Effieciency of a machine is never 100%
1 answer:
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Answer:
(a)Velocity of wind = 35 j
(b)Velocity of jet relative to air = 425 i
(c)True velocity of jet = 425 i + 35 j
(d)True speed of jet = = 426.88 mi/h
Direction of jet is, ∅ = = 5.38°
Explanation:
We can represent east direction by i and north direction by j.
The jet has a relative speed of 425 mi/h relative to the air.
The wind is blowing due north with a speed of 35 mi/h = 35 j
425 mi/h is the relative speed with respect to wind that is
Velocity of jet wrt wind= =
425 i = - 35 j
= 425 i + 35 j
(a)Velocity of wind = 35 j
(b)Velocity of jet relative to air = 425 i
(c)True velocity of jet = 425 i + 35 j
(d)True speed of jet = = 426.88 mi/h
Direction of jet is,
∅ = = 5.38°
Answer:
(A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Explanation:
Given that,
A nucleus undergoes a nuclear decay.
(A). In alpha decay,
We know that,
When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Fr.
(B). In beta-minus decay,
We know that,
When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Po.
(C). In beta-plus decay,
We know that,
When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Ne.
(D). In gamma decay,
We know that,
When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.
We need to calculate the resulting nucleus
Using given data
The resulting nucleus is Tc.
Hence, (A). The resulting nucleus is Fr.
(B). The resulting nucleus is Po.
(C). The resulting nucleus is Ne
(D). The resulting nucleus is Tc.
Answer:
a) v₀ = 32.64 m / s , b) x = 59.68 m
Explanation:
a) This is a projectile launching exercise, we the distance and height of the cliff
x = v₀ₓ t
y = t - ½ g t²
We look for the components of speed with trigonometry
sin 43 = v_{oy} / v₀
cos 43 = v₀ₓ / v₀
v_{oy} = v₀ sin 43
v₀ₓ = v₀ cos 43
Let's look for time in the first equation and substitute in the second
t = x / v₀ cos 43
y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²
y = x tan 43 - ½ g x² / v₀² cos² 43
1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²
v₀² = g x² / [(x tan 43 –y) 2 cos² 43]
Let's calculate
v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]
v₀ = √ (35280 / 33.11)
v₀ = 32.64 m / s
.b) we use the vertical distance equation with the speed found
y = t - ½ g t²
.y = v₀ sin43 t - ½ g t²
25 = 32.64 sin 43 t - ½ 9.8 t²
4.9 t² - 22.26 t + 25 = 0
t² - 4.54 t + 5.10 = 0
We solve the second degree equation
t = (4.54 ±√(4.54 2 - 4 5.1)) / 2
t = (4.54 ± 0.46) / 2
t₁ = 2.50 s
t₂ = 2.04 s
The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled
x = v₀ₓ t
x = v₀ cos 43 t
x = 32.64 cos 43 2.50
x = 59.68 m