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4 years ago

6. Downhill walking or walking down stairs causes _______ contractions of the leg muscles.

Physics

1 answer:

Trava [24]4 years ago

5 0

<span>Downhill walking or walking down stairs causes D. Eccentric contractions of the leg muscles.</span>

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Can any one help pls

Paraphin [41]

Answer:

A and D are correct as they reduce GHG emissions while maintaining people's standard of living.

7 0

3 years ago

Read 2 more answers

Points A, B, and C lie along a line from left to right, respectively. Point B is at a lower electric potential than point A. Poi

arlik [135]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

If the potential at B is lower than A, and the potential at C is lower than B, this means that there is an electric field, directed from A to C.
If a positively-charged particle is released at rest at point B, it will be accelerated by the electric field (which is a force per unit charge, so it produces an acceleration) in the same direction than the field (because it is a positive charge) towards point C.

6 0

3 years ago

A quantity of N2 occupies a volume of 1.4 L at 290 K and 1.0 atm. The gas expands to a volume of 3.3 L as the result of a change

lions [1.4K]

Answer:

$\rho = 0.50 g/L$

Explanation:

As we know that

PV = nRT

here we have

$P = 1.0 atm$

$P = 1.013 \times 10^5 Pa$

so we have

$V = 1.4 \times 10^{-3} m^3$

T = 290 K

now we have

$(1.013 \times 10^5)(1.4 \times 10^{-3}) = n(8.31)(290)$

$n = 0.06$

now the mass of gas is given as

$m = n M$

$m = (0.06)(28)$

$m = 1.65 g$

now density of gas when its volume is increased to 3.3 L

so we will have

$\rho = \frac{m}{V}$

$\rho = \frac{1.65 g}{3.3 L}$

$\rho = 0.50 g/L$

5 0

3 years ago

In this section of a circuit, a current of 2.0 A flows across R1. What is the current across R2? Let R1 = 4.0 ohm, R2 = 8.0 ohm,

Liono4ka [1.6K]

Answer:

I=1A

Explanation:

In parallel combination voltage is same

In R1

V=IR

V=2×4

V=8V

In R2

V=IR

I=V/R

I=8/8

I=1A

3 0

4 years ago

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

3 years ago

Read 2 more answers