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
I think the answer is potential
Answer:
This values shows a right angle triangle
Explanation:
Given;
a vector 4.0 km due East
a 3.0 km due north
the resultant vector is 5.0 km
The resultant vector can be obtained by Pythagoras theorem if the vectors form a right angle triangle.
R² = 4² + 3²
R² = 16 + 9
R² = 25
R = √25
R = 5 km (right angle triangle proved)
Therefore, this values shows a right angle triangle
The choices are:
a. Normal Force
b. Gravity Force
c. Applied Force
d. Friction Force
e. Tension Force
f. Air Resistance Force
Answer:
The answer is letter e, Tension Force.
Explanation:
Force refers to the "push" and "pull" of an object, provided that the object has mass. This results to acceleration or a change in velocity. There are many types of forces such as <em>Normal Force, Gravity Force, Applied Force, Friction Force, Tension Force and Air Resistance Force.</em>
The situation above is an example of a "tension force." This is considered the force that is being applied to an object by strings or ropes. This is a type force that allows the body to be pulled and not pushed, since ropes are not capable of it. In the situation above, the tension force of the rope is acting on the bag and this allows the bag to be pulled.
Thus, this explains the answer.
Explanation:
Given Data:
mass of dog = 12 Kg
dog's center of mass = 0.20m
length of dog = 0.50m
height of dog's jump = ?
Solution:
Work done of gravitational force = Gain in Potential energy
2.1 × mgΔh = mg (h - 0.1)
2.1 × (0.3 - 0.1) = (h - 0.1)
h = 0.52 m
