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3 years ago

What is the weight of an object that has a mass of 3.5kg

Physics

1 answer:

Pepsi [2]3 years ago

5 0

Answer:

<em>Weight = 34.3N</em>

Explanation:

Weight = mass * gravitational acceleration

Gravitational acceleration (on Earth) = 9.8

Weight = 3.5 * 9.8

Weight = 34.3N

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In which areas of the diagram does conduction occur?

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In an experiment, a rectangular block with height h is allowed to float in two separate liquids. In the first liquid, which is w

Amiraneli [1.4K]

Answer:

The relative density of the second liquid is 7.

Explanation:

From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.

Let us assume that the volume of the object is 'V'

Thus for the liquid in which the block is completely submerged

The buoyant force should be equal to weight of liquid

Mathematically

$F_{buoyant}=Weight\\\\\rho _{1}\times V\times g=m\times g\\\\\therefore \rho _{1}=\frac{m}{V}...............(i)$

Thus for the liquid in which the block is 1/7 submerged

The buoyant force should be equal to weight of liquid

Mathematically

$F'_{buoyant}=Weight\\\\\rho _{2}\times \frac{V}{7}\times g=m\times g\\\\\therefore \rho _{2}=\frac{7m}{V}.................(ii)$

Comparing equation 'i' and 'ii' we see that

$\rho_{2}=7\times \rho _{1}$

Since the first liquid is water thus $\rho _{1}=1gm/cm^3$

Thus the relative density of the second liquid is 7.

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3 years ago

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A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the

ioda

Answer:

Force, F = 44 N

Explanation:

Given that,

Initial speed of the football, u = 0

Final speed, v = 15 m/s

The time of contact of the ball, t = 0.15 s

The mass of football, m = 0.44 kg

We need to find the average force exerted on the ball. It is given by the formula as :

$F=ma\\\\F=\dfrac{mv}{t}\\\\F=\dfrac{0.44\times 15}{0.15}\\\\F=44\ N$

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3 years ago

A) the unstretched length of each elastic rope is 24m. The rope obeys hookes law. The vertical distance between P and Q is 35m.

solong [7]

Explanation:

a) The rope obeys Hooke's law, so:

F = k Δx

The elastic energy in the rope is:

EE = ½ k Δx²

Or, in terms of F:

EE = ½ F Δx

Use trigonometry to find the stretched length.

cos 20° = 35 / x

x = 37.25

So the displacement is:

Δx = 37.25 − 24

Δx = 13.25

The elastic energy per rope is:

EE = ½ (3.7×10⁴ N) (13.25 m)

EE = 245,000 J

There's two ropes, so the total energy is:

2EE = 490,000 J

Rounded to one significant figure, the elastic energy is 5×10⁵ J.

b) The elastic energy in the ropes is converted to gravitational energy.

EE = PE = mgh

5×10⁵ J = (1.2×10³ kg) (9.8 m/s²) h

h = 42 m

Rounded to one significant figure, the height is 40 m. So the claim is not justified.

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3 years ago

A student pulls a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185 N and a

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