La velocidad vertical del tanque después de caer 10 m es 14 m/seg .
La velocidad vertical del tanque se calcula mediante la aplicación de la fórmula de velocidad , la componente vertical Vfy, del movimiento horizontal como se muestra a continuación :
Vfy=?
h = 10 m
Fórmula de Velocidad vertical Vfy:
Vfy² = 2*g*h
Vfy= √(2*9.8m/seg2* 10m )
Vfy= 14 m/seg
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
Read more on waves here
brainly.com/question/25699025
#SPJ4
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:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
Answer:
296 N
Explanation:
Draw a free body diagram. The box has two forces on it: tension up and weight down.
Apply Newton's second law:
∑F = ma
T − mg = ma
T = m (g + a)
Given m = 196 N / 9.8 m/s² = 20 kg, and a = +5 m/s²:
T = (20 kg) (9.8 m/s² + 5 m/s²)
T = 296 N
Answer:
B. x - t graph
Explanation:
A position-time (x-t) graph is a graph of the position of an object against (versus) time.
Generally, the slope of the line of a position-time (x-t) graph is typically used to determine or calculate the velocity of an object.
An instantaneous velocity can be defined as the rate of change in position of an object in motion for a short-specified interval of time. Thus, an instantaneous velocity is a quantity that can be found by measuring the slope of a line that is tangent to a point on the graph.
Hence, the x - t graph also referred to as the position-time graph is used for determining the instantaneous velocity from the slope.
<u>For example;</u>
Given that the equation of motion is S(t) = 4t² + 2t + 10. Find the instantaneous velocity at t = 5 seconds.
Solution.
Differentiating the equation, we have;
Substituting the value of "t" into the equation, we have;
S(5) = 42 m/s.
