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LenaWriter [7]
3 years ago
14

An athlete whose mass is 87.0 kg is performing weight-lifting exercises. Starting from the rest position, he lifts, with constan

t acceleration, a barbell that weighs 600 N. He lifts the barbell a distance of 0.65 m in a time of 1.3 s. Use Newton's laws to find the total force that his feet exert on the ground as he lifts the barbell.
Physics
1 answer:
expeople1 [14]3 years ago
6 0

Answer: X = 52,314.12 N

Explanation: Let X be the force the feet of the athlete exerts on the floor.

According to newton's third law of motion the floor gives an upward reaction based on the weight of the athlete and the barbell which is known as the normal reaction ( based on the mass of the athlete and the barbell)

Mass of athlete = 87kg, mass of barbell = 600/ hence total normal reaction from the floor = 87* 61.22/ 9.8 *9.8 = 52,200N.

The athlete lifts the barbell from rest thus making it initial velocity u=0, distance covered = S = 0.65m and the time taken = 1.3s

The acceleration of the barbell is gotten by using the equation of constant acceleration motion

S= ut + 1/2at²

But u = 0

S = 1/2at²

0.65 = 1/2 *a (1.3)²

0.65 = 1.69 * a/2

0.65 * 2 = 1.69 * a

a = 0.65 * 2/ 1.69

a = 0.77m/s²

According to newton's second law of motion

Resultant force = mass * acceleration

And resultant force in this case is

X - 52,200 = (87 + 61.22) * 0.77

X - 52,200 = 148.22 * 0.77

X - 52, 200 = 114.132

X = 114.132 + 52,200

X = 52,314.12 N

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Answer:

R = 9880 yd * 3 ft/yd / 5280 ft/mi = 5.61 mi

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R = 9880 yd * 3 ft/yd = 29640 ft

R = 29640 ft / 5280 ft/mi = 5.61 mi

6 0
3 years ago
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the
Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan m=12 kg

Inclination \theta =30^{\circ}

Tension T=29 N

coefficient of Friction \mu =0.18

Resolving Forces Along x axis

F_x=T\cos \theta -f_r

where f_r=friction\ Force  

F_y=mg-N-T\sin \theta

since there is no movement in Y direction therefore

N=mg-T\sin \theta

and f_r=\mu N

Thus F_x=T\cos \theta -\mu N

F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))                

F_x=25.114-18.558

F_x=6.556 N

Work done by applied Force is equal to change to kinetic Energy

F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2

6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2

v_f^2=\frac{6.556\times 2.7\times 2}{12}

v_f^2=2.95

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8 0
3 years ago
With regard to pneumatic mechanical devices which afford protection against the formation of ice, which of the following stateme
Andreyy89

Answer:

the answer is The pneumatic mechanical device can only be used as a de-icing device.

Explanation:

An ice protection system prevents the formation of ice, or enables the aircraft to shed the ice before it can grow to a dangerous thickness. Ice protection systems are designed to keep atmospheric ice from accumulating on aircraft surfaces such as wings, propellers and engine intakes.

The pneumatic mechanical device is the Pneumatic deicing boots which was invented by the Goodrich Corporation in 1923. The pneumatic boot is usually made of layers of rubber, with one or more air chambers between the layers.

Any design which utilizes either a mechanical means of breaking the bond of ice to the surface, or which operates on a periodic cycle, is necessarily a de-ice system.

4 0
4 years ago
. A laser beam shines along the surface of a block of transparent material (see Fig. E33.8). Half of the beam goes straight to a
Neporo4naja [7]

Answer:

n = 1,875

Explanation:

The speed of light in vacuum is constant (c) and in a material medium it is

          v = d / t

The refractive index of a material is defined by

         n = c / v

         

Let's look for the speed of light in the material, in general the length that light travels is known, this value is high, x = 1, when we place a block on the road, a small amount is lengthened by the length of the block, which in general is despised

These measurements are made on a digital oscilloscope that allows to stop the signals and measure their differences, that is, the zero is taken when the first ray arrives and the time for the second ray is measured,

         

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we calculate the refractive index

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Solution :

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90 revolutions in 1 minute

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                   = 75 rev

Therefore, the ball made 75 revolutions during the 5 seconds.

7 0
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