What is the speed of the object at the end of 10 s?

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

3 years ago

Elements that easily transmit electricity and heat display the property know as ______

andre [41]

It is known as conductivity

7 0

3 years ago

Read 2 more answers

it is determined that a certain light wave has a wavelength of 3.012x10^-12 m. the light travels at 2.99x10^8 m/s. what is the f

Dahasolnce [82]

Answer:

$9.93\times 10^{19}Hz$

Explanation:

Speed of light is the product of its wavelength and frequency, expressed as

S=fw

Where s represent speed, f is frequency while w is wavelength

Making f the subject of the formula then

f=s/w

Substituting 2.99x10^8 m/s for s and 3.012x10^-12 m for w then

$f=\frac {2.99\times 10^{8}}{3.012\times 10^{-12}}=9.926958831341\times 10^{19}\\f\approx 9.93\times 10^{19}Hz$

Therefore, the frequency equals to $9.93\times 10^{19}Hz$

4 0

3 years ago

How can inclined planes change the amount of force to move an object? A. Increase height of the inclined plane B. Increase dista

mash [69]

Answer:

Option ( B ) is correct .

Explanation:

To lift a heavy weight , inclined plane is used . Use of inclined plane , makes the task easier because instead of force mg , force mg sinθ is to be used which is less than mg . Here θ is inclination of inclined plane.

If h be the height by which weight is to be lifted

potential energy acquired by weight = mgh

work done by force mg sinθ = mgsinθ x d where d is displacement required .

mg sinθ x d = mgh ( work done by force = potential energy stored in luggage )

d = h / sinθ

d will be more than h

Hence inclined plane increases the distance to be covered by force applied though it decreases the force itself.

Hence option ( B ) is correct .

4 0

3 years ago

An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first

nordsb [41]

Answer:

a=0.555m/s^2

Explanation:

First we find the distance traveled from the moment the engineer reacts to the car, assuming uniform movement

X=VT

X=(18)(0.45)=8.1m

then we find the distance at which the deceleration begins, which is obtained by subtracting the total distance with the inner result

X=300-8.1=291.9

finally we use the equation for constant acceleration

Vf=0 final speed

Vo=18m/s= initial speed

X=291.9m

(Vf^2-Vo^2)/2X=a

(0-18^2)/(2*291.9)=a

a=0.555m/s^2

8 0

3 years ago

Read 2 more answers