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If the position of a particle on the x-axis at time t is âˆ’5t2, then the average velocity of the particle for 0 â‰¤ t â‰¤ 3 is

Drupady [299]

Answer:

v = 15 m / s

Explanation:

In this exercise we are given the position function

x = 5 t²

and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval

$v= \frac{v_{f} - v_{o} }{t_{f} - t_{o} }$

let's look for the displacements

t = 0 x₀ = 0 m

t = 3 $x_{f}$ = 5 3 2

x_{f} = 45 m

we substitute

$v = \frac{45 -0}{3 - 0}$

v = 15 m / s

3 0

3 years ago

Two polarizers are oriented at 68 ∘ to one another. unpolarized light falls on them. part a what fraction of the light intensity

Radda [10]

Your answer is.07 hope this helped

4 0

3 years ago

Read 2 more answers

A laser beam is incident at an angle of 30.0° from the vertical onto a solution of corn syrup in water. The beam is refracted to

dimaraw [331]

Answer with Explanation:

We are given that

Angle of incidence, $i=30^{\circ}$

Angle of refraction, $r=19.24^{\circ}$

a.Refractive index of air, $n_1=1$

We know that

$n_2sinr=n_1sini$

$n_2=\frac{n_1sin i}{sin r}=\frac{sin30}{sin19.24}=1.517$

b.Wavelength of red light in vacuum, $\lambda=632.8nm=632.8\times 10^{-9} m$

$1nm=10^{-9} m$

Wavelength in the solution, $\lambda'=\frac{\lambda}{n_2}$

$\lambda'=\frac{632.8}{1.517}=417nm$

c.Frequency does not change .It remains same in vacuum and solution.

Frequency, $\nu=\frac{c}{\lamda}=\frac{3\times 10^8}{632.8\times 10^{-9}}$

Where $c=3\times 10^8 m/s$

Frequency, $\nu=4.74\times 10^{14}Hz$

d.Speed in the solution, $v=\frac{c}{n_2}$

$v=\frac{3\times 10^8}{1.517}=1.98\times 10^8m/s$

5 0

4 years ago

A spring is hung from the ceiling. A 0.442-kg block is then attached to the free end of the spring. When released from rest, the

siniylev [52]

Answer:

$K=58.8N/m$

Explanation:

From the question we are told that:

Mass $M=0.442$

Drop distance $d=0.150$

Generally the equation for Spring Constant is mathematically given by

$K=\frac{2mg}{x}$

$K=\frac{2*0.442*9.8}{1.150}$

$K=58.8N/m$

6 0

3 years ago

Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

3 years ago