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2 years ago

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

Physics

1 answer:

Vitek1552 [10]2 years ago

5 0

Answer:

Explanation:

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

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A wagon wheel consists of 8 spokes of uniform diameter, each of mass ms and length L cm. The outer ring has a mass mring. What i

just olya [345]

Answer:

The moment of inertial of the wheel, $I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2$

Explanation:

Given;

8 spokes of uniform diameter

mass of each spoke, = $M_s$

length of each spoke, = L

mass of outer ring, = $M_r$

The moment of inertial of the wheel will be calculated as;

$I = 8I_{spoke} + I_{ring}$

where;

$I_{spoke$ is the moment of inertia of each spoke

$I_{ring$ is the moment of inertia of the rim

Moment of inertia of each spoke $=\frac{1}{3}M_sL^2$

Moment of inertial of the wheel

$I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2$

6 0

2 years ago

Playing with a stress ball you squeeze it as

shutvik [7]

Answer:

it dies

Explanation:

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8 0

3 years ago

Atoms with the same number of protons but with different electrical charges _____.

mash [69]

Answer:

Atoms with the same number of protons but with different electrical charges are different ions

Explanation:

Ions are defined as those atoms or molecules which carry charge

5 0

3 years ago

Read 2 more answers

In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re

Leya [2.2K]

Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

$I=2.6363\ kg.m^2$

6 0

3 years ago

Imagine a system where a block rests on an inclined plane. The block is then given an initial push so that it starts sliding dow

Helen [10]

Answer:

statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.

Explanation:

The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.

The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction

here, the downward direction signifies the downward motion parallel to the inclined plane.

Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.

Hence, for the block to stop sliding the the above statement should be true.

6 0

3 years ago