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aalyn [17]
2 years ago
6

A uniform metal meter-stick is balanced with a 1.0 kg rock attached to the left end of the stick. If the support is located 0.25

m from the left end, what is the mass of the meter-stick? (Hint: meter-stick is 1 m long). a) 0.6 kg b) 2.0 kg c) 1.0 kg d) 3.0 kg
Physics
1 answer:
Dafna1 [17]2 years ago
8 0

Answer:

c) 1.0 kg

Explanation:

The mass of the stick will be located at the centre of the metre rule. Since the rock is located 0.25m from the pivot, the mass of the meter rule is also 0.25m to the Right of the support

According to law of moment

Sum of clockwise moment = sum of anti clockwise moments

Clockwise moment = M×0.25(mass of metre rule is M)

CW moment = 0.25M

Anti clockwise moment = 0.25×1

ACW moments = 0.25kgm

Equate;

0.25M = 0.25

M = 0.25/0.25

M = 1.0kg

Hence the mass of the metre rule is 1.0kg

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B. Direction: In each of the following situations, identify the method of heat transfer that takes place
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Answer:

1) Conduction

2)Covection

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4)Convection (Land breeze one of the application of convection of heat)

5) Convection

6)Radiation

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Hope it helps

4 0
3 years ago
A block of concrete has a mass of 48kg a crane lifts the block to a height of 12m above the ground calculate the gravitational p
Afina-wow [57]

Answer:

5760 J

Explanation:

From the question given above, the following data were obtained:

Mass of block = 48 kg

Height (h) = 12 m

Gravitational field strength (g) = 10 N/Kg

Gravitational potential energy (PE) =?

The gravitational potential energy stored by the block can simply be obtained as follow:

PE = mgh

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PE = 5760 J

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3 0
3 years ago
Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

Thus for the exercise we will have

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

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3 years ago
Which of the following expressions for power or dimensionally correct?
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Power=F.V

dimension: ML^2T^-2


8 0
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