Let's use Newton's 2nd law of motion:
Force = (mass) x (acceleration)
Force = (68 kg) x (1.2 m/s²) = 81.6 newtons .
Answer:
c.
Explanation:
In positron emission, also called positive beta decay (β+-decay), a proton in the parent nucleus decays into a neutron that remains in the daughter nucleus, and the nucleus emits a neutrino and a positron, which is a positive particle like an ordinary electron in mass but of opposite charge.
Does this help? Sorry it's wordy that's just how my teachers taught me :'(
Answer:
47.48 J/g/K or 4.74 × 10⁴J/Kg/K
Explanation:
C(specific heat capacity)= Q(quantity of heat)/M(mass) × ∆T(change in temperature, Kelvin)
Q= 8000J
M= 46g or 46 ×10^-3Kg
T1= 24°C = 297K
T2 = 28°C = 301K
∆T = 301-297
= 4K
C= 8000/46 × 4
= 47.48 J/g/K or 4.74 × 10⁴J/Kg/K
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 = / W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s