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Bogdan [553]
3 years ago
11

a train starting from rest moving with uniform acceleration attains a speed of 36 km/hr in 10 seconds. Find its acceleration​

Physics
1 answer:
Savatey [412]3 years ago
8 0

Answer:

\boxed{\sf Acceleration \ of \ train = 1 \ m/s^{2}}

Given:

Initial speed (u) = 0 km/hr = 0 m/s

Final speed (v) = 36 km/hr

Time taken (t) = 10 sec

To Find:

Acceleration (a) of the train

Explanation:

\sf 1 \ km/hr = \frac{5}{18} \ m/s \\ \therefore \\ \sf 36 \ km/hr = 36 \times \frac{5}{18} \ m/s \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    = 2 \times  \cancel{18} \times  \frac{5}{ \cancel{18}}  \ m/s \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \: = 2 \times 5 \ m/s \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 10  \ m/s

So,

Final speed (v) = 10 m/s

\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ t: \\ \sf \implies 10 = 0 + a(10) \\ \sf \implies 10 = a(10) \\ \sf \implies a \times 10 = 10 \\ \\ \sf Dividing \ both \ sides \ by \ 10: \\ \sf \implies \frac{a \times 10}{\boxed{\sf 10}} = \frac{10}{\boxed{\sf 10}} \\ \\ \sf \frac{\cancel{10}}{\cancel{10}} = 1: \\ \sf \implies a = 1 \:  m/s^2

Acceleration of train = 1 m/s²

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BARSIC [14]

Answer:

D = 43 m

Explanation:

given,

initial velocity = 18 m/s

angle θ = 60°

total horizontal distance covered by the shell is

R = \dfrac{v_0^2sin 2\theta}{g}

applying conservation of momentum in horizontal direction

m v₀ cos θ = m₁v₁ + m₂ v₂

m v₀ cos θ = 0.5 m v₂

v₂ = 2 v₀ cos θ.

distance covered by the shell from point of explosion

R' = v t

  = (2 v_0 cos \theta) (\dfrac{v_0^2sin \theta}{g})

  =(2 \dfrac{v_0^2cos \theta sin \theta}{g})

  = \dfrac{v_0^2sin 2\theta}{g}

  = R

total distance traveled by the shell is

D = \dfrac{R}{2}+R'

   = 1.5 R

   = 1.5\dfrac{v_0^2sin 2\theta}{g}

D = 1.5\dfrac{18^2sin 2\times 60}{9.81}

   = 42.9 ≅ 43 m

D = 43 m

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3 years ago
Hello! Please I need help on 1. (C) I dont know the formula to get the Total height using Mechanical Energy
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4.00m

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Which subatomic particle is NOT found in the nucleus of an atom? *
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3 years ago
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2. Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are +7.0μC, -8.0 μC
Scorpion4ik [409]

Answer:

0.53 N, 25.6°

Explanation:

side of triangle, a = 1.2 m

q = 7 μC

q1 = - 8 μC

q2 = - 6 μC

Let F1 be the force between q and q1

By using the coulomb's law

F_{1}=\frac{Kq_{1}q}{a^{2}}

F_{1}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 8\times 10^{-6}}{1.2^{2}}

F1 = 0.35 N

Let F2 be the force between q and q2

By using the coulomb's law

F_{2}=\frac{Kq_{2}q}{a^{2}}

F_{2}=\frac{9\times 10^{9}\times 7\times 10^{-6}\times 6\times 10^{-6}}{1.2^{2}}

F2 = 0.26 N

Write the forces in the vector form

\overrightarrow{F_{1}}=0.35\widehat{i}

\overrightarrow{F_{2}}=0.26\left ( Cos60 \widehat{i}+Sin60\widehat{j}\right )

\overrightarrow{F_{2}}=0.13 \widehat{i}+0.23\widehat{j}

Net force

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}=0.48 \widehat{i}+0.23\widehat{j}

Magnitude of the force

F=\sqrt{0.48^{2}+0.23^{2}}

F = 0.53 N

Direction of force with x axis

tan\theta =\frac{0.23}{0.48}

θ = 25.6°

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