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rjkz [21]
3 years ago
9

wo ships, one 200200 metres in length and the other 100100 metres in length, travel at constant but different speeds. When trave

lling in opposite directions, it takes 1010 seconds for them to completely pass each other. When travelling in the same direction, it takes 2525 seconds for them to completely pass each other. The speed of the faster ship, in metres per second, is
Physics
1 answer:
shutvik [7]3 years ago
7 0

Answer:

The speed of the faster ship is 22.5 m/s

Explanation:

The length of the ships are;

Ship 1 = 200 m

Ship 2 = 100 m

The time it takes for the ships to completely cross each other when travelling in opposite directions = 10 seconds

The time it takes both ships to cross each other when travelling in the same direction = 25 seconds

Let x represent the speed of the first ship, ship 1, and y represent the speed of the second ship, ship, 2, we have;

(x + y) × 10 = 200 + 100 = 300

10·x + 10·y = 300...(1)

(x - y) × 20 = 200 + 100 = 300

20·x - 20·y = 300...(2)

Multiply equation (1) by 2, to get;

(x + y) × 10 × 2 = 300 × 2

20·x + 20·y = 600...(3)

Adding equation (1) to equation (3) gives;

20·x + 20·y + 20·x - 20·y = 600 + 300

40·x = 900

x = 900/40 = 22.5

x = 22.5

The speed of the first ship, ship 1 = x = 22.5 m/s

From equation (1), we have;

10·x + 10·y = 300

y = (300 - 10·x)/10 = (300 - 10×22.5)/10 = 7.5

y = 7.5

The speed of the second ship, ship 2 = y = 7.5 m/s

Therefore, the faster ship is ship 1 with a speed of 22.5 m/s

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Please awnser and show the ways​
topjm [15]

Answer:

Answers in solutions.

Explanation:

<u>Question 6:</u>

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><u>The density of the mixture:</u></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.

<u>Question 7:</u>

<u />

  • 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

<u>Question 8:</u>

<u />

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

<u>Question 9:</u>

<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

<u>Question 10:</u>

<u />

  • 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 cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative densi
ki77a [65]

Answer:

The mass of the fluid is 72 g.

Explanation:

The following data were obtained from the question:

Height (h) = 10 cm

Area of cross section (A) = 36cm²

Relative density = 0.2

Mass =..?

Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:

Volume = Area x Height

Volume = 36 x 10

Volume = 360 cm³

Next, we shall determine the density of the liquid.

This can be obtained as follow:

Relative density = density of substance/density of water.

Relative density = 0.2

Density of water = 1 g/cm³

Density of fluid =...?

Relative density = density of substance/density of water.

0.2 = density of fluid / 1 g/cm³

Cross multiply

Density of fluid = 0.2 x 1 g/cm³

Density of fluid = 0.2 g/cm³

Finally, we shall determine the mass of fluid as follow:

Volume = 360 cm³

Density of fluid = 0.2 g/cm³

Mass of fluid =...?

Density = mass /volume.

0.2 g/cm³ = mass of fluid /360 cm³

Cross multiply

Mass of fluid = 0.2 g/cm³ x 360 cm³

Mass of fluid = 72 g

Therefore, the mass of the fluid in the jar is 72 g.

6 0
3 years ago
Small evidence is also called what?
siniylev [52]
Small evidence is also called trace evidence.
4 0
3 years ago
Two identical satellites are in orbit about the earth. One orbit has a radius r and the other 2r. The centripetal force on the s
velikii [3]

Answer:

the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

Explanation:

Mass of satellite, m

orbit radius of first, r1 = r

orbit radius of second, r2  = 2r

Centripetal force is given by

F= \frac{mv^{2}}{r}

Where v be the orbital velocity, which is given by

v=\sqrt{gr}

So, the centripetal force is given by

F= \frac{mgr}}{r}}=mg

where, g bet the acceleration due to gravity

g=\frac{GM}{r^{2}}

So, the centripetal force

F= \frac{GMm}}{r^{2}}}

Gravitational force on the satellite having larger orbit

F= \frac{GMm}{4r^{2}} .... (1)

Gravitational force on the satellite having smaller orbit

F'= \frac{GMm}{r^{2}} .... (2)

Comparing (1) and (2),

F' = 4 F

So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.

8 0
3 years ago
A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius R, at what radial distances in
eimsori [14]

Answer:

Electric field at radius r inside the solid sphere is

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:Given

The radius of the solid sphere is

The charge on the solid sphere is

The inner radius of the shell is

The outer radius of the shell is

The total charge on the shell is

PART(A)

The magnitude of electric field at radius r where  \\The volumetric charge density of the solid sphere will be

The charge enclosed by the radius r inside the solid sphere is

rho=

According to gauss law

PART(B)

The electric field at radius r where

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

The charge enclosed by the radius r where

According to gauss law

Read more on Brainly.com - brainly.com/question/13242041#readmore

3 0
3 years ago
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