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natulia [17]
2 years ago
14

A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans

wer to the nearest whole number.
Physics
2 answers:
Anna71 [15]2 years ago
6 0
The Gravitationa potential energy of the mass (PEG) is given by:
U=mgh
where
m is the mass
g is the gravitational acceleration
h is the heigth of the mass above the reference level (the ground)

In this problem, m=250 kg and h=0.5 m, therefore the gravitational potential energy of the mass is:
U=mgh=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J
emmainna [20.7K]2 years ago
3 0

Answer:

i can confirm it is 1,225

Explanation:

based on the website. I just did this question.

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Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly h
Bezzdna [24]
:<span>  </span><span>30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s 

At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg) 

So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road .. 

Fn = mg - mv²/R 
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards) 
(b) 
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero. 

ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)² 

►v = √198 = 14.0 m/s</span>
3 0
2 years ago
13. An aircraft heads North at 320 km/h rel:
AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0
2 years ago
Sea breezes that occur near the shore are attributed to a difference between land and water with respect to what property?
ddd [48]

Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

  • When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.

<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

7 0
2 years ago
How do air mass conditions ahead of the squall line support the development of new cell?
IRISSAK [1]
<span>Storm cells in a squall line typically move from the southwest to the northeast, and as the mature cells in the northeast begin to die off, new ones are formed at the opposite end to advance the line. The air in the southwest corner has strong vertical updrafts that allow new cells to grow and develop into thunderstorms.</span>
7 0
2 years ago
The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. what percentage of the mechanical energy of th
Zanzabum

E = ½KA^2 is the mechanical energy of any oscillator.  It is the sum of elastic potential energy and kinetic energy.  When amplitude A decreases by 3%, then

(E2-E1)/E1 = {½K(A2^2/A1^2) }/ ½K(A1^2)

= {(A2^2 – A1^2) / (A1^2)}

= 97^2 – 100^2/100^2

= 5.91% of the mechanical energy is lost each cycle.

3 0
2 years ago
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