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Stells [14]
3 years ago
6

a 74 kg man jumps from a window 1.6 m above a sidewalk. the acceleration of gravity is 9.81 m/s 2 . a) what is his speed just be

fore his feet strike the pavement? answer in units of m/s
Physics
1 answer:
olasank [31]3 years ago
4 0

Answer:

  5.60 m/s

Explanation:

His initial potential energy is equal to his final kinetic energy.

  mgh = 1/2mv^2

  2gh = v^2 . . . . . . . multiply by 2/m

  v = √(2gh) = √(2(9.81 m/s²)(1.6 m)) = √(31.392 m²/s²)

  v ≈ 5.60 m/s

_____

<em>Additional comment</em>

The mass is irrelevant in the equations we typically use for this purpose.

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5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj
irina [24]

Answer:

jack JACKKK

Explanation:

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3 0
4 years ago
A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?
Natali [406]

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒  Density = \frac{Mass}{Volume}

or,

⇒  Volume=\frac{Mass}{Density}

On substituting the values, we get

⇒               =\frac{6.3 \ g}{19.3 \ g/mL}

⇒               =0.32 \ mL

7 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
A 1,200 kg car rolls down a hill with its brakes off and transmission in neutral. At one moment it is moving 5.0 m/s. A little l
Y_Kistochka [10]
The change in kinetic energy of the car is equivalent to the change in its potential energy. Thus:
K.E  = P.E
1/2 x mΔv² = mgΔh
h = (8.2² - 5²) / 2(9.81)
h = 2.15 meters
4 0
4 years ago
in the 1860s a scientist named ____proposed an arrangement of elements that formed the basis for what we know today as the perio
emmasim [6.3K]

The answer in blank would be " Leo Stevenson"

Hope this helps you! :D

4 0
3 years ago
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