Question

Express 0.1254 BAR in P/Q form.

Answer 1

Step-by-step explanation:

Acceleration = 2 m/s²

Force = 14 kN

Given

A truck starts from rest and rolls down the hill with Constant acceleration

The truck Covers 400 m in 20 Second

To Find

Acceleration

Force acting on it if it's mass is 7 metric tonnes (1 metric tone=1000 kg)

Solution

Initial velocity , u = 0 m/s

[ ∵ start's from rest ]

Distance , s = 400 m

Time , t = 20 s

Apply 2nd equation of motion ,

\begin{gathered}\pink{\bigstar}\ \; \bf s=ut+\dfrac{1}{2}at^2\\\\\to \rm 400=(0)(20)+\dfrac{1}{2}a(20)^2\\\\\to \rm 400=0+200a\\\\\to \rm 200a=400\\\\\to \bf a=2\ m/s^2\ \; \pink{\bigstar}\end{gathered}

★ s=ut+

2

1

at

2

→400=(0)(20)+

2

1

a(20)

2

→400=0+200a

→200a=400

→a=2 m/s

2

★

Now ,

mass , m = 7 metric tonnes

⇒ m = 7 × ( 1000 ) kg

⇒ m = 7000 kg

Acceleration , a = 2 m/s²

Apply Newton Second Law ,

\begin{gathered}\pink{\bigstar}\ \; \bf F=ma\\\\\to \rm F=(7000)(2)\\\\\to \rm F=14000\ N\\\\\to \rm F=14 \times 10^3\ N\\\\\to \bf F=14\ kN\ \; \pink{\bigstar}\end{gathered}

★ F=ma

→F=(7000)(2)

→F=14000 N

→F=14×10

3

N

→F=14 kN ★

Answer 2

Answer:

Hope it helps u...

Step-by-step explanation:

0.1254 bar= 1254-0/9

= 1254/9