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3 years ago

Regular exercise is positively related to wellness

Physics

1 answer:

Harman [31]3 years ago

7 0

Answer:

yes ( true)

Explanation:

positive effects on all the body systems.

You might be interested in

A woman flies 300 kilometers north, 200 kilometers west, and 500 kilometers south. How far is she from her starting point?

maria [59]

She is
$\sqrt{200^2+200^2}= ~282.84$ kilometers away from her starting point

5 0

4 years ago

Which of these materials is permeable?

kogti [31]

Correct answer is letter B. sandstone

8 0

3 years ago

Find the 24th term of the arithmetic sequence 11, 14, 17, … . Express the 24 terms of the series of this sequence using sigma no

Anna71 [15]

Answer:

The 24th term is 80 and the sum of 24 terms is 1092.

Explanation:

Given that,

The arithmetic series is

11,14,17,........24

First term a = 11

Difference d = 14-11=3

We need to calculate the 24th term of the arithmetic sequence

Using formula of number of terms

$t_{n}=a+(n-1)d$

Put the value into the formula

$t_{24}=11+(24-1)\times3$

$t_{24}=80$

$t_{24}=u_{24}=80$

We need to calculate the sum of the first 24 terms of the series

Using formula of sum,

$S_{n}=\dfrac{n}{2}(a+u_{24})$

Put the value into the formula

$S_{n}=\dfrac{24}{2}\times(11+80)$

$S_{n}=1092$

Hence, The 24th term is 80 and the sum of 24 terms is 1092.

5 0

3 years ago

The longest banana split ever made was 7.32 km long (obviously they used more than one banana). If an archer were to shoot an ar

elixir [45]

Answer:

The horizontal displacement of the arrow is not larger than the banana split.

Explanation:

Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.

So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.

y - y₀ = ut - 1/2gt²

0 - y₀ = 0 × t - 1/2gt²

-y₀ = -1/2gt²

t² = 2y₀/g

t = √(2y₀/g)

Substituting the values of the variables, we have

t = √(2y₀/g)

= √(2 × 8848.04 m/9.8 m/s²)

= √(17696.08 m/9.8 m/s²)

= √(1805.72 s²)

= 42.5 s

The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s

So, d = vt

= 100 m/s × 42.5 s

= 4250 m

= 4.25 km

Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.

8 0

2 years ago

A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m

kenny6666 [7]

Answer:

61.76 N.

Explanation:

Given the mass of the car, m = 1.60 kg.

The speed of the car, v = 12.0 m/s.

The radius of the circle, r = 5 m.

As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.

Let N is the normal force.

So, $N - mg = F_c$

$N-mg=\frac{mv^2}{r}$

Now substitute the given values, we get

$N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}$

$N=15.68+46.08$

N = 61.76 N.

Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.

5 0

3 years ago