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2 years ago

9

a 10kg model rocket is fired at a 60 degree angle from horizontal from the football field with a thrust of 200N. what is the acc

eleration of the rocket?

1 answer:

Nikolay [14]2 years ago

5 0

Answer:

the acceleration of the rocket is: a=vemΔmΔt−g a = v e m Δ m Δ t − g .

Explanation:

I answered this before.

hope this helps! :)

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Bjorn is holding a tennis ball outside a second floor window (3.5 meters from the ground) and billie jean is holding one outside

MArishka [77]

The answer is 1.01 x 10^(-11) N. I arrived to this answer through calculating the GPEs of both balls. Bjorn's ball has a GPE of 1.402 x 10^(-11) N. Billie Jean's ball has a GPE of <span>2.503 x 10^(-11) N. I subtracted the two and I found that Billie Jean's tennis ball has a GPE of 1.01 x 10^(-11) more than Bjorn's tennis ball.</span>

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3 years ago

The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the freque

stiv31 [10]

Answer:

Explanation:

Given

wavelength of emissions are

$\lambda _1=589 nm$

$\lambda _2=589.6 nm$

Energy is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

x=velocity of Light

$\lambda$ =wavelength of emission

$E_1=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589\times 10^{-9}}$

$E_1=3.374\times 10^{-19} J$

$E_1 in kJ/mol$

$E_1=203.2 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589\times 10^{-9}}$

$\nu _1=5.09\times 10^{14} Hz$

Energy corresponding to $\lambda _2$

$E_2=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589.6\times 10^{-9}}$

$E_2=3.371\times 10^{-19} J$

$E_2=203.02 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589.6\times 10^{-9}}$

$\nu _1=5.088\times 10^{14} Hz$

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3 years ago

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Rudiy27

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Leviafan [203]

Answer:

papa's donuteria

Explanation:

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