Physics || Shashaank Blogspot

1 ) Two masses of $40 k g$ and $30 k g$ are connected by a weightless string passing over a friction less pulley as shown in the figure. The acceleration of the system would be:

Gravitation force on A, $F_{A} = m g sin 30 = 40 \times 10 \times 0.5 = 200 N$
Gravitation force on B, $F_{B} = M g sin 30 = 30 \times 10 \times 0.5 = 150 N$
$F_{n e t} = F_{A} - F_{B}$
$\Rightarrow (M + m) a = F_{A} - F_{B}$
$\Rightarrow a = \frac{F _{A} - F _{B}}{M + m} = \frac{2 0 0 - 1 5 0}{4 0 + 3 0} = 0.71 m s^{- 2}$
Hence, acceleration of the system is $0.71 m s^{-}^{2}$

Acceleration of the particle when its velocity becomes half of the initial velocity.

$V = - (\frac{2 0}{5 0}) x$
$V = \frac{- 2 x}{5}$
$a = \frac{d v}{d t} = \frac{- 2}{5} \frac{d x}{d t} = \frac{- 2 V}{5}$
$a = \frac{- 2 V}{5}$
initial velocity $(u) = 20 m / s$
$\frac{u}{2} = 10 m / s$
$a = \frac{- 2}{5} (10) = - 4 m / s^{2}$ .

A model airplane with mass $0.750 k g$ is tethered to the ground by a wire so that it flies in a horizontal circle $30.0 m$ in radius. The airplane engine provides a net thrust of $0.800 N$ perpendicular to the tethering wire.
Find the torque the net thrust produces about the center of the circle.

EnergeticTech

Search This Blog

Physics || Shashaank Blogspot

1 ) Two masses of $40 k g$ and $30 k g$ are connected by a weightless string passing over a friction less pulley as shown in the figure. The acceleration of the system would be:

Acceleration of the particle when its velocity becomes half of the initial velocity.

$V = - (\frac{2 0}{5 0}) x$
$V = \frac{- 2 x}{5}$
$a = \frac{d v}{d t} = \frac{- 2}{5} \frac{d x}{d t} = \frac{- 2 V}{5}$
$a = \frac{- 2 V}{5}$
initial velocity $(u) = 20 m / s$
$\frac{u}{2} = 10 m / s$
$a = \frac{- 2}{5} (10) = - 4 m / s^{2}$ .

A model airplane with mass $0.750 k g$ is tethered to the ground by a wire so that it flies in a horizontal circle $30.0 m$ in radius. The airplane engine provides a net thrust of $0.800 N$ perpendicular to the tethering wire.
Find the torque the net thrust produces about the center of the circle.

Labels

Comments

Popular posts from this blog

MY FIRST HTML CODE COVID 19 || Shashaank Blogspot

Lamborghini Cartoon

Data Handling || Shashaank Blogspot

EnergeticTech

Physics || Shashaank Blogspot

1 ) Two masses of 40 kg and 30 kg are connected by a weightless string passing over a friction less pulley as shown in the figure. The acceleration of the system would be:

Gravitation force on A, FA​=mgsin30=40×10×0.5=200NGravitation force on B, FB​=Mgsin30=30×10×0.5=150NFnet​=FA​−FB​⇒(M+m)a=FA​−FB​⇒a=M+mFA​−FB​​=40+30200−150​=0.71ms−2Hence, acceleration of the system is 0.71ms−2

Acceleration of the particle when its velocity becomes half of the initial velocity.

V=−(5020​)xV=5−2x​a=dtdv​=5−2​dtdx​=5−2V​a=5−2V​initial velocity (u)=20m/s2u​=10m/sa=5−2​(10)=−4m/s2.

A model airplane with mass 0.750 kg is tethered to the ground by a wire so that it flies in a horizontal circle 30.0 m in radius. The airplane engine provides a net thrust of 0.800 N perpendicular to the tethering wire.Find the torque the net thrust produces about the center of the circle.

Labels

Comments

Popular posts from this blog

MY FIRST HTML CODE COVID 19 || Shashaank Blogspot

Lamborghini Cartoon

Data Handling || Shashaank Blogspot

1 ) Two masses of $40 k g$ and $30 k g$ are connected by a weightless string passing over a friction less pulley as shown in the figure. The acceleration of the system would be:

$V = - (\frac{2 0}{5 0}) x$
$V = \frac{- 2 x}{5}$
$a = \frac{d v}{d t} = \frac{- 2}{5} \frac{d x}{d t} = \frac{- 2 V}{5}$
$a = \frac{- 2 V}{5}$
initial velocity $(u) = 20 m / s$
$\frac{u}{2} = 10 m / s$
$a = \frac{- 2}{5} (10) = - 4 m / s^{2}$ .

A model airplane with mass $0.750 k g$ is tethered to the ground by a wire so that it flies in a horizontal circle $30.0 m$ in radius. The airplane engine provides a net thrust of $0.800 N$ perpendicular to the tethering wire.
Find the torque the net thrust produces about the center of the circle.