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3 years ago

How often should muscle strengthening activities be performed?

Physics

2 answers:

nevsk [136]3 years ago

8 0

Answer:

B-Once a week

Explanation:

You answer is B-Once a week

LekaFEV [45]3 years ago

3 0

Answer:

once a week

Explanation:

once a week so answer B

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A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi

Darina [25.2K]

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

5 0

3 years ago

Plssss help I need good grade!!!! Plssss and thank u

shepuryov [24]

Answer:

the answer is 'b'

According to me

8 0

2 years ago

Read 2 more answers

A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the

asambeis [7]

EC_1 + EP_1 = EC2 + EP_2

EC_2 = 0

EC_2 = EP_1 - EP_2

EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J

7 0

3 years ago

Read 2 more answers

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

3 years ago

If the energy carried by a wave increases, which other wave property also increases?

damaskus [11]

Answer:

The answer is wavelength

5 0

3 years ago