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ELEN [110]
3 years ago
6

What activity is mainly a strength training excerise?

Physics
1 answer:
Sever21 [200]3 years ago
6 0

Biceps curls & pushups , benchpress

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sound wave-electrical wave-radio wave

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Read this excerpt from Through the Looking-Glass by Lewis Carroll.
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I think it is why is Alice suddenly so thirsty

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One group of students uses different methods to conserve water as listed below.
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The best answer is b) <span>Method 1 and Method 3

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4 years ago
if Gabriel kicks the football 50 meters to the west in 3 seconds , what is the velocity of the football?
Svet_ta [14]

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velocity = 16.67[m/s]

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8 0
3 years ago
A hockey puck is hit on a frozen lake and starts moving with a speed of 12.0 m/s. exactly 5.0 s later, its speed is 6.0 m/s. wha
Anna35 [415]
1) In an uniformly accelerated motion, the acceleration is given by:
a= \frac{v_f-v_i}{\Delta t}
where v_f is the final speed, v_i is the initial speed, and \Delta t is the time interval between the initial and final point of the motion.

Using the data of the problem: v_i = 12.0 m/s, v_f = 6.0 m/s, and the time \Delta t = 5.0 s, the acceleration is
a= \frac{6 m/s-12 m/s}{5.0 s}=-1.2 m/s^2
where the negative sign means that the hockey puck is decelerating.

2) The frictional force F between the puck and ice is responsible for the deceleration of the puck, and for second Newton's law this force is equal to the product between the mass of the puck m and the acceleration:
F=ma (1)
The frictional force can also be written as
F=-\mu_D mg (2)
where \mu_D is the coefficient of kinetic friction, and where the negative sign is due to the fact that the frictional force acts against the direction of motion of the puck. By equilizing (1) and (2) we can find the value of this coefficient:
ma=-\mu_D mg
\mu_D =   -\frac{a}{g}=  -\frac{-1.2 m/s^2}{9.81 m/s^2}=0.12
5 0
4 years ago
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