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4 years ago

Calculate the power of a weight lifter when a 614.7 N barbell is lifted 4.48 m in 27.4 seconds

Physics

1 answer:

ehidna [41]4 years ago

6 0

Answer:

The power of a weightlifter is 100.5 Watt.

Explanation:

Power is the rate at which the work is done. It can also be defined as the amount of energy converted or transferred in to other forms per unit time.
Mathematically,

Power=Workdone/Timetaken

In this case, work is done by lifting the 614.7 N barbell against gravity given by the potential energy.

Potential Energy = mgh= Force* distance= 614.7*4.48=2753.85 J.

Now, Power= Workdone/Timetaken= 2753.85/27.4=100.5 Watts

Therefore, the power of a weight lifter when a 614.7 N barbell is lifted 4.48 m in 27.4 seconds is 100.5 Watt.

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What is the Scientific definition of friction?

seropon [69]

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Hope this is of great help to you, and happy studying~!
~Mistermistyeyed.

8 0

4 years ago

Read 2 more answers

A specimen of aluminum having a rectangular cross section 9.5 mm × 12.9 mm (0.3740 in. × 0.5079 in.) is pulled in tension with 3

ryzh [129]

Answer:

The resultant strain in the aluminum specimen is $4.14 \times {10^{ - 3}}$

Explanation:

Given that,

Dimension of specimen of aluminium, 9.5 mm × 12.9 mm

Area of cross section of aluminium specimen,

$A=9.5\times 12.9=123.84\times 10^{-6}\ mm^2A=122.55\times 10^{-6}\ m^2$

Tension acting on object, T = 35000 N

The elastic modulus for aluminum is, $E=69\ GPa=69\times 10^9\ Pa$

The stress acting on material is proportional to the strain. Its formula is given by :

$\epsilon=\dfrac{\sigma}{E}$

$\sigma$ is the stress

$\epsilon=\dfrac{F}{EA}\\\epsilon=\dfrac{35000}{69\times 10^9\times 122.5\times 10^{-6}}\\\epsilon=4.14\times 10^{-3}$

Thus, The resultant strain in the aluminum specimen is $4.14 \times {10^{ - 3}}$

5 0

4 years ago

Read 2 more answers

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X

Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

$u'=0.625 c$

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0

3 years ago

Imagine a star very similar to our own Sun in both size and mass. This star moves very slightly back and forth in the sky once e

Inessa05 [86]

Answer:

(c) The planet must have a mass about the same as the mass of Jupiter,

(d) The planet must be closer to the star than Earth is to the Sun.

Explanation:

Astrometry is the ideal method to detect high-mass planets that are close to their star. That is because the gravitational effect that it will have the planet over its host star will be greater. This effect can be seen as a wobble in the star as a consequence of how they orbit a common center of mass¹. The center of mass will be closer to the most massive object, So, in the case of an extrasolar planet with masses like Jupiter (Jovian), this point will be a little bit farther from the star, making the wobble more notable than in a system with a low-mass planet.

Key terms:

Astrometry: study of the position of the stars over time in the sky.

¹Center of mass: a geometrical point in which the mass from a whole system is summed.

6 0

3 years ago

A pulley of radius 8.0 cm is connected to a motor that rotates at a rate 7000 rad s-1 and then decelerate uniformly at a rate of

zlopas [31]

Answer:

(a) α = - 1000 rad/s²

Negative sign represents deceleration.

(b) θ = 3581 rotations

(d) a = - 80 m/s²

Explanation:

(a)

using First equation of motion for angular motion:

ωf = ωi + αt

where,

ωf = Final Angular Speed = 2000 rad/s

ωi = Initial Angular Speed = 7000 rad/s

α = Angular Acceleration = ?

t = time = 5 s

Therefore,

2000 rad/s = 7000 rad/s + α(5s)

α = (2000 rad/s - 7000 rad/s)/5 s

α = - 1000 rad/s²

Negative sign represents deceleration.

(b)

Using second equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = No. of Rotations = ?

Therefore,

θ = (7000 rad/s)(5 s) + (1/2)(- 1000 rad/s²)(5 s)²

θ = 35000 rad - 12500 rad

θ = (22500 rad)(1 rotation/2π rad)

θ = 3581 rotations

(c)

Length of String = L = (Circumference of Pulley)(θ)

L = [2π(0.08 m)][3581 rotations]

L = 1800 m

(d)

Tangential Acceleration = a = rα

a = (0.08 m)(-1000 rad/s²)

a = - 80 m/s²

4 0

3 years ago