Muscular strength is BEST described by

Physics

1 answer:

oee [108]3 years ago

7 0

The answer is C. Muscular strength is the force produced in a single maximum effort.

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A pendulum consists of a 1.7-kg block hanging on a 1.6-m length string. A 0.01-kg bullet moving with a horizontal velocity of 82

Rzqust [24]

Answer:

0.42 m

Explanation:

mass of pendulum, M = 1.7 kg

Length of pendulum , l = 1.6 m

mass of bullet, m = 0.01 kg

initial velocity of bullet, u = 828 m/s

final velocity of bullet, v = 340 m/s

initial velocity of pendulum, U = 0

Let the final velocity of pendulum is V.

Use conservation of momentum for bullet and the pendulum

m x u + M x U = m x v + M x V

0.01 x 828 + 1.7 x 0 = 0.01 x 340 + 1.7 x V

8.28 + 0 = 3.4 = 1.7 V

V = 2.87 m/s

Now the kinetic energy of the pendulum is converted into potential energy of pendulum and let it raised to a height of h from the initial level.

Use energy conservation

Kinetic energy of the pendulum = potential energy of the pendulum

0.5 x M x V² = M x g x h

0.5 x 2.87 x 2.87 = 9.8 x h

h = 0.42 m

6 0

3 years ago

A ball is ejected to the right with an unknown horizontal velocity from the top of a pillar that is 50 meters in height. At the

dimulka [17.4K]

Answer:

15.67 m/s

Explanation:

The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.

Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by

$S=\frac{1}{2}gt^2$

where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(50 m)}{9.8 m/s^2}}=3.19 s$

Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:

$v=\frac{d}{t}=\frac{50 m}{3.19 s}=15.67 m/s$