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fomenos
4 years ago
13

A person bends over to grab a 20 kg object. The back muscle responsible for supporting his upper body weight and the object is l

ocated 2/3 of the way up his back (where it attaches to the spine) and makes an angle of 12 degrees with the spine. His upper body weighs 36 N. What is the tension in the back muscle?
A. 1500 N
B. 1700 N
C. 2700 N
D. 3500 N

Physics
1 answer:
ANTONII [103]4 years ago
4 0

Answer:

C) 2700 N

Explanation:

As per FBD of the given situation we can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object

The tension force is acting at an angle of 12 degree

while both weight are acting perpendicular to the length

so here we have

\tau_{clockwise} = \tau_{counterclockwise}

m_1g(\frac{L}{2}) + m_2g(L) = Tsin12(\frac{2L}{3})

now we have

36(9.81)(\frac{L}{2}) + 20(9.81)(L) = T sin12(\frac{2L}{3})

T = 2700 N

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3 years ago
An object with a mass of 10 kg is accelerated upward at 2 m/sec2. What force is required?
telo118 [61]

Answer:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

Explanation:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

6 0
3 years ago
In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a
SOVA2 [1]

Answer:0.084 m/s^2

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance d_1=8.13 km

time taken=25 min =1500 s

initial speed v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s

after 8.13 km mark steve started to accelerate

speed after 60 s

v_2=v_1+at

v_2=5.6+a\times 60

distance traveled in 60 sec

d_2=v_1\times 60+\frac{a60^2}{2}

d_2=336+1800 a

time taken in last part of journey

t_3=1663.6-1560=103.6 s

distance traveled in this time

d_3=v_2\times t_3

d_3=\left ( 5.6+a\times 60\right )103.6

and total distance=d_1+d_2+d_3

10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6

1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6

a=0.084 m/s^2

5 0
4 years ago
How do I do this question?
Mkey [24]

a. 27

after that all you need to do is use the work formula as you will be able to acheive results

5 0
3 years ago
A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in
lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

        A = 1.5 \times 2.30

           = 3.45 in^{2}

The given data is as follows.

          Allowable stress = 14,500 psi

          Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

           P_{max} = \sigma_{a}A

                       = 14500 \times 3.45

                       = 50025 lb

Now, maximum load from shear stress is as follows.

           P_{max} = 2 \times \tau_{a} \times A

                      = 2 \times 7100 \times 3.45

                      = 48990 lb

Hence, P_{max} will be calculated as follows.

       P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})

                  = 48990 lb

Thus, we can conclude that the maximum permissible load P_{max} is 48990 lb.

4 0
4 years ago
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