- Lola applies force to the handle of a jack so that she can change the flat tire on her car. If she uses
1 answer:
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Refer to the free body diagram shown melow.
F = applied force
R = frictional force
m = 15 kg, the mass of the object
The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N
The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.
Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)
Answer:
The most reasonable answer is R = 5.5 N
F = applied force
R = frictional force
m = 15 kg, the mass of the object
The acceleration (to the right) is 16.3 m/s², therefore
F - R = (15 kg)*(16.3 m/s²) = 244.5 N
The normal reaction is
N = mg = (15 kg)*(9.8 m/s²) = 147 N
The frictional force is
R = μN = 147μ N, where μ = coefficient of kinetic friction.
Let us check possible answers:
If R = 5.5 N, then μ = 5.5/147 = 0.0374 (very likely)
If R = 15 N, then μ = 15/147 = 0.102 (possible)
If R = 244.5 N, (Highly unlikely, exceed mg)
If R = 494.5 N, (highly unlikely, exceeds mg)
Answer:
The most reasonable answer is R = 5.5 N
The magnitude of the air drag when the object is traveling at terminal velocity is C. 850 Newtons
Explanation:
There are only two forces acting on the object here:
- The force of gravity, of magnitude , acting downward
- The air drag, , acting upward
Therefore, the equation of motion for the object is
where m is the mass of the object and a its acceleration.
The object in this problem is traveling at terminal velocity: this means that the acceleration is zero, so
a = 0
Therefore the equation becomes
which means that the magnitude of the air resistance is equal to the magnitude of the force of gravity:
Learn more about acceleration and Newton laws of motion here:
#LearnwithBrainly
Answer:
Explanation:
Given
mass of cannon and cart is
Spring constant
mass of Projectile
Launch Velocity of cannon
Launch angle
As the External Force is zero therefore we can conserve momentum
Initially both Projectile and cannon is at rest
Conserving momentum in horizontal direction
i.e. cannon is moving in opposite direction of Projectile