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PtichkaEL [24]
2 years ago
8

a motorcycle begins at rest and accelerates uniformly at 7.9 m/s2. We want to find the time it takes the motorcycle to reach a s

peed of 100 km/h. Which kinetic formula would be most useful to solve for the target unknown?
Physics
2 answers:
gladu [14]2 years ago
6 0

Answer:

It takes the motorcycle about 12.65 seconds to reach 100mph

Explanation:

AURORKA [14]2 years ago
3 0

Answer:3.5 seconds

Explanation:

initial velocity=u=0

Final velocity=v=100km/h

Final velocity=v=(100x5)/18=500/18=27.8 m/s

Acceleration=a=7.9m/s^2

t=(v-u)/a

t=(27.8-0)/7.9

t=27.8/7.9

t=3.5 seconds

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The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the
otez555 [7]

Answer:

0.647 nC

Explanation:

The force experienced by a charge due to the presence of an electric field is given by

F=qE

where

q is the charge

E is the magnitude of the electric field

In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

F=5.50\mu N = 5.50\cdot 10^{-6} N

Therefore, if the electric field magnitude is

E=8.50 kN/C = 8500 N/C

Then the charge on each antenna would be

q=\frac{F}{E}=\frac{5.50\cdot 10^{-6} N}{8500 N/C}=6.47\cdot 10^{-10} C = 0.647 nC

8 0
2 years ago
Miners find a rock that was formed from magma deep underground what type of rock could it be
andreyandreev [35.5K]
It can be an igneous rock
3 0
2 years ago
f a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance D40 compared to the stopping distance
Oksana_A [137]

Answer:

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

Explanation:

given data

speed = 40 miles / hour

distance = D40

speed limit = 25 miles / hour

distance = D25

to find out

express number a multiple of stopping distance @ 25 mph

solution

we know here stopping distance is directly proportional to (speed)²

so here speed ratio is

initial speed = \frac{40}{25}

so initial speed = 1.6

so

stopping distance increase = (1.6)²

\frac{D40}{D25} = (1.6)²

\frac{D40}{D25} = 2.56

so here

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

5 0
2 years ago
By constructing basic electric circuits, you will be able to measure current flow. With Ohm’s law in mind, what broad question a
andrezito [222]

Broad questions that we can answer by doing this experiment are <em><u>the effects of electric current</u></em>

<h3>Further explanation</h3>

Electric current is the amount of electric charge that flows each unit of time

Electric current occurs due to the movement of electrons due to the difference in potential or voltage (from high potential to low potential) between two points

Electrons will flow through the conducting wire that functions as a conductor

Ohm's Law states that

<em>The potential difference is proportional to the electric current as long as the resistance is constant</em>

\displaystyle I=\frac{V}{R}\\\\V=I\times R

A simple electrical circuit consists of a voltage source (battery) and a lamp

Ampermeters to measure the strength of the current, must be installed in series with the load to be measured

By changing the voltage source, with constant resistance from a conductor, different current measurements will be obtained. The greater the voltage, the greater the resulting current.

<h3>Learn more</h3>

electrons flow through the device

brainly.com/question/4438943

Keywords : basic electric circuits, Ohm’s law, experiment

5 0
2 years ago
Read 2 more answers
In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P
tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites r_{sat1} = 48000 \times 10^{3} m

Distance of second satellites r _{sat2} = 64000 \times 10^{3} m

Distance of charon r_{c} = 19600 \times 10^{3} m

Time period of charon T_{c} = 6.39 days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

   T^{2} = r^{3}

   \frac{T}{r^{\frac{3}{2} } } = constant

For first satellites,

  \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} }  }

{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat1} = 24.46 days

For second satellites,

   \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} }  }

{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat2} = 37.67 days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0
2 years ago
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