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2 years ago

7

A 52.3 kg student is standing at rest on roller skates when another student throws a bag of 3.5 kg bag of sand. After the studen

t catches the bag they are pushed a distance of 2 meters in 1 second. What was the initial velocity of the sand bag being thrown?

1 answer:

IRINA_888 [86]2 years ago

4 0

PLEASE HELP, I need help too !! Please !

Explanation:

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PLEASE HELP!! WILL GIVE POINTS

AURORKA [14]

Answer:

49 kg is the mass of the couch.

Explanation:

GPE = mgh

9800 = m * 10 * 20

9800 = 200m

m = (9800/20) = 49 m

Thenks and mark me brainliest :))

3 0

3 years ago

Using the periodic table, choose the more reactive metal. Al or Mg

ziro4ka [17]

Aluminum, and magnesium are metals. For metals, reactivity decreases as you go from left to right across the periodic table. Atomic number of Al is 13 and of Mg is 12. Hence the least reactive of these two is therefore aluminum.

Magnesium is "HIGHLY FLAMMABLE" carefully take a small piece and hit it with a torch. If its Magnesium it will "Caution, very, quickly burn.
Aluminum will not react to simple flame, it will only melt with enough direct heat.

Magnesium
==========
Atomic Number: 12
Atomic Symbol: Mg
Atomic Weight: 24.305
Electron Configuration: 2-8-2

Aluminum
========
Atomic Number: 13
Atomic Symbol: Al
Atomic Weight: 26.9815
Electron Configuration: 2-8-3

Hope this helps some. Any questions please feel free to ask. Thank you

6 0

3 years ago

Read 2 more answers

9. A football punter attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in th

vichka [17]

Answer:

Angle is 55.52°

and Initial Speed is v=26.48 m/s

Explanation:

Given data

$x_{o}=0m\\ y_{o}=1.23m\\a_{oy}=a_{1y}=g=-9.8m/s^{2} \\x_{1}=67.0m\\y_{1}=0m\\t_{o}=0\\a_{ox}=m/s^{2} \\t_{1}=4.50s$

Applying the kinematics equations for motion with uniform acceleration in x and y direction

So

$x_{1}=x_{o}+v_{ox}t_{1}=67.0m\\0+4.50v_{o}Cos\alpha =67.0m\\v_{o}Cos\alpha =14.99\\v_{o}=14.99/Cos\alpha.....(1) \\and\\y_{1}=y_{o}+v_{oy}t_{1}+(1/2)a_{oy}t_{1}^{2} =0m\\ 1+4.50v_{o}Sin\alpha+(-9.8/2)(4.5)^{2}=0\\ v_{o}Sin\alpha=21.828.....(2)$

Put the value of v₀ from equation (1) to equation (2)

So

$\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}$

Put that angle in equation (1) or equation (2) to find the initial velocity

So from equation (1)

$v_{o}=(\frac{14.99}{Cos\alpha } ) \\v_{o}=(\frac{14.99}{Cos(55.52) } ) \\v_{o}=26.48m/s$

7 0

2 years ago

1. Determine the energy released per kilogram of fuel used.

Serga [27]

The formula for energy release per kilogram of fuel burned is energy release per kg=6.702*10-13. and 19. J 1 Mev = 1.602 X 10 T

Calculate the energy in joules per kilogram of reactants given MeV per reaction. Energy is the ability or capacity to perform tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
Think of a mole of plutonium-239 (molar mass: 239 grams) as a mole of "reactions."

Energy used in the US per person annually = 3-5 X 1011
Population (number of people) = 3.108The required mass of the fuel is 3.5x1011 x3-1x10 8x 10)/6.703 X1013 kg. the mass required: 1.62 x 1033 kg Mev in Joules 6 is equal to 101.60*I0-
19. J 1 Mev = 1.602 X 10 T, which translates to 1.602*1013/2.39x10-3 energy release per kilogram, or 6.702*10-13.

To learn more about Energy please visit -
brainly.com/question/27671072
#SPJ1

8 0

1 year ago

Using the picture below, what is the displacement of the triangle?

zavuch27 [327]

Answer:

i think it is 40 kilometers in the positive direction... if not im sorry

Explanation:

7 0

2 years ago