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Mice21 [21]
3 years ago
14

A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. Approximately how m

uch power does the weightlifter exert on the barbell during this time?
Physics
1 answer:
Alex73 [517]3 years ago
6 0

Amount of work done is zero and so power = 0 watts.

<u>Explanation:</u>

Power is the rate at which work is done, or W divided by delta t. Since the barbell is not moving, the weightlifter is not doing work on the barbell.Therefore, if the work done is zero, then the power is also zero.It may seem unusual that the data given in question is versatile i.e. A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. But, still the answer is zero watts , this was a tricky question although conceptual basis of question was good! Power is dependent on amount of work done which is further related to displacement and here the net displacement is zero ! Hence, amount of work done is zero and so power = 0 watts.

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The current theory of the structure of the
Mariana [72]

Answers:

a) 2.82(10)^{21} kg

b) 1410 J

c) 36.62 m/s

Explanation:

<h3>a) Mass of the continent</h3>

Density \rho  is defined as a relation between mass m and volume V:

\rho=\frac{m}{V} (1)

Where:

\rho=2720 kg/m^{3} is the average density of the continent

m is the mass of the continent

V is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}

Finding the mass:

m=\rho V (2)

m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3}) (3)

m=2.82(10)^{21} kg (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy K is given by the following equation:

K=\frac{1}{2}mv^{2} (5)

Where:

m=2.82(10)^{21} kg is the mass of the continent

v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s is the velocity of the continent

K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2} (6)

K=1410 J (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass m=77 kg and the same kinetic energy as that of the continent 1413 J, we can find its velocity by isolating v from (5):

v=\sqrt{\frac{2 K}{m}} (6)

v=\sqrt{\frac{2 (1413 J)}{77 kg}}

Finally:

v=36.62 m/s This is the speed of the jogger

5 0
3 years ago
Pitch is determined by the ____ of a sound wave.
mart [117]
The answer is c .Frequency
8 0
2 years ago
Read 2 more answers
The largest graduated cylinder in my lab holds 2 L and has an inner diamter (the part that holds the water) of 8 cm. When it is
mestny [16]

Answer:

<em>3924 Pa</em>

<em></em>

Explanation:

Volume of cylinder = 2 L = 0.002 m^3  (1000 L = 1 m^3)

diameter of the inner cylinder = 8 cm = 0.08 m  (100 cm = 1 m)

radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m

area of the inner cylinder = \pi r^{2}

where \pi = 3.142,

and r = radius = 0.04 m

area of inner cylinder = 3.142 x 0.04^{2} = 0.005 m^2

<em>height h of the water in this cylinder = volume/area</em>

h = 0.002/0.005 = 0.4 m

<em>pressure at the bottom of the cylinder due to the height of water = pgh</em>

where

p = density of water = 1000 kg/m^3

g = acceleration due to gravity = 9.81 m/s^2

h = height of water within this cylinder = 0.4 m

pressure = 1000 x 9.81 x 0.4 = <em>3924 Pa</em>

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2 years ago
A transformer has input voltage and current of 12 V and 4 A
Advocard [28]

Explanation:

It is given that,

Input voltage, V_i=12\ V

Input current, I_i=4\ A

Output current, I_o=0.8\ A

Number of turns in the secondary side of transformer, N_s=1177

We need to find the number of turns in the primary side of the transformer. The current to the number of turns in the input and output is given by :

\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}

Substituting all the above values

So,

N_p=\dfrac{N_sI_s}{I_p}\\\\N_p=\dfrac{1177\times 4}{0.8}\\\\N_p=5885

So, the number of turns in primary side of the transformer is 5885.

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2 years ago
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