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ad-work [718]
3 years ago
13

A planned high-speed train between Houston and Dallas will travel a distance of 386 kilometers in 5.40 × 10^3 seconds. What is t

he average speed of this train?
Physics
1 answer:
Mazyrski [523]3 years ago
7 0

¡Hellow!

For this problem, first, lets convert the seconds in hours:

5,4x10³\rightarrow 5400

h = sec / 3600

h = 5400 s / 3600

h = 1,5

Let's recabe information:

d (Distance) = 386 km

t (Time) = 1,5 h

v (Velocity) = ?

For calculate velocity, let's applicate formula:

                                                    \boxed{\boxed{\textbf{d = v * t} } }

Reeplace according we information:

386 km = v * 1,5 h

v = 386 km / 1,5 h

v = 257,33 km/h

The velocity of the train is of <u>257,33 kilometers for hour.</u>

<u></u>

Extra:

For convert km/h to m/s, we divide the velocity of km/h for 3,6:

m/s = km/h / 3,6

Let's reeplace:

m/s = 257,33 km/h / 3,6

m/s = 71,48

¿Good Luck?

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An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and
Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is F_{ma}=BqV---(i)

The given formula to find the electric force is F_{el}=qE---(ii)

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

Bqv=qE\\\\v=\frac{E}{B}

v=\frac{187500}{0.125} \\\\v=15\times10^5m/s

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

F_{ce}=\frac{mv^2}{r}---(iii)

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

Therefore,  the particle's charge-to-mass ratio is 958.1\times10^5C/kg

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

8 0
3 years ago
What property of light is suggested by the formation of shadows​
Mandarinka [93]

Answer:

The reflection and rectilinear propagation of light helps in the formation of shadows and also tells light doesn't penetrate opaque materials.

3 0
2 years ago
When u step on the brakes what energy transformation takes place describe the transformation
larisa [96]
Before you step on the brakes, the car has kinetic energy, when you step on the brakes, it turns the kinetic energy into heat (thermal energy). When it stops completely, it has potential energy. Hope this helped :)
5 0
3 years ago
Given the 1-m stick shown below, which is held by a thread at its center. Block 1 is 15 N held at the 10 cm mark, while block 2
sattari [20]
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7 0
2 years ago
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

= 365 d²

This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

= 1/2 x 1.2 x 2.3²

= 3.174 J

Loss of energy

= 365 d² - 3.174  

This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

= .44 x 1.2 x 9.8 x .05

= .2587 J

365 d² - 3.174   = .2587

365 d² = 3.4327

d² = 3.4327 / 365

= .0094

d = .097 m

= 9.7 cm

If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .

5 0
3 years ago
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