Please help<br>........

Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from t

stira [4]

Answer:

(a) $a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) $a_{t} = 2.99\times 10^{- 5} m/s^{2}$

Given:

Time period of Pulsar, $T_{P} = 33.085 ms == 33.085\times 10^{- 3} s$

Equatorial radius, R = 15 Km = 15000 m

Spinning time, $t_{s} = 9.50\times 10^{10}$

Solution:

(a) To calculate the value of the centripetal acceleration, $a_{c}$ on the surface of the equator, the force acting is given by the centripetal force:

$m\times a_{c} = \frac{mv_{c}^{2}}{R}$

$a_{c} = \frac{v_{c}^{2}}{R}$ (1)

where

$v_{c} = \frac{distance covered(i.e., circumference)}{ T}$

$v_{c} = \frac{2\pi R}{Time period, T}$ (2)

Now, from (1) and (2):

$a_{c} = R\frac({2\pi )^{2}}{T^{2}}$

$a_{c} = 15000\frac{2\pi )^{2}}{(33.085\times 10^{- 3})^{2}}$

$a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) To calculate the tangential acceleration of the object :

The tangential acceleration of the object will remain constant and is given by the equation of motion as:

$v = u + a_{t}t_{s} = 0$

where

u = $v_{c}$

$a_{t} = - \frac{2\pi R}{Tt_{s}}$

$a_{t} = - \frac{2\pi 15000}{33.085\times 10^{- 3}\times 9.50\times 10^{10}}$

$a_{t} = 2.99\times 10^{- 5} m/s^{2}$

7 0

3 years ago

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The coefficient of static friction between t

Wewaii [24]

Answer:

a) a = 6.1 m/s^2

b) a = 0.98m/s^2

Explanation:

Mass of slab = 40kg

Mass of block = 10kg

Coefficient of static friction (Us) = 0.60

Kinetic coefficient (UK) = 0.40

Horizontal force = 100N

The normal reaction from 40kg slab on 10 kg block = 10*9.81

= 98.1N

Static frictional force = Us*R

= 98.1*0.6

= 58.86N

This is less than the force applied

If 10 kg block will slide on the 40 kg slab, net force = 100 - kinetic force

Kinetic force (Uk*R) = 0.4*98.1

= 39.28N

= 39N

Net force = 100 -39

= 61N

Recall that F = ma

For 10 kg block

a = F/m

a = 61/10

a = 6.1m/s^2

b) Frictional force on 40 kg slab by 10 kg = 98.1*0.4

= 39.24

= 39N

F = ma

a = F/m

For 40kg slab

a = 39/40

a = 0.98m/s^2

3 0

3 years ago

List the requirements of practical fuel?

rewona [7]

Answer:

Here we have some of the requirement of practical fuel are

1. It must contain large amount of stored energy. So that more amount of power output available to run the engines, motors etc.

2. It must occur in abundance in nature or be easy to produce.

3. The fuel must be made up of elements that combine easily with oxygen. Foe example if hydrogen molecules reacts with oxygen. Then the products are at the reaction of lower energy than the reactants, the result is the explosive release of energy and the product of water.

7 0

3 years ago

To move a suitcase up to the check-in stand at the airport a student pushes with a horizontal force through a distance of 0.95 m

solmaris [256]

Answer:

33.68 N

Explanation:

Data

W= 32J

d- 0.95m

F= ?

W=Fd

They are asking for the magnitude which is the force, so you need to solve for force.

F=W/d

= 32J/ 0.95m

= 33.68 N

6 0

3 years ago

A 2130 kg car is parked on a hill that makes a 15º angle with the horizontal. What is the normal force on the car?

valentinak56 [21]

Answer:

20573.67N

Explanation:

Given;

mass (m) of the car = 2130kg

angle of inclination Θ = 15⁰

The normal force (F) on the car is given by

F = mgcosΘ

where g is the acceleration due to gravity.

Taking g as 10 $m/s^{2}$ and substituting the values of m and Θ into the equation. We have;

F = 2130 x 10 x cos 15⁰

F = 2130 x 10 x 0.9659

F = 20573.67N

Therefore the normal force on the car is 20573.67N

7 0

4 years ago

Please help........​

Please help........