The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
<h3>How to solve for the time interval</h3>
We have y = 0.175
y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.175
sin (1.25x + 99.6t) = 0.5
99.62 = pi/6
t1 = 5.257 x 10⁻³
99.6t = pi/6 + 2pi
= 0.0683
The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.
b. we have k = 1.25, w = 99.6t
v = w/k
99.6/1.25 = 79.68
s = vt
= 79.68 * 0.0683
= 5.02
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complete question
A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?
Answer:
Explanation:
what is the smallest crater that each of these telescopes could resolve on our moon?
For moon ;
s = 3.8 × 10 ⁸ m
y = 1.22 λs/D
where;
λ = 400 nm = 400× 10 ⁻⁹
D = 2.4 m
The smallest crater for the hubble space is calculated as follows:
For Aceribo ;
y = 1.22 λs/D
where :
λ = 75 cm = 0.75 m
D = 305 m
Answer:
k = 26.25 N/m
Explanation:
given,
mass of the block= 0.450
distance of the block = + 0.240
acceleration = a_x = -14.0 m/s²
velocity = v_x = + 4 m/s
spring force constant (k) = ?
we know,
x = A cos (ωt - ∅).....(1)
v = - ω A cos (ωt - ∅)....(2)
a = ω²A cos (ωt - ∅).........(3)
now from equation (3)
k = 26.25 N/m
hence, spring force constant is equal to k = 26.25 N/m
For balancing the lever, force on both the sides shall be equal. so,
Force on 3 m end = m × a = 3 × 98.1 = 294.3
Now, on 6 m end, it would be: = 294.3/6 = 49.05
After rounding-off to the nearest hundredth value, it would be: 49 N
Finally, Option A would be your correct answer.
Hope this helps!
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
