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Kitty [74]
3 years ago
7

if a train starts from rest and attains a velocity of 100m/s in 25 seconds. calculate the acceleration produced by the train.​

Physics
1 answer:
stira [4]3 years ago
5 0
-4 km/s2

Explanation:
0-100/25
-100/25
-4
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Which one of the following is correct option: hydrogen,
Vinil7 [7]

Answer:

hydrogen

Explanation:

pH is a measure of the ion concentration of hydrogen in a solution.

In chemistry, it literally means power of hydrogen ions and it is a measure of the molar concentration of hydrogen ions in a particular solution, thus specifying the acidity, neutrality or basicity of chemical solutions.

Mathematically, the pH of a solution is given by pH = -log 10[H^{+}]

Hence, a solution with a pH of 7 is neutral.

Also, a solution with a pH below 7 is acidic but basic if it's pH is above 7.

3 0
3 years ago
three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a
valentinak56 [21]

Answer:

q_1=+0.375\ {10}^{-9}

Explanation:

Electrostatic Forces

The force exerted between two point charges q_1 and q_2 separated a distance d is given by Coulomb's formula

\displaystyle F=\frac{k\ q_1\ q_2}{d^2}

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

\displaystyle q_3=+5\ 10^{-9}\ c

\displaystyle q_2=-3\ 10^{-9}\ c

The distance between q_3 and q_2 is

\displaystyle d_2=4cm=0.04\ m

The distance between q_3 and q_1 is

\displaystyle d_1=2cm=0.02\ m

We must find the value of q_1 such that

\displaystyle |F_3|=0

Applying Coulomb's formula for q_1 is

\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}

Now for q_2

\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}

If the total force on q_3 is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

\displaystyle F_{13}=F_{23}

\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}

Simplfying and solving for q_1

\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}

\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}

\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}

5 0
3 years ago
Which statement is true of forces?
jasenka [17]
FORCES HAVE  STRENGTH AND DIRECTION 
8 0
3 years ago
An AC voltage source and a resistor are connected in series to make up a simple AC circuit. If the source voltage is given by ΔV
ra1l [238]

Answer:

t = 5.48 × 10⁻³ s

Explanation:

Given:

ΔV = ΔVmax × sin(2πft)

frequency, f = 16.9Hz

thus,

ΔV = ΔVmax × sin(2π×16.9×ft)

Now,

Let 'R' be the resistance

Also according to the ohms law

i = V/R

where,

i = current

V = voltage

hence,

i=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}

also, given at time 't' the current in the circuit is 55.0% of the peak current

thus

i=\frac{55}{100}\times \frac{\Delta V_{max}}{R}=0.55\times \frac{\Delta V_{max}}{R}

thus,

0.55\times \frac{\Delta V_{max}}{R}=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}

or

0.55=sin(2\pi \times 16.9\times t)}

or

0.5823=(2\pi \times 16.9\times t)}

or

t = 5.48 × 10⁻³ s (Answer)

8 0
3 years ago
Use your calculator to evaluate -3.7 meter/second-13.9 meter/second 21.4 second-72 second
cluponka [151]

Answer :

(-3.7 meter/second) - (13.9 meter/second) = -17.6 meter/second

(21.4 second) - (72 second) = -50.6 second

Explanation :

(1) As we are given the expression :

(-3.7 meter/second) - (13.9 meter/second)

Now we have to evaluate this expression, we get:

⇒ -17.6 meter/second

(2) As we are given the expression :

(21.4 second) - (72 second)

Now we have to evaluate this expression, we get:

⇒ -50.6 second

6 0
3 years ago
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