<span>a. KE in electron volts is 1020 eV.
b. KE in Joules is e(1020) = (1.6022E-19)(1020) = 1.634E-16
c. KE = (1/2)mv^2, so v = sqrt[2*KE/m] = 18.94E6 m/s
note: m is the mass of an electron = 9.109e-31 kg
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Answer:
22425 J
Explanation:
From the question,
Applying
Q = cm(t₂-t₁).................. Equation 1
Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.
Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)
Substitute these values into equation 1
Q = 897×1×(25-50)
Q = 897×(-25)
Q = -22425 J
Hence the thermal energy lost by the aluminium is 22425 J
Answer:
t = 39.60 s
Explanation:
Let's take a careful look at this interesting exercise.
In the first case the two motors apply the force in the same direction
F = m a₀
a₀ = F / m
with this acceleration it takes t = 28s to travel a distance, starting from rest
x = v₀ t + ½ a t²
x = ½ a₀ t²
t² = 2x / a₀
28² = 2x /a₀ (1)
in a second case the two motors apply perpendicular forces
we can analyze this situation as two independent movements, one in each direction
in the direction of axis a, there is a motor so its force is F/2
the acceleration on this axis is
a = F/2m
a = a₀ / 2
so if we use the distance equation
x = v₀ t + ½ a t²
as part of rest v₀ = 0
x = ½ (a₀ / 2) t²
let's clear the time
t² = (2x / a₀) 2
we substitute the let of equation 1
t² = 28² 2
t = 28 √2
t = 39.60 s
The area of the circle with radius r is
A = πr²
The rate of change of area with respect to time is

The rate of change of the radius is given as

Therefore

When r = 10 ft, obtain

Answer: - 40π ft²/s (or - 127.5 ft²/s)