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3 years ago

If a man weighs 900 n on the earth, what would he weigh on jupiter, where the acceleration due to gravity is 25.9 m/s2?

1 answer:

4 0

So, if the man weight 900 newtons on Earth then that means, using F=ma, that the mass of the man is approximately 91.84 kg. This is because 900N=m(9.8m/s^2), and so it follows that 900/9.8=91.84. Using the man's found mass we then plug this into F=ma again. It follows that F=(91.84)(25.9)=2378.57N. This means that the man "weighs" 2378.57 Newtons on Jupiter, or about 2.5x as great as his weight on Earth. This makes sense, considering that 25.9/9.8 is approximately equal to 2.64.

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A football field lies so one endzone is to the east and one is to the west. A running back gets the ball and starts running 20o

Jobisdone [24]

Answer:

( 80.87 i - 12.96 j ) m

Explanation:

Running back movements :

20⁰ north of east for 10 meters,

Runs straight east for 60 meters

runs 35⁰ east of south for 20 meters

A) show vector pointing from the starting point to the end ( where he scored )

The final vector displacement : ( 80.87 i - 12.96 j ) m

which is : 81.90 m, 9.10⁰ south of east

attached below is the required diagram

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2 years ago

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Zolol [24]

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3 years ago

Read 2 more answers

Consider the following three objects, each of the same mass and radius: (1) a solid sphere (2) a solid disk (3) a hoop All three

Vinil7 [7]

Answer:

The correct answer is

a) 1, 2, 3

Explanation:

In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus

PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²

The transformation equation fom potential to kinetic energy is =

m×g×h = $\frac{1}{2} mv^{2} + \frac{1}{2} (\frac{2}{5} mr^{2} )(\frac{v}{r}) ^{2}$

$v_{Sphere}$ = $\sqrt{\frac{10}{7} gh}$

$v_{Hoop}$ = $\sqrt{gh}$

$v_{Disc}$ = $\sqrt{\frac{4}{3} gh}$

Therefore the order is with increasing rotational kinetic energy hence

the first is the sphere 1 followed by the disc 2 then the hoop 3

the correct order is a, 1, 2, 3

8 0

3 years ago

Fill in the blanks for the following:

storchak [24]

Answer:

a. 4.21 moles

b. 478.6 m/s

c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = 4.21 moles

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = 478.6 m/s

For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$