The process by which lithospheric plates move apart creating spaces that are filled with hot magma is called ____

vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses

dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

$v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}$

(This is correct because the horizontal motion has acceleration zero). Then:

$v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s$

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

$E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}$

Then, plugging in the given values, we obtain:

$k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m$

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0

2 years ago

A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear

Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

$f_o = f_s\frac {(v+vs)}{v}$ and $f_o = f_s\frac {(v-vs)}{v}$ when moving towards observer

With $f_o$ of 76 then taking speed in air as 343 m/s we have

$76 = f_s\times\frac {(343-vs)}{343}$

$f_s=\frac {343\times 76}{343-v_s}$

Similarly, with $f_o$ of 65 we have

$65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}$

Now

$f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}$

v_s=27.76 m/s

Substituting the above into any of the first two equations then we obtain

$f_s=70.07 Hz$

4 0

2 years ago

Can someone explain how exactly to calculate the net force? Do you subtract the smaller number from the larger number? Any help

aleksley [76]

Explanation:

Let's say right is positive and left is negative.

F₁ = -150 N

F₂ = 220 N

Fnet = F₁ + F₂

Fnet = -150 N + 220 N

Fnet = 70 N

The magnitude of Fnet is 70 N, and since it's positive, the direction is to the right.

And since Fnet isn't 0, the force is unbalanced and the motion is changing.

3 0

2 years ago

g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p

kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × $10^{-5}$ m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

r = $\frac{mv}{qB}$

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × $10^{-5}$ T, v = 121 m/s, Θ = $90^{0}$ (since it enters perpendicularly to the field), q = e = 1.6 × $10^{-19}$ C and m = 9.11 × $10^{-31}$ Kg.