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mario62 [17]
2 years ago
11

Will mark brainlest helpp!!!!!!​

Physics
2 answers:
cestrela7 [59]2 years ago
7 0
  • initial velocity=u=0m/s
  • Final velocity=60km/h=v

Convert to m/s

\\ \Large\sf\longmapsto v=60\times \dfrac{5}{18}=16.6m/s

  • Time=t=5min=5×60=300s

We know

\boxed{\Large{\sf Acceleration=\dfrac{v-u}{t}}}

\\ \Large\sf\longmapsto Acceleration=\dfrac{16.6-0}{300}

\\ \Large\sf\longmapsto Acceleration=\dfrac{16.6}{300}

\\ \Large\sf\longmapsto Acceleration=0.05m/s^2

  • Distance=s

Using second equation of kinematics

\boxed{\Large{\sf s=ut+\dfrac{1}{2}at^2}}

\\ \Large\sf\longmapsto s=0(300)+\dfrac{1}{2}(0.05)(300)^2

\\ \Large\sf\longmapsto s=0.05\times 150^2

\\ \Large\sf\longmapsto s=0.05\times 22500

\\ \Large\sf\longmapsto s=1125m

Luda [366]2 years ago
6 0

Answer:

I don't know if it is correct or not.

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(10%) Problem 5: The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass
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Incomplete Question.The Complete question is

The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants.  Mass of the Earth: 5.97 × 10^24  kg (assume a uniform mass distribution)  Radius of the Earth: 6371 km  Distance of Earth from Sun: 149,600,000 km

(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.

(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?  

Answer:

(i) KE= 2.56e29 J

(ii) KE= 2.65e33 J

Explanation:

i) Treating the Earth as a solid sphere, its moment of inertia about its axis is

I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²

I = 9.69e37 kg·m²

About its axis,

ω = 2π rads/day * 1day/24h * 1h/3600s

ω= 7.27e-5 rad/s,

so its rotational kinetic energy

KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²

KE= 2.56e29 J

(ii) About the sun,

I = mR²

I= 5.97e24kg * (1.496e11m)²

I= 1.336e47 kg·m²

and the angular velocity

ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s

ω= 1.99e-7 rad/s

so  

KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²

KE= 2.65e33 J

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The Force of friction between an object and the surface upon which it is sliding is 12N. The weight of the object is 20N. What i
shtirl [24]

Answer: The coefficient of kinetic friction is μ = 0.6

Explanation:

For an object of mass M, the weight is:

W = M*g

where g is the gravitational acceleration: g = 9.8m/s^2

And the friction force between this object and the surface can be written as:

F = W*μ

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In this case, the weight is:

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And the friction force is:

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Replacing these values in the equation for the friction force we get:

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7 0
3 years ago
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Answer:

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