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VMariaS [17]
3 years ago
8

Which is an example of an unstructured activity that promotes resistance training?

Physics
1 answer:
aleksandrvk [35]3 years ago
3 0
One possible unstructured activity that promotes resistance training would be climbing playground equimpent - A. 

This is by nature a unstructured ctivity. Furthermore, it promotes resistance training because you're forced to move and pull and push yourself. 
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How can you tell there is only one atom of oxygen in H2O?
kow [346]

Answer:

For H2O, there is one atom of oxygen and two atoms of hydrogen. A molecule can be made of only one type of atom. In its stable molecular form, oxygen exists as two atoms and is written O2. to distinguish it from an atom of oxygen O, or ozone, a molecule of three oxygen atoms, O3.

Explanation:

4 0
3 years ago
-Boston tea party-
lisov135 [29]

Answer: The Boston Tea Party was a political protest that occurred on December 16, 1773, at Griffin’s Wharf in Boston, Massachusetts. American colonists, frustrated and angry at Britain for imposing “taxation without representation,” dumped 342 chests of tea, imported by the British East India Company into the harbor. The event was the first major act of defiance to British rule over the colonists. It showed Great Britain that Americans wouldn’t take taxation and tyranny sitting down, and rallied American patriots across the 13 colonies to fight for independence.

Explanation: brainlest please

4 0
3 years ago
A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig
trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

V=\frac{MU+mu}{M+m}

Substituting values:

V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0
3 years ago
A child standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down, the child conti
pochemuha
B. Newton's First Law, I'm pretty sure. The first states that an object in motion stays in motion, and an object at rest stays at rest until an outside force is applied, and that seems pretty relevant.
4 0
4 years ago
Read 2 more answers
The current theory of the structure of the
IRISSAK [1]

1) The mass of the continent is 3.3\cdot 10^{21} kg

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

1)

We start by finding the volume of the continent. We have:

L = 5850 km = 5.85\cdot 10^6 m is the side

t = 35 km = 3.5\cdot 10^4 m is the depth

So the volume is

V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3

We also know that its density is

d=2750 kg/m^3

Therefore, we can find the mass by multiplying volume by density:

m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg

2)

The kinetic energy of the continent is given by:

K=\frac{1}{2}mv^2

where

m=3.3\cdot 10^{21} kg is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s

So, the speed is:

v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s

So, we can now find the kinetic energy:

K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J

3)

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

K=\frac{1}{2}mv^2

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0
3 years ago
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