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Tanya [424]
3 years ago
9

A car increased its velocity to 62m/s in 10s starting from rest. Calculate the distance it covers during this time?

Physics
2 answers:
Roman55 [17]3 years ago
6 0

Answer:

distance= velocity ×time

distance= 62×10

distance=620m

hope it helps you mate please mark me as brainliast

rjkz [21]3 years ago
4 0
  • Answer:

<em>d = 620 m</em>

  • Explanation:

<em>Hi there !</em>

<em><u>Velocity formula</u></em>

<em>v = d/t => d = v×t</em>

<em>d = 62m/s×10s</em>

<em>= 620 m</em>

<em>Good luck !</em>

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A seagull flies at a velocity of 9.00 m/s straight into the wind.
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a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity,\rm v_{SA}  = 9 \ m/sec

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec

Air velocity with reference to the ground is;

\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec

The time the bird takes;

\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7  \ min \  and  \ 42  \ sec

c)\

The total round-trip time compared to what it would be with no wind. is;

\rm  t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec

The time for the round trip is;

\rm  t = \frac{12 \times 10^ 3 }{ 9 \ m/sec }  \\\\ t  = 1333.33 \ sec

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be  4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

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A rocket burns fuel to create hot gases that explode violently out of the rocket engine. This explosion creates thrust. Thrust i
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Answer:

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Explanation:

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Read 2 more answers
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diamong [38]

Answer:

a= 23.65 ft/s²

Explanation:

given

r= 14.34m

ω=3.65rad/s

Ф=Ф₀ + ωt

t = Ф - Ф₀/ω

= (98-0)×\frac{\pi}{180}/3.65

98°= 1.71042 rad

1.7104/3.65

t= 0.47 s

r₁(not given)

assuming r₁ =20 in

r₁ = r₀ + ut(uniform motion)

u = r₁ - r₀/t

r₀ = 14.34 in= 1.195 ft

r₁ = 20 in = 1.67 ft

= (1.667 - 1.195)/0.47

0.472/0.47

u= 1.00ft/s

acceleration at collar p

a=rω²

= 1.67 × 3.65²

a = 22.25ft/s²

acceleration of collar p related to the rod = 0

coriolis acceleration = 2ωu

= 2× 3.65×1 = 7.3 ft/s²

acceleration of collar p

= 22.5j + 0 + 7.3i

√(22.5² + 7.3²)

the magnitude of the acceleration of the collar P just as it reaches B in ft/s²

a= 23.65 ft/s²

4 0
3 years ago
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