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3 years ago

13

Which activity best reflects a muscle movement working against gravity? a lifting the torso during a sit up b lowering the body

during a pushup c doing a squat d swimming?

2 answers:

alexdok [17]3 years ago

6 0

The answer is c) doing a squat.

Hope this helps !

Photon

Ivan3 years ago

4 0

<span>c doing a squat

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</span>

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Question 8 (4 points)

katrin2010 [14]

Answer:

The answer to your question is:

Explanation:

Data

Duane Albert

d = 5 m ; v = 3 m/s v = 4.2 m/s

a) b)

Duane's Albert's

d = 5 + (3)t d = 4.2t

d = 5 + 3t

c) 5 + 3t = 4.2t

4.2t - 3t = 5

1.2t = 5

t = 4.17 s

d)

Duane's

d= 5 + 3(4.17)

d = 17.51 m

Alberts

d = 4.2(4.17)

d = 17.51 m

4 0

2 years ago

Read 2 more answers

A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted

Tpy6a [65]

Answer:

The normal force the seat exerted on the driver is 125 N.

Explanation:

Given;

mass of the car, m = 2000 kg

speed of the car, u = 100 km/h = 27.78 m/s

radius of curvature of the hill, r = 100 m

mass of the driver, = 60 kg

The centripetal force of the driver at top of the hill is given as;

$F_c = F_g - F_N$

where;

Fc is the centripetal force

$F_g$ is downward force due to weight of the driver

$F_N$ is upward or normal force on the drive

$F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N$

Therefore, the normal force the seat exerted on the driver is 125 N.

6 0

2 years ago

A research submarine has a 10-cm-diameter window that is 8.4 cm thick. The manufacturer says the window can withstand forces up

Dmitrij [34]

Pnet = Po + dgh

<span>Density of saltwater = 1030 kg/m^3. </span>

<span>Disregard the thickness. Assuming it's a circular window, then the area is pi(r^2). </span>

<span>d = 20 cm = 0.2 m </span>
<span>r = d/2 = 0.1 m </span>

<span>A = pi(r^2) </span>
<span>A = 3.14159265(.1^2) </span>
<span>A = 0.0314159265 m^2 </span>

<span>p = F/A </span>
<span>p = (1.1 x 10^6) / (0.0314159265) </span>
<span>p = 35,014,087.5 Pa </span>

<span>1 atm = 101,325 Pa </span>

<span>P = Po + dgh </span>
<span>h = (P - Po) / dg </span>
<span>h = (35,014,087.5 - 101,325) / (1030 x 9.81) </span>
<span>h = 3 455.23812 m </span>
<span>h = 3.5 km</span>

8 0

2 years ago

Read 2 more answers

The curvature of the helix r(t)equals(a cosine t )iplus(a sine t )jplusbt k (a,bgreater than or equals0) is kappaequalsStartF

4vir4ik [10]

Answer:

$\kappa = \frac{1}{2 b}$

Explanation:

The equation for kappa ( κ) is

$\kappa = \frac{a}{a^2 + b^2}$

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

$\kappa (a) = \frac{a}{a^2 + b^2}$

Now, the conditions to find a maximum at $a_0$ are:

$\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0$

$\frac{d^2\kappa(a)}{da^2} \left | _{a=a_0} < 0$

Taking the first derivative:

$\frac{d}{da} \kappa = \frac{d}{da} (\frac{a}{a^2 + b^2})$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da} (\frac{1}{a^2 + b^2} )$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1) (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da} (a^2+b^2)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a (\frac{1}{(a^2 + b^2)^2} ) (2* a)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2} - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 - 2 a^2)$

$\frac{d}{da} \kappa = \frac{b^2 - a^2}{(a^2 + b^2)^2}$

This clearly will be zero when

$a^2 = b^2$

as both are greater (or equal) than zero, this implies

$a=b$

The second derivative is

$\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 - a^2}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 - a^2 ) + (b^2 - a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2 a ) + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

$\frac{d^2}{da^2} \kappa = \frac{-2 a}{(a^2 + b^2)^2} + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

We dcan skip solving the equation noting that, if a=b, then

$b^2 - a^2 = 0$

at this point, this give us only the first term

$\frac{d^2}{da^2} \kappa = \frac{- 2 a}{(a^2 + a^2)^2}$

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

$\kappa = \frac{b}{b^2 + b^2}$

$\kappa = \frac{b}{2* b^2}$

$\kappa = \frac{1}{2 b}$

3 0

3 years ago

What force does it take to accelerate a 7.2 kg object 3.0 m/s^2.

grin007 [14]

Answer:

<h2>21.6 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 7.2 × 3 = 21.6

We have the final answer as

<h3>21.6 N</h3>

Hope this helps you

3 0

2 years ago