Given sin ti + cos ti + tk

A person is sitting with one leg outstretched and stationary, so that it makes an angle of θ = 27.5° with the horizontal, as the

dangina [55]

here we will use the torque balance about the knee joint

here we can say that

$\tau_g = \tau_m$

here torque due to weight is given as

$\tau_g = 40.1 cos\theta*(l_1 + l_2)$

$\tau_g = 40.1 cos27.5*(0.105 + 0.150)$

$\tau_g = 9.07 Nm$

now torque due to applied force of muscle

$\tau_m = M*sin\alpha * l_1$

$\tau_m = M*sin30* 0.105$

now by torque balance we will have

$9.07 = M*0.5*0.105$

$M = 173 N$

so here the magnitude of m will be 173 N

7 0

3 years ago

A/an ___ is a machine which tells us about the strength and speed of sisemec waves.

svp [43]

The answer is Seismograph

3 0

4 years ago

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What are the four different types of economic resources?

77julia77 [94]

The factors of production are resources that are the building blocks of the economy; they are what people use to produce goods and services. Economists divide the factors of production into four categories: land, labor, capital, and entrepreneurship. hope that helped

8 0

4 years ago

Suzie skydiver, who weighs 500 n, reaches terminal velocity of 90 km/h. the air resistance on suzie is then

masha68 [24]

Weight of the skydiver m = 500 N
Terminal velocity V = 90 km/h
Here the weight of the person acts as the force, so based on the Newton's third law the applied is the force what we but in the opposite direction making the resistance. So the air resistance exerted on Suzie will be her weight that is 500N

7 0

4 years ago

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

4 years ago

Given sin ti + cos ti + tk ​

Given sin ti + cos ti + tk