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3 years ago

During resistance exercise, muscles are __________.

2 answers:

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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Can somebody please answer this correctly I will give brainliest

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I think what’s wrong is that the paper clip isn’t connecting to the other thing on the bottom

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3 years ago

An electromagnetic wave has a frequency of 1.0 x 10^5. What is the wavelength of the wave?

SashulF [63]

Answer:

The frequency , speed and wavelength of an electromagnetic wave are related by the formula

Speed = frequency x wavelength

frequency = speed / wavelength

substituting the values

frequency = 3 x #10 ^8# m /s / 1 x #10^15# m

= 3 x #10^-7# /s

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3 years ago

Read 2 more answers

Scarcity is not usually a problem in economics true or false

PSYCHO15rus [73]

False because the allocation rule of "first come-first served" promotes productive cooperation.

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3 years ago

Read 2 more answers

A lidar (laser radar) gun is an alternative to the standard radar gun that uses the Doppler effect to catch speeders. A lidar gu

nikitadnepr [17]

Answer:

v = 42.33 m/s

Explanation:

The time difference between two pulses that are emitted by Police is observed in the interval

$\Delta t = 1.27 \times 10^{-7} s$

now the speed of the electromagnetic wave is given as

$c = 3 \times 10^8 m/s$

now the distance traveled by them

$d = c \times t$

$d = (3 \times 10^8)(1.27 \times 10^{-7})$

$d = 38.1 m$

so distance of speeding vehicle covered for round trip of wave is 38.1 m

now for one trip it is given as d = 19.05 m

now the speed of the vehicle is given as

$v = \frac{d}{t}$

$v = \frac{19.05}{0.450}$

$v = 42.33 m/s$

5 0

3 years ago

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$

So the total distance that the car traveled is $d = \frac{v_0^2}{a_0}$

The difference between the train and the car is

$x - d = \frac{2v_0^2}{a_0} - \frac{v_0^2}{a_0} = \frac{v_0^2}{a_0}$

8 0

3 years ago