Answer:
I think it may be decreased, but i'm in 6th grade so-
Explanation:
for this we apply, Heisenberg's uncertainty principle.
it states that physical variables like position and momentum, can never simultaneously know both variables at the same moment.
the formula is,
Δp * Δx = h/4π
m(e).Δv * Δx = h/4π
by rearranging,
Δx = h / 4π * m(e).Δv
Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 5.10*10^-2
Δx = 6.63*10^-34 / 583.9 X 10 ⁻³¹
Δx = 0.011 X 10⁻³
for the bullet
Δx = (6.63*10^-34) / 4 * 3.142 * 0.032*10^-31 * 5.10*10^-2
Δx = 6.63*10^-34 /2.05
Δx =3.23 X 10⁻³² m
therefore, we can say that the lower limits are 0.011 X 10⁻³ m for the electron and 3.23 X 10⁻³² m for the bullet
To know more about bullet problem,
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Answer:
0.265
Explanation:
Draw a free body diagram. There are four forces:
Normal force Fn pushing up.
Weight force mg pulling down.
Tension force T at an angle θ.
Friction force Fn μ pushing left.
Sum the forces in the y direction:
∑F = ma
Fn + T sin θ − mg = 0
Fn = mg − T sin θ
Sum the forces in the x direction:
∑F = ma
T cos θ − Fn μ = 0
Fn μ = T cos θ
μ = T cos θ / Fn
μ = T cos θ / (mg − T sin θ)
Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:
μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)
μ = 0.265
<span>7.7 m/s
First, determine the acceleration you subject the sled to. You have a mass of 15 kg being subjected to a force of 180 N, so
180 N / 15 kg = 180 (kg m)/s^2 / 15 kg = 12 m/s^2
Now determine how long you pushed it. For constant acceleration the equation is
d = 0.5 A T^2
Substitute the known values getting,
2.5 m = 0.5 12 m/s^2 T^2
2.5 m = 6 m/s^2 T^2
Solve for T
2.5 m = 6 m/s^2 T^2
0.41667 s^2 = T^2
0.645497224 s = T
Now to get the velocity, multiply the time by the acceleration, giving
0.645497224 s * 12 m/s^2 = 7.745966692 m/s
After rounding to 2 significant figures, you get 7.7 m/s</span>
