Ask question

Ask question

All categories

3 years ago

11

Select the correct exercise category for each of the specific ways to increase intensity

1 answer:

Alecsey [184]3 years ago

3 0

The answer is cardiovascular.

You might be interested in

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Mila [183]

Answer:

$v=6.65m/sec$

Explanation:

From the Question we are told that:

Mass $m=97.6$

Coefficient of kinetic friction $\mu k=0.555$

Generally the equation for Frictional force is mathematically given by

$F=\mu mg$

$F=0.555*97.6*9.8$

$F=531.388N$

Generally the Newton's equation for Acceleration due to Friction force is mathematically given by

$a_f=-\mu g$

$a_f=-0.555 *9.81$

$a_f=-54455m/sec^2$

Therefore

$v=u-at$

$v=0+5.45*1.22$

$v=6.65m/sec$

4 0

3 years ago

What would be the projectile's initial speed?

zalisa [80]

Math is the process of using the given information, along with
all the general stuff that you know, to find the missing information.
With no given information, we have no way to even guess at
an answer.

8 0

3 years ago

Who is the founding father of modern psychology?

Zepler [3.9K]

Answer:

Sigmund Freud

Explanation:

4 0

3 years ago

Read 2 more answers

By any method you choose, determine your average speed of walking. How do you results compare with those of your classmates?

riadik2000 [5.3K]

joji sanctuary

slow dancing in the dark

happier by olivia

filipino artist like john cena

7 0

3 years ago

Read 2 more answers

The ground state energy of an electron in a one-dimensional trap with zero potential energy in the interior and infinite potenti

ElenaW [278]

Answer:

0.5 eV

Explanation:

$E_1$ = Initial potential energy = $2\ eV$

$E_2$ = Final potential energy

$L_1$ = Initial width

$L_2$ = Final width = $2L_1$

Energy of an electron in a one-dimensional trap is given by

$E=\dfrac{n^2h^2}{8mL^2}$

From the equation we get

$E\propto \dfrac{1}{L^2}$

So,

$\dfrac{E_1}{E_2}=\dfrac{L_2^2}{L_1^2}\\\Rightarrow E_2=\dfrac{E_1L_1^2}{L_2^2}\\\Rightarrow E_2=\dfrac{2L_1^2}{4L_1^2}\\\Rightarrow E_2=0.5\ eV$

The ground state energy will be 0.5 eV

6 0

3 years ago