Answer:
10.13
Explanation:
take note that velocity is distance over time () so you do 456(distance)÷45(time)= 10.13
Hector drew the diagram below to illustrate projectile motion.
2 answers:
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.
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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The required torque at the axle, is given by the difference between the
moments of the applied forces.
The torque required is <u>19.62 N·m counterclockwise</u>
Reasons:
The given parameters are;
Mass of the disk, m = 5.0 kg
Location of the axle = Half the radius of the disk
Diameter of the disk, D = 40 cm = 0.4 m
Applied mass, 0.1 m from the axle = 15 kg
Applied mass, 0.3 m from the axle = 10 kg
Required:
Magnitude of torque at the axle that prevent the disk from rotating
Solution:
Torque needed = Clockwise moment - Counterclockwise moment
Clockwise moment = (10 kg × 0.3 m + 5 kg × 0.1 m) × 9.81 m/s² = 34.335 N·m
Counterclockwise moment = 15 kg × 0.1 m × 9.81 m/s² = 14.715 N·m
τ + Counterclockwise moment = Clockwise moment
τ + 14.715 N·m = 34.335 N·m
Torque required, τ = 34.335 N·m - 14.715 N·m = 19.62 N·m
Torque required, τ = <u>19.62 N·m counterclockwise</u>
Learn more here:
<em>The probable question drawing obtained from a similar question online is attached</em>
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;
where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
Therefore, the mutual force between the two point charges is 319.64 N