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4 years ago

The _______ was an oval track surrounded by grandstands?

1 answer:

5 0

The appropriate response is Hippodrome (A+). The term is utilized as a part of the cutting edge French dialect and some others, with the significance of "speed racecourse". Henceforth, some present-day horse dashing tracks additionally incorporate the word hippodrome in their names, for example, the Hippodrome de Vincennes and the Central Moscow Hippodrome.

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A stone is dropped from the roof of a high building. a second stone is dropped 1.50 s later. how far apart are the stones when t

Aleksandr [31]

2nd stone time = 13.0/9.8 = 1.33 seconds

distance = 0.5*9.8*1.33^2 = 8.67 meters

1st stone time = 1.33+1.5 = 2.83 seconds

distance = 0.5*9.8*2.83^2 = 39.24 meters

39.24-8.67 = 30.57 meters apart

3 0

3 years ago

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

3 years ago

I know that the PE will be the same for all the objects, but other then that I'm confused on what is correct and what isn't.

joja [24]

Answer:

Neither aspects of Dominique are correct.

Explanation:

They will not have the same speed before hitting the ground.

The block C has an initial horizontal velocity v₀ₓ , therefore the total ultimate speed before hitting the ground is v = √ v₀ₓ² + vy²

It didn't matter how the blocks were dropped, they all landed at the same time. If any block has an initial vertical velocity it will have a higher end velocity. If Dominique means dropped from the rest then the previous claim is true. It is important how they were dropped.

The Dominique must said : The vertical speed will be the same because of conservation of energy.

The fact is, they all landed at the same time and that is true. And it is important how they are thrown away.

God is with you!!!

7 0

3 years ago

A canoe floats in a lake. Inside the canoe is a 25 kg steel solid ball. If the ball is thrown into the lake, does the level of t

Rus_ich [418]

Answer:

Explanation:

volume of ball bearing = 4/3 π r³

= 4/3 x 3.14 x 1.5³

= 14.13 cc.

if D be the density of steel , weight of the ball bearing

= 14.13 x D x g

For the first case , water will be displaced to keep it floating

weight of displaced water will be equal to weight of steel

weight of displaced water = 14.13 Dg

mass of displaced water = 14.13 D

volume of displaced water = mass / density of water

= 14.13 D / d ; (where d is density of water) .

Now when the steel ball bearing is dipped in water , it will displace water equal to its volume only and it will be drowned

its volume = 14 .13 cc

14.13 D / d > 14.13 ( because D/d is more than one , since D > d )

volume of water displaced in first case is greater

water level will go up higher in first case .

Hence in the second case water level will go down .

Same will happen in case of 25 kg steel .

4 0

3 years ago

A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a po

Archy [21]

Answer:

(a) Since net charge remains same,after immersion Q is same

(b) I. 14.56pF ii. 3.05V

Explanation:

C = kεA/d

k=1 for air

ε is 8.85x10-12F/m

A = .0025m2

d = .125m

C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF

Q = CV = .177pF * 244V = 43.188pC

Since net charge remains same,after immersion Q is same

C = kεA/d, for distilled water k is approx. 80

Cwater = Cair x k

=0.177pF x 80 = 14.16pF

Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V

c) Change in energy: ΔU = Uwater - Uair

Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ

Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ

ΔU = 5.204nJ

6 0

3 years ago