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ad-work [718]
3 years ago
6

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. J
Physics
2 answers:
morpeh [17]3 years ago
6 0

Answer:

1225 J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

PE=mgh

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

PE=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J

disa [49]3 years ago
4 0

Answer:

1225J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

1225J

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Its true i hope this helps you. Tell me if it is correct. ( =
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3 years ago
2.5 g of helium at an initial temperature of 300 K interacts thermally with 9.0 g of oxygen at an initial temperature of 620 K .
muminat

Answer:

Explanation:

2.5 g of He = 2.5 / 4  mole

= .625 moles

9 g of oxygen = 9/32

= .28 mole of oxygen

C_p of He = 3/2 R

C_p of O₂ = 5/2 R

A ) Initial thermal energy of He = 3/2 n R T

= 1.5 x .625 x 8.32 x 300

= 2340 J

Initial thermal energy of O₂ = 5/2 n R T

= 2.5 x .28 x 8.32 x 620

= 3610.88 J

B ) If T be the equilibrium temperature after mixing

gain of heat by helium

= n C_p Δ T

= .625 x 3/2 R x ( T - 300 )

Loss of heat by oxygen

n C_p Δ T

= .28 x 5/2 R x ( 620 - T )

Loss of heat = gain of heat

.625 x 3/2 R x ( T - 300 ) = .28 x 5/2 R x ( 620 - T )

1.875 T- 562.5 = 868- 1.4 T

3.275 T = 1430,5

T = 436.8 K

Thermal energy of He

= 1.5 x .625 x 8.32 x 436.8

= 3407 J

thermal energy of O₂

= 2.5 x .28  x 8.32 x 436.8

= 2543.92 J

C )

Heat energy transferred

=  .28 x 5/2 R x ( 620 - T )

=  .28 x 5/2 x  8.32 x ( 620 - 436.8 )

1066.95 J

Heat will flow from O₂ to He

Final temperature is 436.8 K

7 0
3 years ago
A 0.150 kg baseball has 118 j of KE. how fast is the ball moving?(unit=m/s)
MrRissso [65]

Answer:

Explanation 118 = (1/2) * 0.15 * v² 118 = 0.075 * v² v² = 1573.33 m/s ... since KE = m/2*V^2 , then : V = √2KE/m = √20*118/1.5 = 39.67 m//sec ( 142.8 km/h ; 88.75 mph).:

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3 years ago
Bird bones have air pockets in them to reduce their weight. This also gives them an average density significantly less than that
mixas84 [53]

Answer:

mas of water displaced = 41.4 g

Explanation:

Weight in air = True weight = 45 g

Apparent weight = 3.6 g

Apparent weight = True weight - Buoyant force

Buoyant force = 45 g - 3.6 g = 41.4 g

Weight of water displaced = Buoyant force

Weight of water displaced = 41.4 g dyne

mas of water displaced = 41.4 g

8 0
3 years ago
You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the s
olga_2 [115]

To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.

Centrifugal acceleration is determined by

a_c = \frac{v^2}{r}

Where,

v = Velocity

r = Radius

From the given data of the moon we know that gravity there is equivalent to

a = 1.62m/s

While the radius of the moon is given by

r = 1.74*10^6m

If we rearrange the function to find the speed we will have to

v = \sqrt{ar}

v = \sqrt{1.6(1.74*10^6)}

v = 1.7km/s

The speed for this to happen is 1.7km/s

3 0
3 years ago
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