A 50.0-kg block is being pulled up a 13.0° slope by a force of 250 N which is parallel to the slope, but the block does not slid
1 answer:
<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>
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Answer:
Net electric field,
Explanation:
Given that,
Charge 1,
Charge 2,
distance, d = 3.2 cm = 0.032 m
Electric field due to charge 1 is given by :
Electric field due to charge 2 is given by :
The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :
So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.