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dsp73
3 years ago
12

During resistance exercise, muscles are __________.

Physics
2 answers:
choli [55]3 years ago
6 0
<span>The answer is contracted or repeatedly contracted. Resistance exercises are a set of exercises that forces the skeletal muscles of an individual's body to contract. Resistance exercises are used to enlarge muscular mass of an individual's body or to increase strength and endurance of a person.</span><span />
gogolik [260]3 years ago
4 0

Answer:

Answer is A. Working against a force

Explanation:

I just took the quiz on E20

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Vilket är det modernaste förvarings materialet?
Zinaida [17]
<span>Det är behållaren med lock.</span>
8 0
3 years ago
The speed of light in a vacuum is 2.99x10^8 m/s. calculate its speed in miles per hour.
34kurt
You'll never get the correct answer without the correct conversion factor. Note carefully that you have no decimal. It should be 
<span>1 km = 0.6214 miles </span>
<span>1000 m = 1 km </span>
<span>60 seconds = 1 minute </span>
<span>60 minutes = 1 hour. </span>
<span>2.998E8 m/s x (1 km/1000m) x (0.6214 miles/km) x (60 sec/min) x (60 min/hr) = ?</span>
6 0
4 years ago
If the ball hits Olaf and bounces off his chest horizontally at 7.90 m/s in the opposite direction, what is his speed vf after t
Reil [10]

Answer:

Velocity = 0.105m/s

Explanation:

Using conservation of momentum:

Momentum before impact= mass of ball × velocity of ball

= 0.400×10.8= 4.32

Momentum after impart= mass of ball × velocity of ball + mass of olaf × velocity of olaf

= 4.35 +3.56 + 74.9Vf

Vf=7.88_74.9 = o.0.105m/s

8 0
3 years ago
Read 2 more answers
A relaxed biceps muscle requires a force of 25.0N for an elongation of 3.0 cm; under maximum tension, the same muscle requires a
Norma-Jean [14]

3.3 x10^4N/m²

6.7 x105N/m²

Explanation:

Let the young modulus of the relaxed biceps be

Y= F¹Lo/ deta L1 x A

= 25 x0.2/ 0.03* 50cm²(1m²

0.0004cm^-²)

= 3.3x10^4N/m²

But young modules of muscle under maximum tension will be

Y= F"Lo/ deta L" x A

= 500x 0.2/ 0.03* 50cm²(1m²

0.0004cm^-²)

= 6.7 x10^5N/m²

4 0
4 years ago
a cannon ball is fired at 40 degrees with an initial velocity of 180 m/s. what is the maximum height reached by the cannonball?
NNADVOKAT [17]

A ball is fired with speed v = 180 m/s at an angle 40 degree

So here we can say the components of velocity is given as

v_{x} = 180 cos40 = 137.9 m/s

v_y = 180 sin40 = 115.7 m/s

now when ball will reach the maximum height its velocity in vertical direction will become zero

so here we can say

v_{fy} = 0

now using kinematics we will have

v_{fy}^2 - v_y^2 = 2 a d

0 - 115.7^2 = 2(-9.8)(H)

so maximum height is

H = 683 m

so it will reach at maximum height of 683 m

8 0
4 years ago
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