HELP ASAP WITH number 6 plz

A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of

jonny [76]

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

7 0

3 years ago

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener

defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s$

(a) the force constant of the spring

$\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m$

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

$A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m$

7 0

3 years ago

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest

mr_godi [17]

Answer and Explanation:

The computation of the shortest wavelength in the series is shown below:-

$\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )$

Where

$\lambda$ represents wavelength

R represents Rydberg's constant

$n_f$ represents Final energy states

and $n_i$ represents initial energy states

Now Substitute is

$1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )$

now we will put the values into the above formula

$= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )$

$= 1218888.889 m^{-1}$

Now we will rewrite the answer in the term of $\lambda$

$\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m$

So, the whole Paschen series is in the part of the spectrum.

8 0

3 years ago

A closed system consisting of 2 lb of a gas undergoes a process during which the relation between pressure and volume is pVn = c

vazorg [7]

Answer:

V1=2.5ft3

V2=1ft3

n=1.51

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=2lb*1.25ft3/lb=2.5ft3

state 2

V2=m.v2

V2=2lb*0.5ft3/lb= 1ft3

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/60)/ ln (1/2.5)

n=1.51

8 0

3 years ago

Please help!?!?!?!?!?!?!

adoni [48]

Answer:

D

Explanation:i think but dont get mad if im wrong

5 0

2 years ago