) Determining the order of events and the relative age of rocks by examining the position of

Which type of telescope does not require darkness in order to be able to use it?

exis [7]

Answer:

A type of telescope that does not require darkness in order to be able to use it is the refracting telescope

Explanation:

A refracting telescope consists of a lens and an eyepiece collects light which is then focused to present a magnified, bright and clear image.

The incident light on a refracting telescope is bent by refraction such that the light is focused to the focal point.

In refracting telescopes, the image is formed by bending light, that is by refraction.

The refracting telescope technology has been applied to binoculars and camera zoom lenses.

5 0

2 years ago

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.4 m/

Bad White [126]

Answer:

1.06 secs

Explanation:

Initial speed of sled, u = 8.4 m/s

Final speed of sled, v = 5.8 m/s

Coefficient of kinetic friction, μ = 0.25

Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:

FΔt = mv - mu

where m = mass of the object

Δt = time interval

F = force applied

The force applied on the sled is the frictional force, which is given as:

F = -μmg

where g = acceleration due to gravity

Therefore:

-μmgΔt = mv - mu

-μmgΔt = m(v - u)

-μgΔt = v - u

Making Δt subject of formula:

Δt = (v - u) / -μg

Δt = (5.8 - 8.4) / (-0.25 * 9.8)

Δt = -2.6/ -2.45

Δt = 1.06 secs

It took the sled 1.06 secs to travel from A to B.

7 0

3 years ago

What voltage produces a current of 50 amps with a resistance of 20 ohms?<br>

PIT_PIT [208]

Answer:

1000 V

Explanation:

Ohm's law:

V = IR

V = (50 A) (20 Ω)

V = 1000 V

5 0

3 years ago

Determine the magnitude and direction of the force on a 200 m power line carrying a current of 5.0-A due west in a magnetic fiel

bogdanovich [222]

Answer:

Explanation:

l = 200 m

i = 5 A west

B = 6 micro Tesla direction 30 degree north of east

angle between length vector and the magnetic field vector = 180 - 30 = 150 degree

Write the length and the magnetic field in the vector form

$\overrightarrow{l}=- 200 \widehat{i}metre$

$\overrightarrow{B}= 6\times 10^{-6}\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )Tesla$

$\overrightarrow{B}=\left ( 5.2\widehat{i}+3\widehat{j} \right )\times 10^{-6}Tesla$

$\overrightarrow{F} = i \overrightarrow{l}\times \overrightarrow{B}$

$\overrightarrow{F} = 5\times \left ( -200\widehat{i} \right )\times \left ( 5.2\widehat{i}+3\widehat{j} \right )\times 10^{-6}$

$\overrightarrow{F} =- 3\times 10^{-3}\widehat{k}Newton$

Thus, the magnitude of force is 3 x 10^-3 newton and it is directed towards negative z axis direction.

7 0

2 years ago

An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g g

HACTEHA [7]

Answer:

Glider it stops just when it reaches the end of the runway

Explanation:

This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy

Po = pf

Ko = Kf

Before crash

Po = m1 Vo1 + 0

Ko = ½ m1 Vo1²

After the crash

Pf = m1 Vif + Vvf

Kf = ½ m1 V1f² + ½ m2 V2f²

m1 V1o = m1 V1f + m2 V2f (1)

m1 V1o² = m1 V1f² + m2 V2f² (2)

We see that we have two equations with two unknowns, so the system is solvable, we substitute in 1 and 2

m1 (V1o -V1f) = m2 V2f (3)

m1 (V1o² - V1f²) = m2 V2f²

Let's use the relationship (a + b) (a-b) = a² -b²

m1 (V1o + V1f) (V1o -V1f) = m2 V2f²

We divide with 3 and simplify

(V1o + V1f) = V2f (4)

Substitute in 3, group and clear

m1 (V1o - V1f) = m2 (V1o + V1f)

m1 V1o - m2 V1o = m2 V1f + m1 V1f

V1f (m1 -m2) = V1o (m1 + m2)

V1f = V1o (m1-m2 / m1+m2)

We substitute in (4) and group

V2f = V1o + (m1-m2 / m1 + m2) V1o

V2f = V1o [1+ + (m1-m2 / m1 + m2)]

V2f = V1o (2m1 / (m1+m2)

We calculate with the given values

V1f = 0.81 (239-513 / 239 + 513)

V1f = 0.81 (-274/752)

V1f = - 0.295 m/s

The negative sign indicates that the planned one moves in the opposite direction to the initial one

V2f = 0.81 [2 239 / (239 + 513)]

V2f = 0.81 [0.636]

V2f = 0.515 m / s

Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,

Eo = Ef

Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point

Eo = K = ½ m2 vf2²

Ef = U = m2 g Y

½ m2 v2f² = m2 g Y

Y = V2f² / 2g

Y = 0.515²/2 9.8

Y = 0.0147 m

At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track

tan θ = Y / x

Y = x tan θ

The crash occurs in the middle of the track whereby x = 1.2 m

Y = 1.2 tan 0.7

Y = 0.147 m

As the two quantities are equal in glider it stops just when it reaches the end of the runway

7 0

3 years ago