Answer:
v₀ = 60.38 mi / h
With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.
Explanation:
For this exercise let's start by using Newton's second law
Y axis
N-W = 0
N = W
X axis
fr = m a
the expression for the friction force is
fr = μ N
we substitute
μ mg = m a
μ g = a
calculate us
a = 0.620 9.8
a = 6.076 m / s²
now we can use the kinematics relations
v² = v₀² - 2 a x
suppose v = 0
v₀ =
Ra 2ax
let's calculate
v₀ = 
v₀ = 27.00 m / s
let's slow down to the english system
v₀ = 27.0 m / s (3.28 ft / 1m) (1 mile / 5280 ft) (3600s / 1h)
v₀ = 60.38 mi / h
With this stopping distance, the starting speed should have been 60.38 mi/h, which is much higher than the maximum speed allowed.

An electron is emitted from a nucleus- State the effect this has on the charge of the nucleus
Answer:
an electron orbiting around the nucleus combines with a nuclear proton to produce a neutron, which remains in the nucleus, and a neutrino, which is emitted. As in positron emission, the nuclear positive charge and hence the atomic number decreases by one unit, and the mass number remains the same.
hope this helps!!!
-xXxAnimexXx-
False,it could be false information like Wikipedia
Answer:
The velocity when the ball hits the ground is obtained using v2. 2 = v1. 2 + 2 g Dy with v1=0 and Dy=h. Thus solving for v2 yields 17.1 m/s v2 = 2 g h =.
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