there are 28 neutrons present
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
<h3>Magnitude of net force on the crate</h3>
F(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
<h3>Net work done on the crate</h3>
W = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
<h3>Acceleration of the crate</h3>
a = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
<h3>Speed of the crate</h3>
v² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
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The accuracy of a series of measurements is best understood by looking at the known value.
The values obtained in a series of measurement can be precise or accurate.
The precision of the measurement is obtained by looking at the trend of the values obtained.
- If the values are close to each other, the measurement is precise but if the values are far from each other, the measurement is not precise.
The accuracy of a measurement is determine by comparing the determine value against a known value.
- If the measured value is close to a <em>known value</em>, then the measurement is accurate, but if the measured value is from from a known value then the measurement is not accurate.
Thus, the accuracy of a series of measurements is best understood by looking at the known value.
Learn more here: brainly.com/question/13688896
The net force performs a total amount of work equal to
(45 N) (0.80 m) = 36 J
on the bullet, and this is in turn is equal to the change in the bullet's kinetic energy by the work-energy theorem. So we have
W = ∆K = 1/2 mv²
since the bullet starts at rest, where m = its mass and v = its final velocity.
Solve for v :
36 J = 1/2 (0.0050 kg) v² ⇒ v = 120 m/s
I think it would be atomic energy, which would most likely be in very short waves to affect the protons.
