Show the following inequalities on a number line. (i) x > 4 and x < 9 (ii) x _< –1
Answers
Answer:
Answer:
Given :-
A mass of 1000 kg having a speed of 10 m/s is brought to rest over a distance of 100 m.
To Find :-
What is the retardation.
What is the retardation force.
Formula Used :-
\clubsuit♣ Third Equation Of Motion Formula :
\longrightarrow \sf\boxed{\bold{\pink{v^2 - u^2 =\: 2as}}}⟶v2−u2=2as
where,
v = Final Velocity
u = Initial Velocity
a = Acceleration
s = Distance Covered
\clubsuit♣ Force Formula :
\longrightarrow \sf\boxed{\bold{\pink{F =\: ma}}}⟶F=ma
where,
F = Force
m = Mass
a = Acceleration
Solution :-
First we have to find the retardation :
Given :
Final Velocity = 0 m/s
Initial Velocity = 10 m/s
Distance Covered = 100 m
According to the question by using the formula we get,
\implies \sf (0)^2 - (10)^2 =\: 2a(100)⟹(0)2−(10)2=2a(100)
\implies \sf (0 \times 0) - (10 \times 10) =\: 200a⟹(0×0)−(10×10)=200a
\implies \sf 0 - 100 =\: 200a⟹0−100=200a
\implies \sf - 100 =\: 200a⟹−100=200a
\implies \sf \dfrac{- 100}{200} =\: a⟹200−100=a
\implies \sf - 0.5 =\: a⟹−0.5=a
\implies \sf\bold{\purple{a =\: - 0.5\: m/s^2}}⟹a=−0.5m/s2
Now, we have to find the retardation :-
\leadsto \bf Retardation =\: - (Acceleration)⇝Retardation=−(Acceleration)
\leadsto \sf Retardation =\: - (- 0.5)⇝Retardation=−(−0.5)
\leadsto \sf\bold{\red{Retardation =\: 0.5\: m/s^2}}⇝Retardation=0.5m/s2
\therefore∴ The retardation is 0.5 m/s² .
Now, we have to find the retardation force :-
Given :
Mass = 1000 kg
Acceleration = 0.5 m/s²
According to the question by using the formula we get,
\dashrightarrow \sf Force =\: 1000 \times 0.5⇢Force=1000×0.5
\dashrightarrow \sf\bold{\red{Force =\: 500\: N}}⇢Force=500N
\therefore∴ The retardation force is 500 N .