Which conversion can take place in a transformer

A see-saw is balanced on a pivot with two children on it. One child is sitting 1.5 m to the left of the pivot and has a mass of

Lorico [155]

Answer:

Explanation:

Remark

This is a second class lever. It is much more efficient than the fishing pole problem. All distances are measured from the pivot in these kinds of questions.

Givens

d1 = 1.5

d2 = ?

m1 = 50 kg

m2 = 30 kg

The lighter child will have to sit further away from the pivot to make the two conditions equal.

Formula

d1*m1 = d2*m2

1.5*50 = d2 * 30

75 = 30 * d2

75/30 = d2

d2 = 2.5

Remark

Notice that the distance is longer for the lighter child. The fact that these are masses and not forces does not matter, but you should take note of it. There is a difference between masses and forces. See the fishing pole problem.

Answer to the multiple Choice question. No motion on this kind of problem means equal moments. The answer is D

Problem 2

1) The wheels are further apart making B more stable. The wider the distance the wheels are apart, the harder it would be to tip the concrete mixer over

2) The center of gravity is lower. The higher the force is the more chance you have of exerting an external force to tip the mixer over.

7 0

3 years ago

The dimensions of the disk of our milky way galaxy are:.

Katen [24]

Answer:

According to studies, the milky way is approximately, "170,000–200,000 light-years (52–61 kpc) in diameter and, on average, approximately 1,000 ly (0.3 kpc) thick."

With that being said, it is safe to say that the dimensions are somewhere around 100,000 by 1,000

6 0

2 years ago

Please awnser and show the ways

topjm [15]

Answer:

Answers in solutions.

Explanation:

Question 6:

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

The density of the substance in Crown A;

Density = mass ÷ volume = $\frac{1930}{100}$ = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3>The density of the mixture:</h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to $\frac{29.8}{2}$ = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

Question 7:

An empty masuring cylinder has a mass of 500 g.

Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.

The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³

Question 8:

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

$\frac{0.02 * 1}{0.001}$ = 20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

$\frac{0.02 * 1,000,000}{1}$ = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

Question 9:

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

Question 10:

A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
in a displacement can
The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; $\frac{20 * 1}{1}$ = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density × volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

7 0

3 years ago

A silver bar 0.125 meter long is subjected to a temperature change from 200 C to 100 C . What will be the length of the bar afte

dimulka [17.4K]

\Delta L= \alpha L_0 (T_f-T_i)

= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)

= -0.00225 m

New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
So your answer is B.

8 0

4 years ago

Read 2 more answers

Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters

Vesnalui [34]

Answer:

C. $\frac{3F}{8}$