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Levart [38]
3 years ago
11

A 60 kg pupil runs for 600m in 1 minute uniformly calculate kinetic energy​

Physics
1 answer:
AURORKA [14]3 years ago
8 0

velocity = traveled distance ÷ time of the traveled distance is seconds

velocity = 600 ÷ 60

velocity = 10 m/s

_________________________________

Kinetic Energy = 1/2 × mass × ( velocity )^2

KE = 1/2 × 60 × ( 10 )^2

KE = 30 × 100

KE = 3000 j

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An objects mechanical energy never changes it simply changes between
tatiyna
Actually moving and not. It is the sum of potential and kinetic energy.
4 0
3 years ago
A 3.2-kg thin, circular hoop with a radius of 5.4 m is rotating about an axis through its center and perpendicular to its plane.
elena-14-01-66 [18.8K]

Answer:

Torque = –207.4 Nm

Explanation:

Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)

Torque = I × α

α = angular acceleration

I = moment of inertia

I = MR² for a circular hoop

Torque = 3.2×5.4×(– 12)

Torque = –207.4 Nm

6 0
3 years ago
Read 2 more answers
A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object
Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

    Em_{f} = U = m g h

Initial mechanical energy, in the lower part of the track

    Em₀ = K = ½ m v²

    Em=   Em_{f}

    ½ m v² = m g h

    v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and  final velocity v

Initial before the crash

    p₀ = M v₁₀ + 0

Final after the crash

      p_{f} = M v1f + m v

   p₀ =   p_{f}

   M v₁₀ = M v_{1f}+ m v

As the shock is elastic the kinetic energy is conserved

     K₀ = K_{f}

    ½ M v₁₀² = ½ M v_{1f}² + ½ m v²

Let's write the system of equations

    M v₁₀ = M  v_{1f} + m v

    M v1₁₀² = M v_{1f}² + m v²

We cleared v1f in the first we replaced in the second

   v_{1f} = (M v₁₀ - mv) / M

    M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

    M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

     v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

     2 m v₁₀ - v (m + 1) m/ M = 0

     v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

     v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

    v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2  5.41)

    V₁₀ = 7.668 (2.68) / 10.82

   v₁₀ = 1.90 m / s

5 0
3 years ago
Watt (w) is a drived unit why​
ValentinkaMS [17]

Answer:

because it is from a mathematical combination of SI base units

Explanation:

4 0
3 years ago
How many seconds of space should you keep between you if you're driving a 100 foot truck 30 mph?
Marrrta [24]

Answer:

<em> The space in seconds that will be kept = 2.27 seconds</em>

Explanation:

S = d/t..................... Equation 1

making t the subject of formula in the equation above,

t = d/S.................... Equation 2

Where S = speed, d = distance, t = time.

<em>Conversion: (i)if 1 mph = 0.44704 m/s,</em>

<em>                 then, 30 mph = 30×0.44704    </em>

<em>                = 13.41 m/s</em>

<em>               (ii) If 1 foot = 0.3048 m</em>

<em>            then, 100 foot = 30.48 m.</em>

<em>Given: S = 30 mph = 13.41 m/s, d = 100 foot = 30.48 m</em>

<em>Substituting these values into equation 2</em>

<em>t = 30.48/13.41</em>

<em>t = 2.27 seconds.</em>

<em>Therefore the space in seconds that will be kept = 2.27 seconds</em>

7 0
3 years ago
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