How do we know that pulsars must be neutron stars?

A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an

posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle $\theta=30°$

We need to calculate the angle in radian

$\theta=30\times\dfrac{\pi}{180}$

$\theta=0.523\ rad$

We need to calculate the path length

Using formula of path length

$Path\ length =angle\times radius$

$Path\ length=0.523\times5.9$

$Path\ length =3.09\ m$

(b). Angle = 30 rad

We need to calculate the path length

$Path\ length=30\times5.9$

$Path\ length=177\ m$

(c). Angle = 30 rev

We need to calculate the angle in rad

$\theta=30\times2\pi$

$\theta=188.4\ rad$

We need to calculate the path length

$Path\ length=188.4\times5.9$

$Path\ length =1111.56\ m$

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0

3 years ago

Arocket with an initial velocity of 20 m/s another engine that gives it an acceleration of 4 m/s ^ 2 over 10 secondsHow far did

ExtremeBDS [4]

Answer:

50m

Explanation:

Given parameters:

Initial velocity = 20m/s

Acceleration = 4m/s²

Time = 10s

Unknown:

Distance traveled by the rocket = ?

Solution:

To solve this problem use the expression below;

v² = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

final velocity = 0

Insert the parameters and solve;

0² = 20² + 2 x 4 x s

-400 = 8s

s = 50m

Disregard the negative sign because distance cannot be negative.

3 0

2 years ago

What does the electric field strength tell about the electric firld?

Solnce55 [7]

Answer:

Explanation:

The region around a charged particle where another charged particle experiences a force of attraction or repulsion is called electric field.

The strength of electric field is defined as the force experienced by the unit positive test charge.

E = F / q

Electric field strength is a vector quantity and it is measured in newton per coulomb.

Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.

The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.

8 0

3 years ago

Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission

bija089 [108]

Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law P = σ A e T⁴

Wien displacement law λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

P₁ = σ A T₁⁴

T = 12000K

P₂ = σ A T₂⁴

P₂ / P₁ = T₂⁴ / T₁⁴

P₂ / P₁ = (12000/3000)⁴

P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

λ T = 2,898 10-3

λ₁ = 2.89810-3 / T

λ₁ = 2,898 10-3 / 3000

λ₁ = 0.966 10-6 m

λ₁ = 966 nm

T = 12000K

λ₂ = 2,898 10-3 / 12000

λ₂ = 0.2415 10-6 m

λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

5 0

3 years ago

a ball is shot from the ground straight up into the air with initial velocity of 50 50 ft/sec. assuming that the air resistance

Anna [14]

The height attained by the ball is 11.86m

a ball is shot from the ground straight up into the air its initial and final velocity is

initial velocity, u = 50 ft/s = 50×0.305 = 15.25m/s

final velocity ,v = 0 m/s

gravity =-9.8 m/s²

( negative sign shows acceleration in opposite direction)

height =?

using the newton motion of equation

v² = u² + 2as

where

a= acceleration due to gravity(g)

s = height

v² = u² + 2gs

(0)² = (15.25)² + 2×(-9.8)×s

0 = (15.25)² - 19.6 × s

s= - (15.25)²/ 19.6

s = 11.86m

after ignoring the air resistance the maximum height of the ball is 11.86m

To learn more about motion under gravity -

brainly.com/question/27962354

#SPJ4

3 0

1 year ago