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2 years ago

John and Caroline go out for a walk one day. This graph represents their distance from home.

Physics

2 answers:

Serga [27]2 years ago

3 0

Answer:

Option C is the correct answer.

Explanation:

We have velocity of a body = Change in position/ Time.

Considering first portion of graph,

Change in position = 30 - 0 = 30 m

Time = 0.75 hours = 45 minutes = 2700 seconds

Velocity = 30/2700 = 0.011 m/s

Considering second portion of graph,

Change in position = 30 - 30 = 0 m

Time = 0.5 hours = 30 minutes = 1800 seconds

Velocity = 0/1800 = 0 m/s

Considering third portion of graph,

Change in position = 0 - 30 = -30 m

Time = 0.75 hours = 45 minutes = 2700 seconds

Velocity = -30/2700 = -0.011 m/s

So firstly they walked in one direction(positive direction), then they were still(velocity is zero), then they walked in the opposite direction( velocity is negative).

Option C is the correct answer.

Yuri [45]2 years ago

3 0

Answer:

I had this same question the answer is C 100% sure

Explanation:

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1 year ago

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