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kozerog [31]
1 year ago
10

According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f

lash in miles. Assuming that the flash of light arrives in essentially no time at all, estimate the speed of sound in m/s from this rule. Express your answer to one significant figure and include the appropriate units. What would be the rule for kilometers?
Physics
1 answer:
Bond [772]1 year ago
3 0

Answer:

S_{s}=300 m/s

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

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m_a_m_a [10]

Answer:

F = 10.788 N

Explanation:

Given that,

Charge 1, q_1=6\ \mu C=6\times 10^{-6}\ C

Charge 2, q_2=2\ \mu C=2\times 10^{-6}\ C

Distance between charges, d = 0.1 m

We know that there is a force between charges. It is called electrostatic force. It is given by :

F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N

So, the force applied between charges is 10.788 N.

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2 years ago
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A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma
Nesterboy [21]

Answer:

T = g μ_s ( M+m )

78.4 N

Explanation:

When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .

Net force on the lower box

= T - mg μ_s = M a    ( a is the acceleration created by net force in M )

Considering force on the upper box

mg μ_s = ma

a = g μ_s

Put this value of a in the equation above

T - m gμ_s = M g μ_s

T = mg μ_s + M g μ_s

=  g μ_s ( M+m )

2 )

Largest tension required

T = 9.8 x  .50 x ( 10+6 )

= 78.4 N

5 0
2 years ago
Find the mass of a person walking west at a speed of 0.8 m/s with a momentum of 52.0 kg.m/s west.
KIM [24]

Answer:

mass of the person walking to west is 65 kg.

Given:

Momentum = 52 \frac{kg m}{s}

Speed = 0.8 \frac{m}{s}

To find:

Mass of the person = ?

Formula used:

Momentum is given by,

P = m × v

Where, P = momentum

m = mass

v = speed

Solution:

Momentum is given by,

P = m × v

Where, P = momentum

m = mass

v = speed

Mass = \frac{P}{v}

m = \frac{52}{0.8}

m = 65 kg

Thus, mass of the person walking to west is 65 kg.

5 0
1 year ago
A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.
Airida [17]

Answer:

Part A :  E =   \frac{1}{4\pi}ε₀ Q₁/R₁² Volt/meter

Part B :  V =  \frac{1}{4\pi}ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

<em>Part A</em>

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.  

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

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Then the electric field at that point is

                                   E =  F/1

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Part B

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Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

                           V =  \frac{1}{4\pi}ε₀ Q₁/R₁  Volt

4 0
1 year ago
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UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity.  If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 0
2 years ago
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