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Vitek1552 [10]
4 years ago
8

An athlete does one push-up period in the process, she moves half of her body weight, 250 N, a distance of 20 cm, this distance

is the distance her gravity moves when she fully extend her arms how much work did you do after one push-up
Physics
1 answer:
ioda4 years ago
3 0
In the push-up, the athlete's weight  (or force) of 250 N moves through a distance of 20 cm.

By definition,
Work = force * distance.

Therefore, work done in one push-up is
W=(250 \, N)*(20 \, cm)*(10^{-2} \,  \frac{cm}{m}) = 50 \, J

Note that 1 J = 1  N-m.

Answer:  1 J
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The block A and attached rod have a combined mass of 50 kg and are confined to move along the guide under the action of the 796-
Luba_88 [7]

Answer:

The bending moment is 459.16 N.m

Explanation:

From the given information;

Let's assume that the angle is 66°

Then, the free body diagram is draw and attached in the file below.

Now, the calculation of the acceleration from the first part of the free body diagram is:

\sum F_x = ma_x \\ \\ 796 - 50(9.81) sin 66=50a \\ \\ 796 - 448.094 = 50 a  \\ \\ a = \dfrac{347.906}{50} \\ \\ a  = 6.96 \ m/s^2

Bending moment M:

From the second part of the diagram:

\sum M_B = mad \\ \\ M - (15 \times 9.81) (1.5) = (25 \times 6.96)(1.5 sin 66) \\ \\ M - 220.725 = 238.435  \\ \\  M = 238.435 + 220.725 \\ \\  \mathbf{M = 459.16 \ N.m}

6 0
3 years ago
A light rope is attached to a block with mass 4.10 kg that rests on a frictionless, horizontal surface. The horizontal rope pass
musickatia [10]

Answer:

a)  please find the attachment

(b) 3.65 m/s^2

c) 2.5 kg

d) 0.617 W

T<weight of the hanging block

Explanation:

a) please find the attachment

(b) Let +x be to the right and +y be upward.

The magnitude of acceleration is the same for the two blocks.  

In order to calculate the acceleration for the block that is resting on the horizontal surface, we will use Newton's second law:  

∑Fx=ma_x

   T=m1a_x

  14.7=4.10a_x

 a_x= 3.65 m/s^2

c) <em>in order to calculate m we will apply newton second law on the hanging   </em>

<em>    block</em>

<em> </em>∑F=ma_y

T-W= -ma_y

T-mg= -ma_y

T=mg-ma_y

T=m(g-a_y)

a_x=a_y

14.7=m(9.8-3.65)

 m = 2.5 kg

<em>the sign of ay is -ve cause ay is in the -ve y direction and it has the same magnitude of ax</em>

d) calculate the weight of the hanging block :

W=mg

W=2.5*9.8

  =25 N

T=14.7/25

 =0.617 W

T<weight of the hanging block

6 0
3 years ago
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