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denis-greek [22]
3 years ago
12

On a journey of 600 km, a train was delayed 1 hour and 30 min after having covered 1 4 of the way. To arrive on time at the dest

ination, the engine driver had to increase the speed by 15 km/hour. How long did the train travel for
Physics
1 answer:
mars1129 [50]3 years ago
5 0

Answer:

The train travelled 10hours.

Explanation:

Using speed= distance/time ...eq1

Let the time taken by train to cover the journey be t.

Let the speed of train be s

Time= distance/speed ...eq2

Time t =600/s ...eq3

The train is delayed for 1 1/2 hours=3/2 hours

Train increased by 15km/hr.

Train travelled 1/4 of 600k.= 150km.

Speed increased by s + 15 to cover the remaining 450km

t = 150/s + 400/(s + 15) + 3/2 ..eq4

Equating eq3 and 4

600/s= 150/s + 460/(s + 15) + 3/2

450/s = 3/2 + 450/(s + 15)

3s^2 + 45s -13500=0

Solving the quadratic equation

S= -45 +-sqrt(45^2-4(3×13500)/2×3

S= 60 , -75

Hence speed of train is 60km/hr

Using eq 2 distance/speed=time

600/60=10

t= 10hours

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Answer:

A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin.

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As we know that arc length is given as

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length of chord is given as

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so here

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\frac{\theta}{2} > sin(\frac{\theta}{2})

6 0
2 years ago
if a bus has a mass of 5000 kg and a velocity of 10 m/s, what must the velocity of a 1500 kg can be to have the same momentum
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6 0
2 years ago
Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir le
kap26 [50]

Answer:

Explanation:

1 ha = 10⁴ m²

1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²

In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³

Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m

Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³

Let Q be the withdrawal in m³

Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶

Q = 26.20 x 10⁶ m³

rate of withdrawal per second

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= 26.20 x 10⁶ / 2.592 x 10⁶

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3 years ago
If a 42kg rolling object slows from 11.5m/s to 3.33m/s how much work did friction do
geniusboy [140]

Answer: 1608.39 J

Explanation: Given that the

mass M = 42kg

U = 11.5m/s

V = 3.33m/s

how much work did friction do

Work done = Force × distance

Work done = Ma × distance

But acceleration a = V/t

Work done = M × V/t × d

Work done = M × V × d/t

Where d/t = velocity

Therefore,

Work done = M × U × V

Work done = 42 × 11.5 × 3.33

Work done = 1608.39 J

8 0
3 years ago
a 280 nm thin film with index of refraction 1.6 floats on waterwhat is the largest wavelength of reflected light for which const
aksik [14]

Answer:

Inside the film the wavelength will be λ/n

For constructive interference to occur the film must be λf/4 thick where λf is the wavelength of the light in the film - there will be a 180 degree phase shift at the water/film interface since the index of refraction of the film is greater than that of water - and the light has to travel λ/2 inside the film for constructive interference to occur

280 nm / 1.6 * 4 = 700 nm is the greatest wavelength allowed

Note that 700 nm is also the upper wavelength of the visible spectrum

3 0
2 years ago
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