find the ratio of the pressure exerted by a block of 200 and when placed on the top of the table along its two different sides with dimensions 20 centimetre x 15 centimetre and 30 centimetre x 15 centimetre ??
Answers
Answered by
1
i am answering the question if the mass is 200 kg ok because u have not mentioned in the question....
p1 = 200kg × 10ms^-1/ (0.2m × 0.15m)
–– –––––––––––––––––––––––––
p2 = 200kg × 10 ms^-1/(0.3m × 0.15m)
p1/p2 = 0.3 / 0.2
p1/p2 = 3:2.
p1 = 200kg × 10ms^-1/ (0.2m × 0.15m)
–– –––––––––––––––––––––––––
p2 = 200kg × 10 ms^-1/(0.3m × 0.15m)
p1/p2 = 0.3 / 0.2
p1/p2 = 3:2.
Answered by
2
Force= pressure*area
Also
Force=mass*gravitational force
Where,mass=200(gms)(say)
g=9800 cm/sec^2
Area1= area of dimensions 1
=20*15=300sq.cm
Area2=area of dimensions 2
=30*15=450sq.cm
=>200*9800=p1*area1
=>1960000=P1*300------(1)
Now, similarly,
=>200*9800=P2*450-----(2)
Comparing (1) and (2),
P1*300=P2*450
=>P1:p2=450:300
=> P1:p2=3:2
Also
Force=mass*gravitational force
Where,mass=200(gms)(say)
g=9800 cm/sec^2
Area1= area of dimensions 1
=20*15=300sq.cm
Area2=area of dimensions 2
=30*15=450sq.cm
=>200*9800=p1*area1
=>1960000=P1*300------(1)
Now, similarly,
=>200*9800=P2*450-----(2)
Comparing (1) and (2),
P1*300=P2*450
=>P1:p2=450:300
=> P1:p2=3:2
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