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3 years ago

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10. A person is about to kick a soccer ball. Consider the leg and foot to be a single rigid body and that it rotates about a fix

ed axis through the knee joint center. Immediately prior to impact the leg and foot are in the vertical direction and the distal end of the foot has an acceleration of 10 g in the horizontal direction. The muscle force vector makes an angle of 15 degrees with the vertical and has a moment arm of 5 cm from the knee joint center. Assume the person is 1.7 m tall and has a mass of 75 kg. Find the force in the muscle

1 answer:

statuscvo [17]3 years ago

4 0

Answer:

i don't know but i hope you get it right

Explanation

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You are approaching a police car at 68. 1 mph and the police car is approaching you at 94. 8 mph. Assume that the speed of sound

Veronika [31]

We will hear the sound of siren of frequency 1553.4606 Hz.

<h3>What is Doppler Effect?</h3>

The apparent change in wave frequency brought on by the movement of a wave source is known as the Doppler effect. When the wave source is coming closer and when it is moving away, the perceived frequency changes. The Doppler effect explains why we hear a passing siren's sound changing in pitch.

according to Dopplers Effect,

$f'=[\frac{v + v_{0} }{v - v_{s} } ]f$

$f'= [\frac{700+68.1}{700-94.8} ]* 1224$

$f'= 1553.4606 Hz$

the frequency would be 1553.4606 Hz.

to learn more about Doppler Effect go to - brainly.com/question/9165991

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7 0

2 years ago

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st

scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

$x(t) = v_0 cos \theta t$

where

$v_0$ is the initial speed

t is the time

$\theta=32.0^{\circ}$ is the angle below the horizontal

We can rewrite this equation as

$t=\frac{x(t)}{v_0 cos \theta}$ (1)

The vertical displacement instead is given by

$y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2$ (2)

where

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting (1) into (2),

$y(t) = -x(t) tan \theta - \frac{1}{2}gt^2$

We know that for t = time of flight, the horizontal displacement is

$x(t) =50.8 m$

We also know that the vertical displacement is

$y(t) = -45 m$

Substituting everything into the equation, we can find the time of flight:

$\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s$

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

$x(t) = v_0 cos \theta t$

where we have

x = 50.8 m

$\theta=32.0^{\circ}$

Substituting the time of flight,

t = 1.64 s

We find:

$v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s$

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

$v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s$

The initial vertical velocity is instead

$u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s$

And it changes according to the equation

$v_y = u_y -gt$

So at t = 1.64 s (when the ball hits the ground),

$v_y = -19.3 - (9.8)(1.64)=-35.4 m/s$

So the impact speed is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$

8 0

3 years ago

When I first joined Brainly i had a problem with drinking

Ludmilka [50]

Were you able to fix that problem?

4 0

3 years ago

A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow

Vinvika [58]

Answer:

The answer is 1.87nm/s.

Explanation:

The $110g/hr$ water loss must be replaced by $110g/hr$ of sap. 110g of sap corresponds to a volume of

$110g \div \dfrac{1040*10^3g}{1*10^6cm^3} = 106cm^3$

thus rate of sap replacement is

$106cm^3/hr = 106*10^{-6}m^3/3600s = 2.94*10^{-8}m^3/s$

The volume of sap in the vessel of length $x$ is

$V = Ax$ ,

where $A$ is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

$V = 2000Ax$

taking the derivative of both sides we get:

$\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}$

on the left-hand-side $\dfrac{dx}{dt}$ is the velocity $v$ of the sap, and on right-hand-side $\dfrac{dV}{dt} = 2.94*10^{-8}m^3/s$ ; therefore,

$2.94*10^{-8}m^3/s=2000Av$

and since the cross-sectional area is

$A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2$ ;

therefore,

$2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v$

solving for $v$ we get:

$v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}$

$\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}$

which is the upward speed of the sap in each vessel.

4 0

3 years ago

Dock jumping In the sport of dock jumping, dogs run at full speed off the end of a dock that sits a few feet above a pool of wat

melamori03 [73]

To develop this problem we will use the linear motion kinematic equations. To find the time, we will use the information given on the y axis, which will allow us to find the time it takes to complete the route. With this time and the speed given in the horizontal direction, we can calculate the total distance traveled in that direction. Consider the motion of dog along the Y-direction

$V_{0y} = \text{Initial velocity} = 0 m/s$

$a = \text{Acceleration} = 9.8m/s^2$

$y = \text{displacement} = 0.61 m$

$t = \text{time of travel}$

Using the equation,

$y = V_{0y} t + \frac{1}{2} a t^2$

$0.61 = 0*t + \frac{1}{2} (9.8) t^2$

$t = 0.35s$

Along the x-direction

$V_{0x} = \text{Constant velocity along X-direction} = 8.5 m/s$

Horizontal Displacement,

$x = V_{0x}t$

$x = (8.5)(0.35)$

$x = 2.975 m$

Therefore the dog goes before splashing down a distance of 2.975m

5 0

3 years ago