Answer:
F = 69.3 N
Explanation:
For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by
fr = μ N
We define a reference system parallel to the floor
block B ( lower)
Y axis
N - W₁-W₂ = 0
N = W₂ + W₂
N = (M + m) g
X axis
F -fr = M a
for block A (upper)
X axis
fr = m a (2)
so that the blocks do not slide, the acceleration in both must be the same.
Let's solve the system by adding the two equations
F = (M + m) a (3)
a =
the friction force has the formula
fr = μ N
fr = μ (M + m) g
let's calculate
fr = 0.34 (2.0 + 0.250) 9.8
fr = 7.7 N
we substitute in equation 2
fr = m a
a = fr / m
a = 7.7 / 0.250
a = 30.8 m / s²
we substitute in equation 3
F = (2.0 + 0.250) 30.8
F = 69.3 N
Answer:
145.8 cm³ of paint
Explanation:
d₁ = Smaller diameter paintball = 5 cm
d₂ = Larger diameter paintball = 9 cm
V₂ = Volume of larger diameter paintball
Volume of smaller diameter paintball
Similarly
Dividing the above two equations, we get
∴ The larger one hold 163.296 cm³ of paint
Answer:
0.0002°, 0.1691°, 0.338°
Explanation:
Difference between the two line = 5.97 * 10-⁸m
d = 1 / N
N = 5.0 * 10³
d = 2.0 * 10⁴m
nL = Nsin¤
For first order
588.995 * 10-⁹ = 2.0 * 10-⁴ sin ¤
Sin¤ = 2.944*10-³
¤ = sin-¹ 0.002944
¤ = 0.1687°
First order ¤ =
Sin-¹(589.592*-⁹ / 2.0 * 10-⁴)
Sin-¹ (0.002947) = 0.1689°
Angular separation = 0.1689 - 0.1687 = 0.0002°
Second order ¤ = sin-¹ [2 (589.59*10-⁹ / 2.0*10-⁴)] = sin-¹ (0.005895)
Second order ¤ = 0.3378°
Angular difference = 0.3378° - 0.1687° = 0.1691°
Third order ¤ = sin-¹ [3(589.59*10-⁹ /2.0*10-⁴] = 0.5067°
Angular difference = 0.5067° - 0.1687° = 0.338°
Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m
Formula:
Solving:
Answer:
<span>
B.159 N</span>
Answer: A
Explanation:
Well the high and lows effect the humidity the more humidity the more hot it is so the high brings higher temperatures.
