Fitness results come from hard work, not an instant fix Question 1 options: True False

1 answer:

5 0

<span>This is a true statement. Fitness is typically a long-term solution, not a short-term repair. When looking for fitness results, it will usually take weeks to months to see any tangible results, even though there could be some short-term benefits such as increased energy. This is why fitness is sometimes seen as a lifetime endeavor and not one that occurs overnight.</span>

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Please just give me an answer, thanks.

Colt1911 [192]

$\\ \rm\dashrightarrow P=VI\implies V=\dfrac{P}{I}$

$\\ \rm\dashrightarrow V=\dfrac{0.4\times 10^3}{33\times 10^{-3}}$

$\\ \rm\dashrightarrow V=0.012\times 10^{-6}$

$\\ \rm\dashrightarrow V=12mV$

$\\ \rm\dashrightarrow R=\rho \dfrac{\ell}{A}$

$\\ \rm\dashrightarrow \rho =\dfrac{RA}{\ell}$

$\\ \rm\dashrightarrow \rho=\dfrac{86.3(0.4\times 10^6)}{69}$

$\\ \rm\dashrightarrow \rho=0.5\times 10^{6}$

It should be silicon

Ohms law

$\\ \rm\dashrightarrow R=\dfrac{V}{I}$

$\\ \rm\dashrightarrow R=\dfrac{7(10^3)}{6(10^{-6})}$

$\\ \rm\dashrightarrow R=1.167\times 10^9\Omega$

7 0

2 years ago

What should be the spring constant k of a spring designed to bring a 1200 kg car to rest from a speed of 85 km/h so that the occ

borishaifa [10]

Answer:

k = 5178.8 N/m

Explanation:

As we know that spring mass system will oscillate at angular frequency given as

$\omega = \sqrt{\frac{k}{m}}$

now we have

$\omega = \sqrt{\frac{k}{1200}}$

now the maximum acceleration of the spring block system is at its maximum compression state which is given as

$a = \omega^2 A$

here A= maximum compression of the spring

so here in order to find maximum compression of the spring we will use energy conservation as we know that initial total kinetic energy of the car will convert into spring potential energy

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2$