

20.96 m/s

Explanation:

Apply the kinematic equation:

Vf=Vi+at

Vi=8m/s

a=1.8m/s^2

t=7.2s

Putting this all in should give you your answer of 12.96m/s

6 0

3 years ago

How are pendulums used in society? Give an example of a real-world pendulum that transfers a lot of energy.

mart [117]

Answer:

An old clock that has a swinging pendulum

Explanation:

7 0

3 years ago

3.The space heater in a room does not have a fan, but after a few minutes you feel warm air moving by you. What type of energy t

Natali [406]

Heat transfer

Have a good day

6 0

2 years ago

What momentum of a 50kilogram ice skater gliding across the ice at a speed of 5m/s?

Ivenika [448]

Momentum (P) = Mass (kg) * Velocity (m/s)

P = M * V
P = 50 * 5
P = 250

So momentum is 250 kgm/s

5 0

3 years ago

Read 2 more answers

A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial

Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

$qvB=m\frac{v^2}{r}$

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

$r=\frac{mv}{qB}$

For the ion of oxygen-16, we have:

$m_A=2.66\cdot 10^{-26}kg$

$q_A = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_A=2.90\cdot 10^6 m/s$

$B_A=1.30 T$

So the radius of its orbit is

$r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m$

For the ion of oxygen-18, we have:

$m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg$

$q_B = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_B=2.90\cdot 10^6 m/s$

$B_B=1.30 T$

So the radius of its orbit is

$r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m$

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

$d=2(r_B-r_A)=2(0.417-0.371)=0.092 m$

3 0

3 years ago