Answer:
0.2289
Explanation:
Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then
P= 700*9.81*2.5=17167.5 W= 17.1675 kW
To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%
Explanation:
1-How many moles of NazCOs are in 10.0 ml of a 2.0 M solution?
2-How many moles of NaCl are contained in 100.0 ml of a 0.20 M solution?
3- What weight (in grams) of H2SO4 would be needed to make 750.0 ml of
2.00 M solution?
4-What volume (in ml) of 18.0 M H2SO4 is needed to contain 2.45 g H2S04?
Answer:
1185 N
Explanation:
From Newton’s second law of motion,
F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force
Net force acting on motorcycle
is given by
Where F is force acting on motorcycle and f is frictional force
Substituting F-f for
hence ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided
245*3= F- 450
F=245*3 +450= 1185 N
That’s an atom
I hope that helped
Answer:
Explanation:
The mass of that science book...wow. In pounds that would be 35.2! Yikes!
Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is
Δx where
v is the final velocity, our unknown,
v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,
a is the pull of gravity which is always -9.8 m/s/s, and
Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:
so
and
v = 48 m/s
As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:
Δx and
144 = -19.6Δx so
Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:
120 - 7.3 = 112.7 m off the ground.