All categories

3 years ago

Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest.

2 answers:

7 0

5.7 km/h north and 5.8 km/h west are instantaneous velocities, while 8.1 km/h northwest is the average velocity.<span>

The answer choice above is correct.

The instantaneous velocities are the actual </span>velocities while traveling ( the velocity during that instant ). The average velocity is the average of the instantaneous velocities ( the speed in one direction equivalent to the two speeds <span>in different directions ).

I used speed in the explanation because velocity is speed with direction.</span>

Katyanochek1 [597]3 years ago

3 0

the correct answer is letter D

You might be interested in

Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face

Sunny_sXe [5.5K]

Answer:

$\mu = 1.645$

Explanation:

By Snell's law we know at the left surface

$\theta_i = 19^o$

$\theta_r = ?$

$\mu_1 = 1$

$\mu_2 = \mu$

now we have

$1 sin19 = \mu sin\theta_r$

$0.33 = \mu sin\theta_r$

now on the other surface we know that

angle of incidence = $\theta_r'$

$\theta_e = 90$

so again we have

$\mu sin\theta_r' = 1 sin90$

so we have

$\theta_r = sin^{-1}\frac{0.33}{\mu}$

$\theta_r' = sin^{-1}\frac{1}{\mu}$

also we know that

$\theta_r + \theta_r' = 49$

$sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49$

By solving above equation we have

$\mu = 1.645$

3 0

3 years ago

Read 2 more answers

A force of 20 N is executed to raise a rock weighing 30 N. What is the actual mechanical advantage?

xxMikexx [17]

I am absolutely sure its 1.5

5 0

3 years ago

Read 2 more answers

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 m

kramer

Answer:

The focal length of eyepiece is 3.68 cm.

Explanation:

Given that,

Distance = 19 cm

Focal length = 5.5 mm

Magnification = 200

Object distance = -25 cm

We need to calculate the focal length

Using formula of magnification

$m=\dfrac{d}{f_{o}}+\dfrac{-25}{f_{0}f_{e}}$

Put the value into the formula

$f_{e}=\dfrac{19\times(-25)}{.55(-200-\dfrac{19}{.55})}$

$f_{e}=3.68\ cm$

Hence, The focal length of eyepiece is 3.68 cm.

3 0

3 years ago

A body of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2

erastova [34]

You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2

Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.

-F=25(-1.8)
F=45N

Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.

45=u(25g)
45=u(25*10)

Therefore, the coefficient of friction is 0.18

Hope that helps

5 0

3 years ago

For thermal equilibrium at temperature Tan appropriate measure of energy is kT where k is Boltzmann's constant. Convert the foll

Schach [20]

Answer:

1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K

Explanation:

The relationship between temperature and thermal energy is

E = K T

The relationship of the speed of light

c =λ f = f / ν 1/λ= ν

The Planck equation is

E = h f

Let's start the transformations

c = f λ = f / ν

f = c ν

E = h f

E = h c ν

E = KT

h c ν = K T

T = h c ν / K =( h c / K) ν

Let's replace the constants

h = 6.63 10⁻³⁴ J s

c = 3 10⁸ m / s

K = 1.38 10⁻²³ J / K

v = 1 cm-1 (100 cm / 1 m) = 10² m-1

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²

A = h c / K = 1,441 10⁻²

T = 1.44K

ν = 103 cm⁻¹ = 103 10² m

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²

T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

T = (h c / K) 1 /λ

T = 1,441 10⁻² 1 / 1,097 10⁷

T = 1.3 10⁻⁹ K

E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J

E = KT

T = E/K

T = 1.6 10⁻¹⁹ /1.38 10⁻²³

T = 1.16 10⁴ K

3 0

4 years ago