All categories

4 years ago

Which type of exercise is weightlifting?

Physics

2 answers:

expeople1 [14]4 years ago

6 0

B- anaerobic. It is a type of weight lifting

MatroZZZ [7]4 years ago

5 0

The answer is option B "anaerobic." Weightlifting deals with stress to the muscles when lifting weights and due time the muscles will begin to adapt and get stronger. Other examples of anaerobic exercise are things like: weight training, sprinting, cycling, and jumping anything that has short exertion, and high-intensity movement is an anaerobic exercise.

Hope this helps!

Nonportrit

You might be interested in

All chemical reactions occur at the same rate true or false

SSSSS [86.1K]

Answer:

false

Explanation:

7 0

3 years ago

Read 2 more answers

At a baseball game, Bill observes from the bleachers Jack throwing a ball toward home at 15 km/s while Kevin runs toward home fr

sattari [20]

Answer:bill 5 m/s. Jack:10 m/s

Explanation:

Cuz I took it

7 0

3 years ago

A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60

mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

$z(t) = \sqrt{x^{2} + y(t)^{2}}$

$z(t) = \sqrt{70^{2} + 60t^{2}}$

Now, differentiate the above eqn w.r.t 't':

$\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600$

$\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t$

For t = 4 s:

$\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s$

4 0

3 years ago

A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in th

algol [13]

Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

$v^2=u^2+2as$

where

'v' is the final speed of the ball

'u' is the initial speed of the ball

'a' is the acceleration that the ball is under which in this case equals 9.81 $m/s^{2}$

's' is the distance it covers

Thus for maximum height applying the values in the equation we get

$0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters$

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

$v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s$

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

$v=u+at\\\\t=\frac{v-u}{a}\\\\t=\frac{10.28-20}{-9.81}=0.991seconds\approx 1second$

4 0

4 years ago

If you went to a planet with an acceleration due to gravity of 20 m/s^2 and twirled a bucket of water, what happens to the angul

wel

Answer:

Explanation:

It's Scientifically proven

7 0

3 years ago