Answer:
37357 sec
or 622 min
or 10.4 hrs
Explanation:
GIVEN DATA:
Lifting weight 80 kg
1 cal = 4184 J
from information given in question we have
one lb fat consist of 3500 calories = 3500 x 4184 J
= 14.644 x 10^6 J
Energy burns in 1 lift = m g h
= 80 x 9.8 x 1 = 784 J
lifts required 
= 18679
from the question,
1 lift in 2 sec.
so, total time = 18679 x 2 = 37357 sec
or 622 min
or 10.4 hrs
Answer: Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside
Explanation:
Given that;
friction force of ground box = 68 N
student of 7th grade = n
Whitmore can apply a force of 25 N
every other 7th grade student can apply a force of 6 N.
now
friction force = forced applied by whitmore + total force ny 7th grade student
we substitute
68 = 25 + 6n
6n = 68 - 25
6n = 43
n = 43/6
n = 7.17
Therefore Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside
Answer:
The work done on the suitcase is, W = 600 J
Explanation:
Given,
The average force exerted by Jose on his suitcase, F = 60 N
Jose carried the suitcase to a distance, S = 10 m
The work done on the suitcase is given by the relation
<em>W = F x S</em>
Substituting the given values in the above equation,
W = 60 N x 10 m
= 600 J
Hence, the work done on the suitcase is, W = 600 J
On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s
Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.
Maximum velocity of a road with friction is given by the formula,
v = μRg
where, v is the maximum velocity
μ is the coefficient of static friction
R is the radius of the circle road
g is the acceleration due to gravity
Given,
μ = 0.20
R = 75m
g = 9.8m/s²
On substituting the given values in the above formula,
v = 0.20× 75 ×9.8
v = 147m/s
So, the Maximum velocity of the wet road is 147m/s.
Learn more about Velocity here, brainly.com/question/18084516
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