A 10n falling object encounters 4n of air resistance. what is the net force on the object?

Physics

1 answer:

Margaret [11]3 years ago

5 0

It would be 6n down.

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A thin film of oil rests on an inclined plane

dolphi86 [110]

What is the question?

3 0

3 years ago

When 224-nm light falls on a metal, the current through a photoelectric circuit is brought to zero at a stopping voltage of 1.84

OLga [1]

Answer:

3.71 eV

Explanation:

λ = Wavelength of light = 224 nm = 224 x 10⁻⁹ m

c = speed of electromagnetic wave = 3 x 10⁸ m/s

V₀ = stopping potential = 1.84 volts

W₀ = Work function of the metal = ?

Using the equation

$\frac{hc}{\lambda } = eV_{o} + W_{o}$

$\frac{(6.63\times 10^{-34})(3\times 10^{8})}{224\times 10^{-9} } = (1.6\times 10^{-19})(1.84) + W_{o}$

$W_{o}$ = 5.94 x 10⁻¹⁹

$W_{o}$ = 3.71 eV

7 0

3 years ago

Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro

luda_lava [24]

Answer:

c. $4.582\times10^{21} kg$

Explanation:

$r_{i}$ = Initial distance between asteroid and rock = 7514 km = 7514000 m

$r_{f}$ = Final distance between asteroid and rock = 2823 km = 2823000 m

$v_{i}$ = Initial speed of rock = 136 ms⁻¹

$v_{f}$ = Final speed of rock = 392 ms⁻¹

$m$ = mass of the rock

$M$ = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

$(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg$