The acceleration of the box up the ramp is 9.65 m/s².
<h3>
What is the magnitude of acceleration of the box?</h3>
The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;
F(net) = ma
where;
- m is the mass of the box
- a is the acceleration of the box
The net force on the box is calculated as follows;
F(net) = F - Ff
F(net) = F - μmgcosθ
where;
- θ is the inclination of the plane
- μ is coefficient of friction
F(net) = 170 - (0.3 x 15 x 9.8 x cos55)
F(net) = 144.7
The acceleration of the box is calculated as;
a = F(net) / m
a = (144.7) / (15)
a = 9.65 m/s²
Thus, the acceleration of the box up the ramp is 9.65 m/s².
Learn more about acceleration here: brainly.com/question/14344386
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The position of the centre of gravity of an object affects its stability. The lower the centre of gravity (G) is, the more stable the object. The higher it is the more likely the object is to topple over if it is pushed. Racing cars have really low centres of gravity so that they can corner rapidly without turning over.
Increasing the area of the base will also increase the stability of an object, the bigger the area the more stable the object. Rugby players will stand with their feet well apart if they are standing and expect to be tackled.
Answer:
4
Explanation:
We are given that
K.E at x=0 m=20 J
K.E at x=3 m=11 J
We have to find the value of c.
By work energy theorem
Work done=Change in kinetic energy
W=
![W=[\frac{cx^2}{2}-x^3]^{3}_{0}](https://tex.z-dn.net/?f=W%3D%5B%5Cfrac%7Bcx%5E2%7D%7B2%7D-x%5E3%5D%5E%7B3%7D_%7B0%7D)
<span>The magnitude of a is 1.5
The magnitude of b is -3
The magnitude of the vector is
</span>√(1.5² + (-3)²) = 3.35
<span>The angle is
</span>θ = tan⁻¹ (-3/1.5) = 63.43°
<span>The vector is drawn with a magnitude of 3.35 and an angle of 63.43</span>°.
Answer: 21 N
Explanation:
Force=mass x Accceleration
So 21=3x7
