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viktelen [127]
3 years ago
6

A house brick has a volume of 1900 cm³ and a weight in air of 80N.What is its apparent weight in water?The density of water is 1

.00 g cm³?​
Physics
1 answer:
Shtirlitz [24]3 years ago
3 0

Answer:

61 N

Explanation:

We'll begin by calculating the mass of the brick when placed in water. This can be obtained as follow:

Volume of brick = 1900 cm³

Density of water = 1 g/cm³

Mass of brick in water =…?

Density = mass / volume

1 = mass of brick in water / 1900

Cross multiply

Mass of brick in water = 1 × 1900

Mass of brick in water = 1900 g

Next, we shall convert 1900 g to Kg.

1000 g = 1 Kg

Therefore,

1900 g = 1900 g × 1 Kg / 1000 g

1900 g = 1.9 Kg

Next, we shall determine the weight in water. This can be obtained as follow:

Mass (m) = 1.9 Kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

W = m × g

W = 1.9 × 10

W = 19 N

Thus, the weight of the brick in water is 19 N.

Finally, we shall determine the apparent weight of the brick in water. This can be obtained as follow:

Weight in air = 80 N

Weight in water = 19 N

Apparent weight =?

Apparent weight = weight in air – weight in water

Apparent weight = 80 – 19

Apparent weight = 61 N

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A:

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Calculate specific heat of liquid if a 450. g iron block is heated to 98.0 C and then droped into a calorimeter with 125 g of un
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Explanation:

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If my mass is 196 lbm and I tackle one of my teammates - while decelerating from a velocity of 6.7 m/s to 0 m/s in 0.5 s, how mu
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Explanation:

given,

mass = 196 lbm

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time taken for deceleration = 0.5 sec        

F = mass × acceleration

acceleration = \dfrac{0-6.7}{0.5}              

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hence, the force acting on the team mate is 1.19 kN.

8 0
3 years ago
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