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3 years ago

Choosing to run so that you can become a better runner is an example of which fitness principle?

2 answers:

7 0

The answer is specificity. You are selecting a specific exercise option that is directed to be relevant to improve your goal.

Nitella [24]3 years ago

5 0

Specificity, hopefully this helps

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In order to sail through the frozen Arctic Ocean, the most powerful icebreaker ever built was constructed in the former Soviet U

professor190 [17]

Answer:

955.36 seconds ≈ 16 minutes

Explanation:

Power(P) is the rate of doing work(W)

That is, P = W/t, where t is the time.

multipying both sides with 't' and dividing with 'P', we get: t=W/P

Here, W = 5.35 x 10^10 J and P = 5.6 x 10^7 W ( 1 W = 1 J/s).

Therefore , on dividing W with P, we get 955.36 seconds.

6 0

3 years ago

is this true or false? an object with a mass of 9.0kg has an acceleration of 3m/s2. the force acting on it is 3N

ludmilkaskok [199]

The answer is "False". The force acting on the object is 27 N.

According to Newton's second law, when a force F acts on am object of mass m, it produces an acceleration a. The force is given by the expression,

$F=ma$

Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,

$F=ma\\ =(9.0kg)(3m/s^2)\\ =27N$

Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.

5 0

3 years ago

What is nuclear fission? (Points : 1)

mixas84 [53]

The splitting of the atomic nucleus into parts

4 0

3 years ago

Read 2 more answers

The horizontal surface on which the block (mass 2.0 kg) slides is frictionless. The speed of the block before it touches the spr

ch4aika [34]

Answer:3.67 m/s

Explanation:

mass of block(m)=2 kg

Velocity of block=6 m/s

spring constant(k)=2 KN/m

Spring compression x=15 cm

Conserving Energy

energy lost by block =Gain in potential energy in spring

$\frac{m(v_1^2-v_2^2)}{2}=\frac{kx^2}{2}$

$2\left [ 6^2-v_2^2\right ]=2\times 10^3\times \left [ 0.15\right ]^2$

$v_2=3.67 m/s$

7 0

3 years ago

A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.

LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

$\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}$

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

$W = \frac{1}{2}(50)(5.61^2) = 786.8025 J$

Now, we can define work:

$\large\boxed{W = Fdcos\theta}}$

Now, plug in the values:

$786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta$

Solve for theta:

$cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}$

4 0

2 years ago