Answer:
(35 N - 10 N)/8kg = 3.125 m/s^2
Explanation:
The formula for Force is:
Force = Mass*Acceleration
(Force is equal to Mass times Acceleration)
Since we're told to find the acceleration of the box. We make acceleration the subject of the equation:
Acceleration = Force/Mass
(Acceleration equal to Force divided by Mass)
We know that the force are 35 N forward and 10 N backward, and the weight of the box is 8kg.
= (35 N - 10 N)/8kg
The reason that 35 N minus 10 N is because the 10 N is pushing the box backward.
= 25 N/8kg
= 3.125 m/s^2
Hope it helps :DD
Main sequence stars are characterised by the source of their energy.They are all undergoing fusion of hydrogen into helium within their cores. The mass of the star is the main element for such process or phenomenon to take place for it is a determinant of both the rate at which they perform the said activity and the amount of fuel available.
To answer the question, the lower mass limit for a main sequence star is about 0.08. If the mass of a main sequence star is lower than the above-mentioned value, there would be a deficit or insufficiency of gravitational force to generate a standard temperature for hydrogen core fusion to take place and the underdeveloped star would form into a brown dwarf instead.
The largest resultant amplitude would be that created by constructive interference, basically when the two waves are of the same phase, so it would be 0.36m+0.22m= 0.58 m.
Answer:
True.
Explanation:
If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero.
The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.
Hence, the given statement is true.
The work done by the applied force on the block against the frictional force is 15.75 J.
<h3>
Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
Learn more about work done here: brainly.com/question/25573309
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