<span>Weight of block, Wb = mass*gravity = 50*9.8 = 490 N</span>
Since block is being pulled up by a 13-degree slope
Therefore, Force which is acting parallel to the slop:
<span> F p =490 Sin </span>= 110.2N
Force which is acting perpendicular to the slope:
<span> Fv =490 Cos</span> = 477.4 N
Net force can be given as follows:
<span>F n = (250 - 110.2 - 0.2*</span>477.4) N
<span>Fn=44.3N</span>
Now acceleration is given by the ratio of force to mass
<span>a = Fn/m</span>
<span>=44.3/50 = 0.89 ms^<span>-2</span></span>
Uhh? Do you have any questions or need help?
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The question appears to be incomplete.
I assume that we are to find the coefficient of static friction, μ, between the desk and the book.
Refer to the diagram shown below.
m = the mass of the book
mg = the weight of the book (g = acceleration due to gravity)
N = the normal reaction, which is equal to
N = mg cos(12°)
R = the frictional force that opposes the sliding down of the book. It is
R = μN = μmg cos(12°)
F = the component of the weight acting down the incline. It is
F = mg sin(12°)
Because the book is in static equilibrium (by not sliding down the plane), therefore
F = R
mg sin(12°) = μmg cos(12°)
Therefore, the static coefficient of friction is
μ = tan(12) = 0.213
Answer: μ = 0.21 (nearest tenth)
Explanation:
Let is the frequency of oscillation of a block that undergoes simple harmonic motion. The frequency of oscillation is given by :
...............(1)
Where
k is the spring constant of the spring
m is the mass of the block
It is clear from equation (1) that the frequency of the oscillation depends on the spring constant and the mass of the block. So, on increasing the spring constant, the block oscillate with a greater frequency.