Answer:
R (120) = 940Ω
Explanation:
The variation in resistance with temperature is linear in metals
ΔR (T) = R₀ α ΔT
where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC
let's calculate
ΔR = 1000 (-0,0005) (120-0)
ΔR = -60
Ω
ΔR = R (120) + R (0) = -60
R (120) = -60 + R (0)
R (120) = -60 + 1000
R (120) = 940Ω
Answer:
33.2 m
Explanation:
For the first object:
y₀ = 81.5 m
v₀ = 0 m/s
a = -9.8 m/s²
t₀ = 0 s
y = y₀ + v₀ t + ½ at²
y = 81.5 − 4.9t²
For the second object:
y₀ = 0 m
v₀ = 40.0 m/s
a = -9.8 m/s²
t₀ = 2.20 s
y = y₀ + v₀ t + ½ at²
y = 40(t−2.2) − 4.9(t−2.2)²
When they meet:
81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²
81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)
81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716
81.5 = 61.56t − 111.716
193.216 = 61.56t
t = 3.139
The position at that time is:
y = 81.5 − 4.9(3.139)²
y = 33.2
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
Answer:
F = 2 * 30 / 5 = 12 N to stop forward motion
F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees
(12^2 + 16^2)^1.2 = 20 N average force applied
Answer:
The density of gold is of 18 grams per cm3.
Explanation:
The mass density of a homogeneous material expresses how much mass of that material is present in a given volume. Since the density of an object is obtained by dividing its mass by its volume, to obtain the density of gold, its 90 grams of mass must be divided by its 5 cm3 volume, performing the following calculation:
90/5 = X
18 = X
Thus, the density of gold is 18 grams per cm3.