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MaRussiya [10]
3 years ago
15

the weight of elephant is 40000N. The area of its feet in contact with the ground is 0.25 m2. Show your working and give the uni

t
Physics
1 answer:
horrorfan [7]3 years ago
3 0

Answer:

P=160000Pa

Explanation:

You haven't specified what to calculate

But looking at the parameters I think the question is asking to calculate the pressure exerted.

Pressure=Force/Area

P=40000/0.25m²

P=160000Pa

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An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction  between the object and th
Gnom [1K]

Answer:

8m * (μg/v)^2

Explanation:

k, the spring constant = ?

(k in terms of μ, m, g, and v.)

Frictional force = μmg

Note: lost KE is converted to work done against the friction + PE of the spring

1/2mv2 = μmgx + 1/2kx^2....equation i

Cancel the 1/2 on both sides

mv^2 = μmgx + kx^2

Lets recall that:

Due to frictional effect, further enegy will be lost when the spring recoils backward

Therefore

1/2kx^2 = μmgx..... equation ii

Let's substitute 1/2kx^2 in equation I for ii

So we can say that:

1/2mv^2 = (μmgx)+ μmgx

1/2mv^2 = 2 (μmgx)

1/4mv^2 = μmgx

Cancel out m on both sides

1/4v^2 = μgx

Make x subject of the formula

x = (1/4v^2) / (μg)...... equation iii

substitute x to equation ii

But first make k in equation ii subject of the formula

1/2kx^2 = μmgx

k = 2μmg/x

Now substitute x

k = 2μmg / ((1/4v^2) / (μg))

k = 2μmg * ((μg) / (1/4v^2))

k = 8m * (μg/v)^2

8m * (μg/v)^2

7 0
3 years ago
9.
julsineya [31]
I don’t know just needed points im sorry
8 0
3 years ago
What if someone refers to a crystal's " habit" what are they referring to?
Alenkinab [10]
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7 0
3 years ago
At what time does the box have the greatest kinetic energy
Y_Kistochka [10]

Answer:

therefore, more kinetic energy! An object has the MOST kinetic energy when it's movement is the GREATEST.

Explanation:

3 0
3 years ago
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Vesnalui [34]

Answer:

C. \frac{3F}{8}

Explanation:

Let initial charges on both spheres be,q

F=\frac{Kq^2}{d^2}   \ \ \  \ \ \  \ \ \  \ \_i

When the sphere C is touched by A, the final charges on both will be,\frac{q}{2}

#Now, when C is touched by B, the final charges on both of them will be:

q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\

Now the force between A and B is calculated as:

F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}

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3 years ago
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