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2 years ago

Hector drew the diagram below to illustrate projectile motion.

Physics

2 answers:

Serjik [45]2 years ago

6 0

<u>Answer </u>

He should make the arrow for “Path” curve downward.

<u>Explanation </u>

The force of gravity is usually directly downward. So from the diagram, it is correctly labelled.

If the object was given a horizontal force, the direction of the inertial is also correct. Inertial is the force the resist the change of state of motion.

What Hector should change is the path followed by the object. It will be a curve not a straight line as it is drawn.

The correct answer is He should make the arrow for “Path” curve downward.

Eduardwww [97]2 years ago

5 0

Let us review the answers given.
1. He should switch the labels "Gravity" and "Inertia".
These labels are correct, so no change is required.

2. He should make the arrow for "Path" curve downward.
This statement is correct.

3. He should switch the labels "Inertia" and "Path".
No change is required.

4. He should make the arrow for "Inertia" curve downward.
This statement is incorrect.

Answer:
Hector should make the arrow for "Path" curve downward as shown in the figure.

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Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

Masja [62]

Answer:

Not possible

Explanation:

$E_{cl}$ = longitudinal modulus of elasticity = 35 Gpa

$E_{ct}$ = transverse modulus of elasticity = 5.17 Gpa

$E_m$ = Epoxy modulus of elasticity = 3.4 Gpa

$V_{\rho l}$ = Volume fraction of fibre (longitudinal)

$V_{\rho t}$ = Volume fraction of fibre (transvers)

$E_f$ = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

$E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764$

Transverse modulus of elasticity is given by

$E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148$

$V_{\rho l}\neq V_{\rho t}$

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0

3 years ago

The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain riv

kow [346]

Answer:

1.122 m/s

Explanation:

So usually a river with a speed of 1 meters per second can transport particle that weighs:

$1^6 = 1 kg$

If the particle is twice as massive as usual, then its weights would be 1 * 2 = 2kg

This means the river must be flowing at a speed of

$2^{\frac{1}{6}} = 1.122 m/s$

5 0

3 years ago

A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.

Tcecarenko [31]

Answer:

a) 1.89 m/s b) 172.1 N

Explanation:

Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
This work, is just 4,475 J.
So we can write the following equation:

$\Delta K = \frac{1}{2} * m*v^{2} = 4,475 J$