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An athlete is working out in the weight room. he steadily holds 50 kg above his head for ten seconds which statement is true abo

Andrews [41]

Hi. The answer to your question is the first option.

The athlete isn’t doing any work because he doesn’t move the weight.

Hope this helps :))

5 0

3 years ago

A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff

Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0

2 years ago

A 153 g mass is attached to the end of an unstressed vertical spring (of constant 24.7 N/m) and then dropped. The acceleration o

Arte-miy333 [17]

Answer:

The answer to the question is

Its maximum speed is 1.54 m/s

Explanation:

Work done = Kinetic energy

0.5·m·v² = 0.5·k·x²

Where

m = mass

v = velocity

k = spring constant

x = extension of the spring

We note that Force F is given by

F = m·a

Where

a = acceleration due to gravity

= 0.153×9.8 = 1.4994 N

Equating the work done by the force to the work done on the spring gives

Work done = Force × Distance = 1.4994×x = 0.5×k÷x² = 0.5×24.7×x²

x = 1.4994÷12.35 = 0.121 m

Substituting the value of x into the equation below gives

0.5·m·v² = 0.5·k·x²

0.5×0.153×v² = 12.35×0.121²

v² = 0.182÷0.0765 = 2.379

v = 1.54 m/s

6 0

3 years ago

A horizontal force of 400.0 N is required to pull a 1760 N truck across the floor at a constant speed. Find the coefficient of s

Tomtit [17]

Data given:

Fh=400N

Ftruck=1760N

Data needed:

u=?

Formula needed:

Fh=Ftruck×u

Solution:

u=Fh/Ftruck

u=400N/1760N

u=0,227272727

4 0

3 years ago

For an increase in the bulk modulus of a material but without any change in the density, what happens to the speed of sound in t

vazorg [7]

The speed of sound, c, is given by the Newton-Laplace formula
$c = \sqrt{ \frac{K}{\rho} }$
where
K = bulk modulus
ρ = density

Because the density is constant, the speed of sound is proportional to the square root of the bulk modulus.

Therefore when the bulk modulus increases, the speed of sound increases by the square root of the bulk modulus.

For example, if K is doubled, then
$c = \sqrt{2K} = \sqrt{2} \sqrt{K}$

Answer:
If the bulk modulus increases by a factor of n, then c increases by a factor of √n.

7 0

3 years ago