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LenaWriter [7]
3 years ago
15

Determine the acceleration of the box of mass 140 kg, when a net force of 900 N acts on it.​

Physics
1 answer:
Georgia [21]3 years ago
8 0

Answer:

a = 6.4 m/s²

Explanation:

a = F/m = 900 / 140 = 6.428571... m/s²

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Which strategies maximize recovery following an intense bout of resistance training?
RoseWind [281]

Answer:

The answers are:

a) drinking 20–24 ounces of water for every pound of body mass lost,

b) stretching muscle groups, holding each stretch for 30 seconds or more, and

c) replenishing low blood sugar with carbohydrates

Explanation:

I just took the exam and got this correct.

8 0
3 years ago
I need help with these questions
Feliz [49]
7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J

8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m

9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333


6 0
4 years ago
Which is not a way to conserve existing energy resources?
zimovet [89]
For the answer to the question above, the answer is simple.<span> among</span> the choices given the only way on not to conserve energy is by using the available fossil fuel.
I hope my answer helped you. feel free to ask more questions. Have a nice day!
8 0
3 years ago
Read 2 more answers
Please help me do this problem
Digiron [165]

Answer:

i) acceleration from B to D is 0, because the velocity is constant (stays the same)

ii) whatever units of distathat might be, we can calculate the number:

for 4 time-steps (2 to 6) the velocity is 6 per time step, that makes 24 distance units in these 4 time steps. it's the same the area underneath the graph.

there is also the vertical line from 0 to 2. we can calculate that distance like the area of a triangle with 2*6 / 2 = 6

the total distance from 0 to D is therefore 30

4 0
3 years ago
A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H
Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

\theta =44.8{\circ}

using v^2-u^2=2as

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration   (gsin\theta )

s=distance traveled

0-(1.72)^2=2(-9.81\times sin(44.8))s

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

0=1.72-gsin(44.8)\times t

t=\frac{1.72}{9.8\times sin(44.8)}

t=0.249 s

(c)Speed of the block at bottom

v^2-u^2=2as

Here u=0 as it started coming downward

v^2=2\times gsin(44.8)\times 0.214

v=\sqrt{2.985}

v=1.72 m/s

3 0
3 years ago
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