Question

solve this questions plz

Answer 1

1.
Force along horizontal = FCos60
= 72 × 1/2
= 36 dyne

Acceleration along horizontal = Force along horizontal / Mass
= 36 dyne / 9 g
= 4 cm/s^2

Answer 2

hello,
1st has been answered
(2)
F=ma
1=5a  (according to the question)
a=1/5m/s²
now using equations of motion
1st equation
v=u+at
body starts from rest hence,u=0
v=at
v=1/5×10
v=2m/s²
now,3rd equation to find the distance
2aS=v²-u²
2·1/5·S=2²-0
2/5S=4
S=10m

(3)
F=ma
15=150/1000×a  (SI Unit is kg)
15×100/15=a
100=a
hence a=100m/s²

(4)
F=ma
a=(v-u)/t   (1st equation of motion)
here u=0  (see the language of the question)
a=v/t
a=20/5
a=4m/s²
F=100N
100=m4
m=25
therefore,m=25kg

(5)
the first force is 5√2N at 60°
Fcosθ=Fcos60°=5√2×1/2
=5√2/2N
second force is 6√2N at 60°
force=6√2/2N
total force=11√2/2N
F=ma
11√2/2=1000a
a=5.5√2/1000
a=0.055√2m/s²

(6)
we know that 
F=ma
F=(m₁+m₂)a
F=(2+3)a
5=5a
a=1m/s²
now (a)bigger block m₂
F₂=m₂a   (acceleration will be same)
F₂=3×1
F₂=3N
force on area of contact when force is applied on bigger box is 3N
(b)small box
F₁=m₁a
F₁=2×1
F₁=2N

(7)
(a)when the elevator is descending with constant velocity there will be no change in acceleration as it will remain 0
hence weight read by scale will 72.2kg×10=722N
(b)when the elevator is ascending with acceleration of 3.2m/s²,the weight will increase as it more acceleration will be there than the normal i.e acceleration becomes 10+3.2=13.2m/s²
therefore,F=72.2×13.2=953.04N
hope this helps,if u like please mark it as brainliest