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garri49 [273]
3 years ago
13

A cube of metal has a mass of 11 grams and a volume of 1 cm . When fully submerged in water this metal cube hanging from an accu

rate spring scale will weigh what amount?
Physics
1 answer:
Anarel [89]3 years ago
7 0

Answer:

0.098 N

Explanation:

From the question,

Spring scale reading = W-U............... Equation 1

Where W = weight of the cube, U = upthrust.

W = mg

Where m =  mass of the cube, g = acceleration due to gravity.

Given: m = 11 g = 0.011 kg, g = 9.8 m/s².

W = 0.011(9.8)

W = 0.1078 N.

From Archimedes principle,

Upthrust = weight of water displaced.

U = (Density of water×volume of metal cube)×acceleration due to gravity.

U = (D×V)g

Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²

U = 1000(9.8)(10⁻⁶)

U = 0.0098 N.

Substitute the value of W and U into equation 1

Reading of the spring scale = 0.1078-0.0098

Reading of the spring scale = 0.098 N

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Answer:

find out the le

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3 0
3 years ago
For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long)
Lesechka [4]

The question is missing a diagram of the ray reflection. I attached a diagram which comes from a similar question in the answer section. The full question should be as follows:

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point d = 10.0cmfrom their point of intersection, as shown in the figure. For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is s=29.0cm long) after reflecting from the first mirror?

Answer:

34.6°

Explanation:

To strike the midpoint of the second mirror, the ray light will have to travel half of the distance vertically

i.e. 29/2 = 14.5

We can solve this through trigonometry.

Let the angle between the ray and the vertical plane mirror is known as α

tan α = 10/14.5

α = tan^{-1} (10/14.5) = 34.6°

The angle of incidence is the angle between the ray and the normal line of the mirror.

Let angle of incidence of first mirror be β

β = α = 34.6

6 0
3 years ago
You, Archimedes, suspect that the king’s crown is not solid gold but is instead gold-plated lead. To test your theory, you weigh
hjlf

Answer:

a) 16675.75 Kg/m³ b) 77.6%

Explanation:

the weight of the crown = 60 N, density of gold = 19300 Kg/m^3, density of lead = 11340 kg/m^3, density of water = 1000kg/m^3 and acceleration due to gravity = 9.8 m/s^2

upthrust on the crown = weight in air - weight when fully submerged in water = 60 - 56.4 = 3.6 N

mass of water displaced = 3.6 / 9.8  since weight = mass × g

mass of water displaced = 0.367 Kg

density of water = mass / volume

1000 = 0.367 / volume

cross multiply and find volume

volume of the crown = 0.367 / 1000 = 0.000367 m³ since the crown will displace water of equal volume according to Archimedes principle

Let V1 represent the volume of Gold and let V2 represent the volume of lead

Total volume of the crown = V1 + V2

also

density of gold = mass of gold / V1 and density of lead = mass of lead / V2

19300 = mass of gold in the crown / V1 and 11340 = mass of lead in the crown / V2

19300 V1 = mass of gold and 11340 V2 = mass of lead

add the two together

19300 V1 + 11340 V2 = weigth of the crown / 9.8

19300 V1 + 11340 V2 = 6.12 also

V1 + V2 = 0.000367

make V1 subject of the formula in equation 2

V1 = 0.000367 - V2

substitute for V1 in equation 1

19300 (0.000367 - V2) + 11340 V2 = 6.12

open the bracket

7.083 - 19300 V2 + 11340 V2 = 6.12

rearrange the equation

-7960 V2 = 6.12 - 7.083  

-7960 V2 = -0.963

V2 = -0.963 / -7960 = 0.000121 (volume of lead in the crown)

substitute V2 into equation 2

V1 + 0.000121  = 0.000367m³

V1 = 0.000367 - 0.000121 = 0.000246m³ (volume of gold in the crown)

so mass of gold in the crown = 19300 × 0.000246 = 4.748 kg

and mass of lead = 11340 × 0.000121 = 1.372 kg

average density of the crown = (mass of gold + mass of lead) / total volume = 6.12 / 0.000367 = 16675.75 kg/ m³

b) percentage make of gold = mass of gold / total mass × 100 = 77.6 % approx

4 0
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astra-53 [7]

Answer:

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3 0
3 years ago
how does the number of chromosomes in a persons sex compare with the number of chromosome in the body cell
koban [17]

no it doesn't why because I think that it is not the same but different.

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