Answer:
In one rotation, the large wheel turns 4m.
Explanation:
The given values are:
Input distance,
= 0.64 m
Mechanical advantage,
= 0.16
As we know,
⇒
On putting the values, we get
⇒
⇒
A student in an undergraduate physics lab is studying Archimede's principle of bouyancy. The student is given a brass cylinder a
1 answer:
Answer:
V = 2.64 10⁻⁴ m³, T = 21.85 N , T = 19.26 N
Explanation:
To calculate the cylinder volume we use the density equation
ρ = m / V
V = m / ρ
ρ = 8.44 g / cm³ (1 kg / 1000g) (10² cm / 1m)³ = 8.44 10³ kg / m³
V = 2.23 / 8.44 10³
V = 2.64 10⁻⁴ m³
We calculate the tension with Newton's second law
In the air
T-W = 0
T = mg
T = 2.23 9.8
T = 21.85 N
In water
In this case we have the push up
T + B -W = 0
T = W -B
The push formula is
B = g V
T = m g - g V
T = 2.23 9.8 - 1.00 103 9.8 2.64 10-4
T = 21.85 - 2.5872
T = 19.26 N
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Answer:
The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.
Explanation:
The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.
To solve this problem, start with setting up the net force equations for both block A and B:
where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:
Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:
The force of friction acting on block B is approximately 26.7N.
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