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Ronch [10]
2 years ago
11

How much work is done if a force of 20 N moves an object a distance of 6 m?​

Physics
1 answer:
Nesterboy [21]2 years ago
3 0

Explanation:

W=F×s

w=20N×6m=120J

W=120J

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The wheels of an automobile are locked as it slides to a stop from an initial speed of 30.0 m/s. If the coefficient of kinetic f
Amiraneli [1.4K]

Answer:

  x = 76.5 m

Explanation:

Let's use Newton's second law at the point of contact between the wheel and the floor.

     fr = m a

     fr = miy N

     N-W = 0

     N = W

     μ mg = m a

     a = miu g

    a = 0.600 9.8

    a = 5.88 m / s²

Having the acceleration we can use the kinematic relationships to find the distance

     v_{f}² = v₀² + 2 a x

    v_{f} = 0

    x = -v₀² / 2 a

Acceleration opposes the movement by which negative

   x = - 30²/2 (-5.88)

   x = 76.5 m

8 0
3 years ago
If someone walkes 1000m <br> in 20min, what is their speed?
slava [35]

Answer:

Distance - 1000m

Time - 20min

Speed - ?

Use the formula of distance ÷ time = speed.

s = d/t

s = 1000m/20min

s = 50 m/min

Hope this helps, thank you !!

6 0
2 years ago
Part A Determine the absolute pressure on the bottom of a swimming pool 30.0 m by 8.7 m whose uniform depth is 1.8 m . Express y
Sphinxa [80]

Answer:

17.66 kPa

Explanation:

The volume of water in the swimming pool is the product of its dimensions

V = 30 * 8.7 * 1.8 = 469.8 cubic meters

Let water density \rho = 1000 kg/m^3, and g = 9.81 m/s2 we can calculate the total weight of water in the swimming pool

W = mg = \rho V g = 1000 * 469.8 * 9.81 = 4608738 N

The area of the bottom

A = 30 * 8.7 = 261 square meters

Therefore the pressure is its force over unit area

P = F/A = 4608738  / 261 = 17658 N/m^2 or 17.66 kPa

7 0
3 years ago
A sprain ?
saw5 [17]
The answer is C. I<span>s a stretch and/or tear of a ligament.</span>
5 0
3 years ago
Read 2 more answers
Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm
Gre4nikov [31]

Here since both children and merry go round is our system and there is no torque acting on this system

So we will use angular momentum conservation in this

I_1\omega_1 = I_2\omega_2

now here we have

I_1 = \frac{MR^2}{2} + m_1R^2 + m_2R^2

I_1 = \frac{100(1.60)^2}{2} + (22 + 28)(1.60)^2

I_1 = 256

Now when children come to the position of half radius

then we will have

I_2 = \frac{MR^2}{2} + m_1(\frac{R}{2})^2 + m_2(\frac{R}{2})^2

I_2 = \frac{100(1.6)^2}{2} + (28 + 22)(0.8)^2

I_2 = 160

now from above equation we have

256 (20.0 rpm) = 160(\omega_2)

\omega_2 = 32 rpm

8 0
3 years ago
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