Ask question

Ask question

All categories

3 years ago

9

What is a resistance training method that involves performing a set with light load followed by increasing the load and decreasi

ng the repetitions with each continuing set?

1 answer:

Luden [163]3 years ago

4 0

Answer:

<em>Pyramid system</em>

<em></em>

Explanation:

A <em>pyramid system </em>resistance training is a collection of sets of the same exercise, that start with light weights and higher repetitions, building up to heavier weights and fewer repetitions.

A full pyramid training is an extension of the above mentioned statement, reducing the weight after you have reached the peak until you complete the pyramid.

You might be interested in

A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an e

Gala2k [10]

Explanation:

Given that,

Charge acting on the object, $q=-4\ nC=-4\times 10^{-9}\ C$

Force acting on the object, $F=19\ nC=19\times 10^{-9}\ C$ (in downward direction)

(a) The electric force acting in the electric field is given by :

$F=qE$

E is the electric field

$E=\dfrac{F}{q}$

$E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}$

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, $q=1.6\times 10^{-19}\ C$

The force acting on the proton is :

$F=qE$

$F=1.6\times 10^{-19}\times 4.75$

$F=7.6\times 10^{-19}\ N$

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

8 0

2 years ago

Can someone help me please

makvit [3.9K]

Answer:

Time, I believe. Pretty sure it's time lol

3 0

2 years ago

A woman lifts her 100-newton child up one meter and carries her for a distance of 50 meters to the child's bedroom. How much wor

tankabanditka [31]

100 J

Please mark me brainliest it would be greatly appreciated haha

5 0

3 years ago

Read 2 more answers

Suppose you read in the newspaper that a new planet has been found. Its average speed in its orbit is 33 kilometers per second (

Harlamova29_29 [7]

Answer:

E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.

Explanation:

We can answer this question by using Kepler's second law of planetary motion, which states that:

"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"

This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).

In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is

E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.

5 0

2 years ago

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

<u>Known Data</u>

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

<u>Time of the dropped Rock</u>

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

<u>Time of the Thrown Rock</u>

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$

6 0

3 years ago