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____ [38]
2 years ago
7

A bus starts from village A at 7:00 am and reaches village B at 8:30 am. If the distance between the villages A and B is 60 km.

Find the average speed of the bus​
Physics
2 answers:
aev [14]2 years ago
6 0

Answer:

The answer is 40 km/h

Explanation:

<h3><u>Given</u>;</h3>
  • distance (s) covered by the bus = 60 km
<h3><u>To</u><u> Find</u>;</h3>
  • The average speed of the bus
<h3><u>Formula</u>;</h3>
  • Average velocity (v) = Distance ÷ Time

Now,

Time (t) taken = 8:30 am – 7:00 am = 1 ½ hour = 3/2 hour

Here,

v = s/t

v = 60 ÷ 3/2

v = 60 × 2 ÷ 3

v = 20 × 2

v = 40 km/h

Thus, The average speed of the bus is 40 km per hour

 

<u>-TheUnknownScientist 72</u>

SVEN [57.7K]2 years ago
5 0

1.1111111111

Explanation:

v=s/t

t=8:30-7:00=1:30

convert to second=5400second

s=60km covert to meter=6000

6000/5400=1.11111111

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A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel
larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\      (1)

R1 =  13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}            (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

R_{eq}=\frac{23V}{4A}=5.75\Omega

Now, you can calculate the unknown resistor R2 by using the equation (1):

\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega

hence, the resistance of the unknown resistor is 10.31Ω

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Answer:

1. Largest force: C;  smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

F=K\dfrac{ q_{1}q_{2}}{r^{2}}

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write  

F=\dfrac{k}{r^{2}}

1. Net force on each particle

Let's

  • Call the distance between adjacent charges d.
  • Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(b) Force on B

\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}}  - \dfrac{k}{d^{2}}  + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(C) Force on C

\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(d) Force on D

\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

F_{A} : F_{B} : F_{C} : F_{D}  =  \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\

2. Ratio of largest force to smallest

\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}

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