What is the force of an object with a mass of 65 kg and an an unknown acceleration?

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

Which one of the following explains how light energy helps us see all kinds of objects around us?

vampirchik [111]

Answer:

light reflects off the objects and enter our eyes

6 0

3 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

3 years ago

A tall building is swaying back and forth on a gusty day. The wind picks up and doubles the amplitude of the oscillation. By wha

xz_007 [3.2K]

Answer:

by a factor of 2

Explanation:

Maximum speed of a body in simple harmonic motion relate to the amplitude by the following formula:

v ( maximum speed in m/s ) = x ( amplitude in meters ) √K /m where K is in N/m and m is kg

v is directly proportional to the amplitude and increases as the amplitude increases by a factor of 2

4 0

3 years ago

A motor transfers 12 kJ of energy in 30 s. Calculate its power.

Diano4ka-milaya [45]

Answer:

power=400Watt

Explanation:

work done =12kJ=12×10³=12000j

time taken=30s

power=?

as we know that

power=work done/time taken

power=12000J/30s

power=400Watt

i hope this will help you :)

4 0

3 years ago

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