Answer:
a) a = 7.72 m / s², N = 19.9 N and b) F = 25.5 N
Explanation:
To solve this problem we will use Newton's second law, let's set a reference system with an axis parallel to the plane and gold perpendicular axis. Let's break down the weight (W)
sin52 = Wx / W
cos52 = Wy / W
Wx = W sin52
Wy = w cos 52
Let's write them equations
X axis
Wx = ma
Y Axis
N-Wy = 0
N = Wy
a) Let's calculate the acceleration
a = W sin52 / m = mg sin 52 / m
a = g sin 52
a = 9.8 sin52
a = 7.72 m / s²
The force of the ramp is normal
N = Wy = mg cos 52
N = 3.3 9.8 cos 52
N = 19.9 N
b) For the block to move at constant speed the sum of force on the axis must be zero,
F - Wx = 0
F = Wx
F = mg sin52
F = 3.3 9.8 sin 52
F = 25.5 N
Parallel to the plane and going up
Answer:
unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]
Explanation:
Given:
v = (-23.2, -104.4, 46.4) m/s
Above expression describes spacecraft's velocity vector v.
Find:
Find unit vector in the direction of spacecraft velocity v.
Solution:
Step 1: Compute magnitude of velocity vector.
mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)
mag (v) = 116.58 m/s
Step 2: Compute unit vector unit (v)
unit (v) = vec (v) / mag (v)
unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58
unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]
Answer:
Option (C)
Explanation:
Einsteinium is an element of the periodic table grouped in the Actinide series, with atomic number 99. They are dense element and highly electro-positive. <u>They are highly radioactive</u>, i.e the atoms within the element are unstable and constantly decay until they reach a stable environment. It has 99 number of electrons and protons, 153 number of neutrons.
Due to its high radioactivity, they are health hazardous and can used in making nuclear weapons but their uses are very limited and unknown.
Thus, the correct answer is option (C).
Answer:
8.75
Explanation:
First, find the force of friction.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.9 m/s)² = F (1.4 m)
F = 11.7 N
Next, find the distance at the new velocity.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d
d = 8.75 m
If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.
Given,
Tension force in the rope is (T) = 160 N
Displacement of the skier (S) = 270 m
The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.
Therefore, work done by the rope on the skier is,
⇒
Hence work done by the rope is - 43200 J.
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