Please help!! I'm stuck on this, could someone also explain if they can!! Will give brainiest if I can!!

4) A drag racer starts her car from rest and accelerates at 10.0 m/s² for a distance of 400 m (1/4 mile). (a) How long did it ta

mel-nik [20]

Answer:

A) s=1/2at^2

t=√(2s/a)=√(2x400)/10.0)=9.0s

B) v=at

v=10.0x9=90m/s

3 0

2 years 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

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

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

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

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago

In freefall, heavier objects fall with a greater acceleration than lighter objects.

Rom4ik [11]

Answer:

That is not true all objects fall at the same speed excepts things like feathers or paper.

6 0

2 years ago

Read 2 more answers

What direction does an applied force move an object

monitta

The object will move in the direction of the applied force.

4 0

2 years ago

Read 2 more answers

Planet a and planet b are in circular orbits around a distant star. planet a is 7.8 times farther from the star than is planet

gogolik [260]

To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula:

V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting.

So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb

Va/Vb=[ √( {G*M}/{7.8*Rb} ) ] / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b)

Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:

Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.

6 0

2 years ago