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maria [59]
3 years ago
14

A wave on a string is reflected from a fixed end. The reflected wave 1. Is in phase with the original wave at the end. 2. Has a

larger speed than the original wave. 3. Is 180◦ out of phase with the original wave at the end. 4. Cannot be transverse. 5. Has a larger amplitude than the original wave.
Physics
1 answer:
MrRa [10]3 years ago
5 0

Answer:

3. Is 180◦ out of phase with the original wave at the end.

Explanation:

Here when wave is reflected by the rigid boundary then due to the rigidly bounded particles at the end or boundary they have tendency not to move and remains fixed at their position.

Due to this fixed position we can say when wave reach at that end the particles will not move and they apply equal and opposite force at the particles of string

Due to this the reflected wave is transferred back into the string in opposite phase with respect to the initial wave

so here correct answer will be

3. Is 180◦ out of phase with the original wave at the end.

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Answer:

x=7227

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(7227,1678)

Explanation:

Ok, check the picture I attached you, because of the problem don't give us aditional information, let's asume that the projectile is fired from an initial position x=0 and y=0. Now let's use projectile motion equations, but firs let's find the initial velocity components in x-axis and y-axis:

v_ox=v_o*cos(\theta_o)=300*cos(55)=172.0729309m/s

v_oy=v_o*sin(\theta_o)=300*sin(55)=245.7456133m/s

Now, let's find the x coordinate with this equation:

x-x_o=x-0=x=v_ox*t=172.0729309*42=7227.063098m

Finally asumming a gravity constant g=9.8, let's find the y coordinate with the next equation:

y-y_o=y-0=y=v_oy*t-\frac{1}{2}*g*t^{2} =(245.7456133*42)-\frac{42^{2} *9.8}{2}

y=10321.31576-8643.6=1677.71576m

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3 years ago
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