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eimsori [14]
3 years ago
14

A train of mass 50000 kg starts from rest and moves with uniform acceleration 5 km/h2 then, find its kinetic energy after 30 min

utes.
Physics
1 answer:
r-ruslan [8.4K]3 years ago
7 0

Answer:

12 kJ

Explanation:

Given:

v₀ = 0 km/h

a = 5 km/h²

t = 30 min = 0.5 h

Find: v

v = at + v₀

v = (5 km/h²) (0.5 h) + 0 km/h

v = 2.5 km/h

v = 2.5 km/h × (1000 m/km) × (1 h / 3600 s)

v = 0.694 m/s

KE = ½ mv²

KE = ½ (50,000 kg) (0.694 m/s)²

KE = 12,000 J

KE = 12 kJ

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joja [24]

Answer: Boyle found that when the pressure of a gas at a constant temperature is increased, the volume of the gas decreases. When the pressure of a gas is decreased, the volume increases. This relationship between pressure and volume it's called Boyle's law.

Explanation: In the 1600s, Boyle measured the volumes of gases at different pressures. Boyle found that when the pressure of a gas at a constant temperature is increased, the volume of the gas decreases. When the pressure of a gas is decreased, the volume increases. This relationship between pressure and volume it's called Boyle's law.

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2 years ago
5. What is the speed of a wave in a spring if it has a wavelength of 10 cm and a period of 0.2s
eimsori [14]

Explanation:

speed of wave

v = wavelength x frequency

since frequency is f = 1/Period then

v = wavelength : Period

v = 10 cm/ 0.2 s = 50 cm/s

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3 years ago
Five 60 ohm resistors are connected in parallel. What is their equivalent resistance?
Mnenie [13.5K]

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7 0
3 years ago
Read 2 more answers
A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

x_{1} = 0\\h_{s} = 4ft

F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

here

h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

F = 2187937.5 lb

7 0
3 years ago
Hàng ngày ta thấy Mặt trời , Mặt trăng quay quanh Trái Đất, khi đó ta đã lấy vật mốc là
marshall27 [118]
I’m sorry I don’t understand this language
5 0
2 years ago
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