Ice does not need to melt into liquid water before it can return to the atmosphere as water vapor.

What is the object’s velocity, in meters per second, at time t = 2.9? Calculate the object’s acceleration, in meters per second

madreJ [45]

Answer:

the question is incomplete, below is the complete question

"An object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 5.5 cos(4.4t - 1.8), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.

a.What is the object's velocity, in meters per second, at time t = 2.9?

b.Calculate the object's acceleration, in meters per second squared, at time t = 2.9.

c. What is the magnitude of the object's maximum acceleration, in meters per second squared?

d.What is the magnitude of the object's maximum velocity, in meters per second?"

a. $v(t)==24.1m/s$

b. $a(t)=3.79m/s^{2}$

c. $a_{max}=106.48m/s^{2}$

d. $v_{max}=24.2m/s$

Explanation:

the gneral expression for the displacement of object in simple harmonic motion is represented by

$x(t)=Acos(wt- \alpha)\\$

while the velocity is express as

$v(t)=-Awcos(4.4t-1.8)\\$

and the acceleration is

$a(t)=-aw^{2}cos(wt- \alpha )\\$

Note: the angle is in radians

The expression for the displacement from the question is $x(t)=5.5cos(4.4t-1.8)\\$

comparing, A=5.5, w=4.4,α=1.8

a.To determine the object velocity at t=2.9secs,

we substitute for t in the velocity equation

$v(t)=-5.5*4.4sin(4.4*2.9-1.8)\\v(t)=-24.2sin(10.96)\\$

$v(t)=-24.2*(-0.9993)\\v(t)==24.1m/s$

b.To determine the object acceleration at t=2.9secs,

we substitute for t in the acceleration equation

$a(t)=-5.5*4.4^{2} cos(4.4*2.9-1.8)\\a(t)=-106.48cos(10.96)\\$

$a(t)=-106.48*0.0356\\a(t)=3.79m/s^{2}$

c. The acceleration is maximum when the displacement equals the amplitude. hence magnitude of the object acceleration is

$a_{max}=-w^{2}A\\ a_{max}=-4.4^{2}*5.5\\ a_{max}=106.48m/s^{2}$

d.The maximum velocity is expressed as

$v_{max}=wA\\v_{max}=4.4*5.5\\v_{max}=24.2m/s$

5 0

3 years ago

What is the process when thermal energy is being added

kap26 [50]

That process is called "warming" or "heating".

6 0

3 years ago

Magnitude of force F experienced by a certain object moving with speed v is given by F=Kv race to 2, where K is constant. Find t

My name is Ann [436]

<h2>Answer:</h2>

MT⁻¹

<h2>Explanation:</h2>

Given equation:

F = Kv --------------(i)

Where;

F = Magnitude of force experienced by the object

v = speed at which the object is moving

K = constant.

To get the dimension of K, follow the following steps:

i. Make K subject of the formula in equation(i)

K = F / v

ii. Get the dimension of force F and speed v

Dimension of force = MLT⁻²

Dimension of velocity = LT⁻¹

iii. Substitute these dimensions into the result of (i) above;

K = MLT⁻² / LT⁻¹

iv. Simplify the result in (iii)

First, the Ls in the numerator and denominator will cancel out

K = MT⁻² / T⁻¹

Next, the Ts will be expressed as follows

K = MT⁻² x T¹ [using the law of indices]

K = MT⁻²⁺¹

K = MT⁻¹

Therefore, the dimension of K is MT⁻¹

6 0

3 years ago

1. Identify this mystery element: I am very reactive, I desperately want to get rid of my electron. When I make a compound with

Lapatulllka [165]

1 ) Sodium (Na)

2) Neon (Ne)

8 0

3 years ago

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? (e = 1.60 × 10-19

Nata [24]

Answer:

So speed of electron will be $5.89\times 10^{-7}m/sec$

Explanation:

We have given potential difference V = 9.9 KV

Charge on electron $e=1.6\times 10^{-19}$

So energy of electron $E=eV=1.6\times 10^{-19}\times 9.9\times 10^3=15.84\times 10^{-16}J$

This energy of electron will be equal to kinetic energy of electron

So $\frac{1}{2}mv^2=15.84\times 10^{-16}J$

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=15.84\times 10^{-16}$

$v=5.89\times 10^{-7}m/sec$

So speed of electron will be $5.89\times 10^{-7}m/sec$

8 0

3 years ago