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son4ous [18]
3 years ago
15

Define work done against friction(helpp)​

Physics
1 answer:
77julia77 [94]3 years ago
7 0

Answer:

The work done against friction is the work done on an object that overcomes this frictional force allowing the object to move - it doesn't include any extra work used to accelerate an object for example - only the work used to beat the frictional force.

Explanation:

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A tree is fixed relative to Earth a tree is blank relative to the Sun
RUDIKE [14]

moving,

the sun is a different "observer"

8 0
3 years ago
A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3
Advocard [28]

Answer:

r = 5.07 m

Explanation:

given,

velocity of the man , v = 3.43 m/s

centripetal acceleration, a = 2.32 m/s²

magnitude of position of = ?

using centripetal acceleration formula

a_c =\dfrac{v^2}{r}

2.32 =\dfrac{3.43^2}{r}

r =\dfrac{3.43^2}{2.32}

   r = 5.07 m

The magnitude of the position vector relative to rotational axis is equal to 5.07 m.

5 0
3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
2 years ago
How fast does a 2 MeV fission neutron travel through a reactor core?
Artemon [7]

Answer:

The answer is 1.956 \times 10^7\ m/s

Explanation:

The amount of energy is not enough to apply the relativistic formula of energy E = mc^2, so the definition of energy in this case is

E = \frac{1}{2}m v^2.

From the last equation,

v= \sqrt{2E/m}

where

E = 2 MeV = 3.204 \times 10^{-13} J

and the mass of the neutron is

m = 1.675\times 10^{-27}\ Kg.

Then

v = 1.956 \times 10^7\ m/s

the equivalent of 0.065 the speed of light.

5 0
3 years ago
A 0.450 kg soccer ball has a kinetic energy of 119 J.
Anastaziya [24]

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2  = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

Square root both sides

Sq rt v² = sq rt 528.89

V = 22.998m/s

V is approximately = 23m/s

I hope this was helpful, please rate as brainliest

8 0
3 years ago
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