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3 years ago

Which formula describes acceleration? m/s^2 m/s s/m m2

Physics

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

m/s^2

Explanation:

Force = mass × acceleration

kgm/s^2 = kg × acceleration

where acceleration = Force ÷ mass

= kg m/s^2 ÷ kg

:Acceleration = m/s^2

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The distance, in feet, a moving object has traveled after t seconds is given by 2t/(4 + t). find the acceleration of the object

lianna [129]

Hi! Let me help you!
a = (Vf - Vi)/t ; where distance d = [2(t)]/(4+t), t = 5secs, and Vi = 0
a = [(2t)/(4+t)]/t <---- working equation
a = {[2(5)]/9}/5 <---- cancel 5
a = 2/9 ft/s^2 <---- Answer

3 0

2 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

2 years ago

The bond enthalpy value for a carbon to hydrogen bond is 413 kJ. What does this mean?

AlladinOne [14]

Breaking bond requires energy. The bond between the carbon and hydrogen is broken when the energy is absorbed. The enthalpy is defined to be the energy taken to break the one mole of the stated carbon and hydrogen bond. Thus a should be the correct answer

6 0

3 years ago

Read 2 more answers

After a model rocket reached its maximum height, it then took 5.0 seconds to return to the launch site. what is the approximate

kifflom [539]

Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.

Let h = the maximum height reached.
Let u = the vertical launch velocity.

Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0
u = 49 m/s

The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
= 122.5 m

Answer: 122.5 m

3 0

2 years ago

The moon Phobos orbits Mars

dlinn [17]

Answer:

The moon Phobos orbits Mars

(mass = 6.42 x 1023 kg) at a distance

of 9.38 x 106 m. What is its period of

orbit?

Explanation:

Answer: 27.9816 x 10^3 is the period of orbit

8 0

2 years ago