All categories

3 years ago

How much physical activity should an adult have each week?

Physics

2 answers:

il63 [147K]3 years ago

5 0

Answer:

D. 30 minutes per day, 5 days a week

Explanation:

As per World Health Organization (WHO) physical activity includes the following :

- Leisure time physical activity like running

- Occupational work

- Household chores

- Sports

An adult (aged 18-64) should do at least 150 minutes of physical activity per week. That can be divided over 5 days a week as well with 30 minutes of physical activity per day.

Mars2501 [29]3 years ago

4 0

d 30 minutes a day , 5 days a week

You might be interested in

Amadeo the Archer is practicing his archery skills. He shoots both arrows into a field. His first arrow (arrow A) is embedded ha

attashe74 [19]

Answer:

The arrow with greater impart is Arrow B

The both arrows will feel the same impulse

Explanation:

1. Arrow B since $KE= W = F * d$ it used more force to stop itself in a shorter distance.

2. They should feel the same impulse since the both had the same momentum $m * v= F * t$

4 0

3 years ago

A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor

Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force $F_{1}=6\ N$

Second force $F_{2}=10\ N$

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

$\tau=Force\times lever\ arm$

So, Here,

$\tau_{2}=5 \tau_{1}$

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

$F_{2}\times d_{2}=5\times F_{1}\times r_{1}$

$r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}$

Where, $F_{1}$ =First force

$F_{1}$ =First force

$F_{2}$ =Second force

$r_{1}$ = distance

Put the value into the formula

$r_{2}=\dfrac{5\times6\times0.2}{10}$

$r_{2}=0.6\ m$

Hence, The moment arm is 0.6 m

6 0

3 years ago

Chanice drives her scooter 7 kilometers north. she stops for lunch and then drives 5 kilometers and then 1 km east again. what d

lord [1]

Distance covered is given as follows

1). 7 km North

2). 5 km North

3). 1 km East

Now total distance covered will be given as

$d = d_1 + d_2 + d_3$

$d = 7 km + 5 km + 1 km$

$d = 13 km$

Now in order to find the displacement we will show all with their directions

$\vec d_1 = 7 + 5 = 12 km$ towards North

$\vec d_2 = 1 km$ towards East

So total displacement is

$\vec d = \vec d_1 + \vec d_2$

$\vec d = 12 \hat j + 1 \hat i$

so net displacement will be

$d = \sqrt{12^2 + 1^2} = 12.04 km$

so displacement is 12.04 km

6 0

3 years ago

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

Gwar [14]

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

3 0

3 years ago

An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul

PIT_PIT [208]

Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

= 100 N .

weight at height = 100 N

5 0

3 years ago