Answer:
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your answer is here
SI (International System of Units) (meter-kilogram-second-ampere-kelvin-mole-candela)
FPS (foot-pound-second)
MKS (meter-kilogram-second)
CGS (centimeter-gram-second)
EMU (Electromagnetic) (centimeter-gram-second-abampere)
ESU (Electrostatic) (centimeter-gram-second-abcoulomb)
Atomic (bohr-electron mass-atomic second-electron)
MTS (meter-tonne-second)
Explanation:
hope it may help you !!
Answer:
F = 385.56 N
Explanation:
Given that,
Mass, m = 37.8 kg
He applies a horizontal force and cross the wooden floor.
We need to find the force that must be applied to move a dog with a constant speed of 1 m/s.
The coefficient of kinetic friction between the dog in the floor is 1.02.
The net force acting on it is given by :
F = f- μmg
f is force due to constant speed, f = 0 (since, a = 0)
F = μmg
= 1.02 × 37.8 × 10
= 385.56 N
Hence, the required force is 385.56 N.
Yes there called stars because when a star in space goes nova it shines in the sky like a little twinkle because it far away the sun is also a star too.<span />
OK. So you're pushing on the small box, and on the other side of it, the small
box is pushing on the big box. So you're actually pushing both of them.
-- The total mass that you're pushing is (5.2 + 7.4) = 12.6 kg.
-- You're pushing it with 5.0N of force.
-- Acceleration of the whole thing = (force)/(mass) = 5/12.6 = <em>0.397 m/s²</em> (rounded)
-- Both boxes accelerate at the same rate. So the box farther away from you ...
the big one, with 7.4 kg of mass, accelerates at the same rate.
The force on it to make it accelerate is (mass) x (acceleration) =
(7.4 kg) x (5/12.6 m/s²) = <em>2.936 N.</em>
The only force on the big box comes from the small box, pushing it from behind.
So that same <em>2.936N</em> must be the contact force between the boxes.
Answer:
13.2m
Explanation:
Step one:
given data
Energy= 5610J
Force F= 425N
Required
The distance traveled
Step two:
We know that work done is given as
WD= force* distance
so
5610=425*d
divide both sides by 425
d= 5610/425
d=13.2m
