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3 years ago

5

Newtons third lawWhat action-reaction forces are involved when a rocket engine fires? Why doesnt a rocket need air to push on? A

tction force:_______ Reaction Force:________

1 answer:

dangina [55]3 years ago

4 0

Answer: action forc roketorce

reaction force is engine fires

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if you drive your car at an average speed of 105km/h, how far (in km) will you go in 3hours 30minutes ?

irinina [24]

Hi ;-)

1 hour ⇒ 105 km

0,5 hour (30 minutes) ⇒ 105 km : 2 = 52,5 km

3 hours 30 minutes ⇒ 3 · 105 km + 52,5 km =

= 315 km + 52,5 km = <u>367,5 km</u>

5 0

3 years ago

What kind of waves occur when the motion of the medium is parallel the the wave direction

Nata [24]

Answer:

A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.

Explanation:

hope this helped

8 0

2 years ago

A child’s toy rake is held so that its output arm is 0.75 meters. If the mechanical advantage is 0.33, what is

Masja [62]

Input= ?
Output= 0.75
MA= 0.33

So I just kept on Dividing numbers until I was close to "0.33" & I cam up with the answer of .......... 0.25.

So when you divide 0.25 by 0.75 you get the MA of 0.33

4 0

3 years ago

A magnetic field is applied to a freely floating uniform iron sphere with radius R = 2.00 mm. The sphere initially had no net ma

grigory [225]

Answer:

$\omega=4.2704*10^-^5 rad/s$

Explanation:

Angular Momentum Formula For atoms= $L_{atom}=0.12Nm_{s}h$

Where:

m_{s}h is the momentum for one atom (m_s is the spin quantum number)

N is the number of atoms= $\frac{N_{A}*m}{M}$

Where:

N_A is Avogadro Number

m is the mass of sphere

M is the molar mass of iron

Angular Momentum Formula For atoms will be= $L_{atom}=0.12\frac{N_{A}m}{M} m_{s}h$

Angular Momentum of Sphere= $L_{sphere}=I\omega$

where:

$I=\frac{2mR^{2}}{5}$

So,Angular Momentum of Sphere= $L_{sphere}=\frac{2mR^{2}}{5}\omega$

Angular Momentum of sphere=Angular Momentum of atoms

$L_{sphere}$ = $L_{atom}$

$\frac{2mR^{2}}{5}\omega$ = $0.12\frac{N_{A}m}{M} m_{s}h$

For iron, $m_s =\frac{1}{2}$ . So above equation will become:

$\omega=\frac{0.12*5*N_{A}h}{4*M*R^{2} }$

Where R=2mm, M=0.0558Kg/mol (Molar Mass of iron),h=Planck's Constant/2π

$\omega=\frac{0.12*5*(6.022*10^{23})(6.63*10^{-34}/2*\pi)}{4*0.0558*(2*10^{-3})^{2}}$

$\omega=4.2704*10^-^5 rad/s$

5 0

3 years ago

In a real isothermal expansion, the temperature of the surroundings must be________the temperature of the gas.

Lady_Fox [76]

Must be isolated from the temperature of the gas

7 0

3 years ago