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Ulleksa [173]
3 years ago
11

. A 1.50kg mass on a spring has a displacement as a function of time given by the equation: x(t) = (7.40cm)cos[(4.16s-1)t – 2.42

]. Find, a) the time for one complete vibration; b) the force constant of the spring; c) the maximum speed of the mass; d) the maximum force on the mass; e) at t= 1.00s, find the mass is a. the position, b. the speed, c. the acceleration d. the force on the mass
Physics
1 answer:
rusak2 [61]3 years ago
5 0

Answer:

Solution:

we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]

Position, velocity and acceleration are all based on the equation of motion.

The equation represents the position.  The first derivative gives the velocity and the 2nd derivative gives the acceleration.

x(t)=8sint

x'(t)=8cost

x"(t)=-8sint

now at time t=2pi/3,

position, x(t)=8sin(2pi/3)=4*squart(3)cm.

velocity, x'(t)=8cos(2pi/3)==4cm/s

acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2

so at present the direction is in y-axis.

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A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
3 years ago
Imagine that you have a 6.50 l gas tank and a 4.50 l gas tank. you need to fill one tank with oxygen and the other with acetylen
Margarita [4]
V₁(O2) = 6.50<span>  L                
</span>p₁(O2) = 155 atm          
V₂(acetylene) = <span>4.50 L                   
</span>p₂(acetylene) =?                             

According to Boyle–Mariotte law (At constant temperature and unchanged amount of gas, the product of pressure and volume is constant) we can compare two gases that have ideal behavior and the law can be usefully expressed as:

V₁/p₁ = V₂/p₂

6.5/155 = 4.5/p₂

0.042 x p₂ = 4.5

p₂ = 107.3 atm



7 0
3 years ago
Donde se hizo el “siluetazo”, <br>Me aydan​
brilliants [131]

Answer:

La palabra silueta se deriva del nombre de Étienne de Silhouette, una ministra de finanzas francesa que, en 1759, se vio obligada por la crisis crediticia de Francia durante la Guerra de los Siete Años a imponer severas demandas económicas al pueblo francés, particularmente a los ricos.

Explanation:

3 0
3 years ago
1. What distance is required for a train to stop if its initial velocity is 23 m/s and its
Irina-Kira [14]

Answer:

x=?

dt=?

vi=23m/s

vf=0m/s (it stops)

d=0.25m/s^2

time =

vf=vi+d: 0=23m/s+(0.25m/s^2)t

t=92s

displacement=

vf^2=vi^2+2a(dx)

23^2=0^2+2(0.25m/s^2)x =-1058m

Explanation:

you can find time from vf = vi + a(Dt): 0 = 23 m/s + (0.25 m/s/s)t so t = 92 s and you can find the displacement from vf2 = vi2 + 2a(Dx) and find the answer in one step: 232 = 02 + 2(0.25 m/s/s)x so x = -1058 m

6 0
3 years ago
Objects 1 and 2
sattari [20]

Answer:

12 units

Explanation:

Original F = C q1q2/r^2    now change the parameters

  New F = C  1/3 q1  2 q2  / (2r)^2

              =  2/12  C q1q2/r^2    <=======the new force is 1/6 of original

1/6 * 72 = 12 units  

3 0
2 years ago
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