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MArishka [77]
3 years ago
11

Many fitness programs include weight training, which A. can be used in place of aerobic exercise, when done regularly. B. helps

to increase cardiorespiratory capacity and endurance. C. helps to increase muscular strength and endurance. D. is a form of aerobic exercise involving the muscles.
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0
The answer would be C) because weight lifting is used to increase your muscular strength, and that's why you try and add more weight during sessions, so it's not too easy for you and you can build a higher endurance.
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A(n) 10.1 g bullet is fired into a(n) 2.41 kg ballistic pendulum and becomes embedded in it. The acceleration of gravity is 9.8
fomenos

Answer:

v = 186.90\,\frac{m}{s}

Explanation:

The motion of ballistic pendulum is modelled by the appropriate use of the Principle of Energy Conservation:

\frac{1}{2}\cdot (m_{p}+m_{b})\cdot v^{2} = (m_{p}+m_{b})\cdot g \cdot h

The final velocity of the system formed by the ballistic pendulum and the bullet is:

v = \sqrt{2\cdot g\cdot h}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}

v\approx 0.78\,\frac{m}{s}

Initial velocity of the bullet can be calculated from the expression derived of the Principle of Momentum:

(0.0101\,kg)\cdot v = (2.41\,kg + 0.0101\,kg)\cdot (0.78\,\frac{m}{s} )

v = 186.90\,\frac{m}{s}

5 0
3 years ago
A block of mass m is pushed a horizontal distance D from position A to position B, along a horizontal plane with friction coeffi
Wewaii [24]

Answer:

The total work done by friction is -2 · μ · m · g · D

Explanation:

Hi there!

The work done by a force is calculated as follows:

W = F · d · cos θ

Where:

W = work.

F = force that does the work.

d = displacement.

θ = angle between the displacement and the force.

If the force is horizontal, as in this case, cos θ = 1

The friction force is calculated as follows:

Ffr = μ · m · g

Where:

μ = friction coefficient.

m = mass of the object.

g = acceleration due to gravity.

Then, in this case, the work done by friction when pushing the block from A to B will be:

W AB = -Ffr · D

W AB = - μ · m · g · D

Notice that the friction force is negative because it is opposite to the pushing force P.

When the block is pushed from B to A, the work done by friction will be:

W BA = Ffr · (-D)

W BA = -μ · m · g · D

Now, the displacement is negative and the friction force is positive (in the opposite direction to -P).

The total work done by friction will be:

W AB + W BA = - μ · m · g · D  - μ · m · g · D  = -2 μ · m · g · D

5 0
3 years ago
What’s a good hypothesis and and experiment for water?
Lelu [443]

Answer: You could do something like, "how does water react to being mixed with baking soda"...or something along those lines

Explanation:

6 0
3 years ago
If, while standing on the bank of a stream, you wished to spear a fish swimming in the water out in front of you, would you aim
Serggg [28]

Answer:

<em>a) below the observed position</em>

<em>b) directly at the observed position</em>

<em></em>

Explanation:

If I'm standing on the bank of a stream, and I wish to spear a fish swimming in the water out in front of me, I would aim below the observed fish to make a direct hit. This is because the phenomenon of refraction of light in water causes the light coming from the fish is refract away from the normal as it passes  into the air and into my eyes.

If I'm to zap the fish with a taser, I would aim directly at the observed fish because the laser (a form of concentrated light waves) will refract into the water, taking the same path the light from the fish took to get to my eyes.

3 0
3 years ago
In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b
lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

Diameter of ride=5 m

radius(r)=2.5 m

Static friction coefficient range=0.60-1

Here Frictional force will balance weight

And limiting  frictional force is provided by Centripetal force

f=\mu N=\mu m\omega ^2\cdot r

weight of object=mg

Equating two

f=mg

\mu m\omega ^2\cdot r=mg

\omega ^2=\frac{g}{\mu r}

\omega =\sqrt{\frac{g}{\mu r}}

\omega =2.55 rad/sec

8 0
3 years ago
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