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3 years ago

Which kind of inclined plane pushes up more? Steeper or flatter?

Physics

1 answer:

Viktor [21]3 years ago

7 0

Answer:

They should've sensed that it was easier as the board got steeper. A flatter inclined plane has to apply more upward force to support something (or someone) because gravity, pulling toward the center of the Earth, pulls the object more into the table.Explanation:

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You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3

LenaWriter [7]

Answer:

0 Joules

Explanation:

The work done is given by

$W=F\times s\times cos\theta$

where,

F = Force applied

s = Displacement of the object = 0 m

$\theta$ = Angle between the force applied and the horizontal = 0

$W=F\times 0\times cos0\\\Rightarrow W=0\ J$

Work is only observed when there is a displacement.

The work done by me is 0 Joules as I was unable to move it.

6 0

3 years ago

Can someone see if it's right? thank you.

bearhunter [10]

Answer:

it's right you did a great job

8 0

2 years ago

Oil having a density of 922kg/m^3 floats on water. A rectangular block of wood 3.97 cm high and with a density of 963 kg/m^3 flo

Blizzard [7]

Explanation:

For the equilibrium:

\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ

wood

gh−ρ

oil

g(h−x)−ρ

water

gx=0

\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ

wood

h−ρ

oil

(h−x)−ρ

water

x=0

(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0

x=2.54\ cmx=2.54 cm

3 0

3 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