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3 years ago

Why losing is an important aspects of sports

Physics

2 answers:

g100num [7]3 years ago

5 0

It is important to realize that it is ok to be defeated during the game and that it’s not just all about winning

dlinn [17]3 years ago

3 0

Answer:

It reminds us that we need to work harder. It allows us to make adjustments in the way and manner in which we train and practice. In a loss, we are able to identify our vulnerabilities and weaknesses, and work to improve.

Explanation:

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Which of the following are not units of frequency: Hertz. m/s. waves/s

ra1l [238]

The answer is waves.

8 0

3 years ago

The seismic activity density of a region is the ratio of the number of earthquakes during a given time span to the land area aff

Natalija [7]

Answer:

0.0059

Explanation:

According to the question the seismic activity density is given by

$\text{Seismic activity density}=\frac{\text{Number of Earthquakes over a given time span}}{\text{The land area affected}}$

Here,

Number of Earthquakes over a given time span = 424

The land area affected = 71300 mi²

So,

$\text{Seismic activity density}=\frac{424}{71300}\\\Rightarrow \text{Seismic activity density}=0.0059$

The seismic activity density is 0.0059

8 0

3 years ago

Read 2 more answers

A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air

fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

$u_y=0$ is the initial vertical velocity

t is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

$v_x = 10 m/s$

Therefore, the horizontal distance covered after a time t is

$d=v_x t$

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

$d=(10)(2.02)=20 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

3 years ago

A first order reaction, A -> products, has a rate reaction of .00250 Ms-1 when [A] = . 484 M. (a) What is the rate constant,

tamaranim1 [39]

Answer: a) The rate constant, k, for this reaction is $0.00516s^{-1}$

b) No $t_{\frac{1}{2}}$ does not depend on concentration.

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$A\rightarrow products$