Answer:
0.000000113 or 1.13*
meters
Explanation:
One nanometer is 
meters. So 113 nanometers would be 113*, or 1.13* meters. That's expessed on "cientific notation." On the "usual" notation, it will be 0.000000113 meters.
<u>Answer:</u>
The force F applied to the handle = 330.03 N
<u>Explanation:</u>
The force can be resolved in to two, horizontal component and vertical component. If θ is the angle between horizontal and applied force we have
Horizontal component of force = F cos θ
Vertical component of force = F sin θ
In this problem normal force exerted on the suitcase is 160 N, that is vertical component of force = 160 N and angle θ = 29⁰.
So, F sin 29 = 160
F = 330.03 N
The force F applied to the handle = 330.03 N
By Newton's Law of Universal Gravitation.
F = GMm/r²
Where F is Force of Gravitation, M = Mass of first object, m = mass of second object, r = distance of separation
From the formula, you can see that if the masses, M and m, increased, the value of F would definitely increase as well.
And if r increased the value of F would be reduced because you would be dividing by a bigger number, but when the value of r is decreased the value of F would be increased, because you would then be dividing by something smaller. Note the r is at the denominator of the formula.
So F would increase if there was increase in Masses and decrease in distance.
So the answer is C. a and b.
Answer:
6.4 J
Explanation:
m = mass of the bullet = 10 g = 0.010 kg
v = initial velocity of bullet before collision = 1.8 km/s = 1800 m/s
v' = final velocity of the bullet after collision = 1 km/s = 1000 m/s
M = mass of the block = 5 kg
V = initial velocity of block before collision = 0 m/s
V' = final velocity of the block after collision = ?
Using conservation of momentum
mv + MV = mv' + MV'
(0.010) (1800) + (5) (0) = (0.010) (1000) + (5) V'
V' = 1.6 m/s
Kinetic energy of the block after the collision is given as
KE = (0.5) M V'²
KE = (0.5) (5) (1.6)²
KE = 6.4 J
