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Ivahew [28]
3 years ago
13

Work out the kinetic energy of a 2.5 kg remote-controlled car that is moving at 2 m/s.

Physics
1 answer:
lbvjy [14]3 years ago
8 0

Answer: 5 joules

Explanation:

mass=m=2.5kg

Velocity=v=2m/s

Kinetic energy=ke

ke=(m x v x v)/2

ke=(2.5 x 2 x 2)/2

Ke=10/2

Ke=5

Kinetic energy=5 joules

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This type of water occurs as a liquid resource that is dispersed through numerous holes, pores, fractures, and cavities in bodie
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Answer:

C

Explanation:

Although this may seem surprising, water beneath the ground is commonplace. Usually groundwater travels slowly and silently beneath the surface, but in some locations it bubbles to the surface at springs. The products of erosion and deposition by groundwater were described in the Erosion and Deposition chapter.

Groundwater is the largest reservoir of liquid fresh water on Earth and is found in aquifers, porous rock and sediment with water in between. Water is attracted to the soil particles and capillary action, which describes how water moves through a porous media, moves water from wet soil to dry areas.

Aquifers are found at different depths. Some are just below the surface and some are found much deeper below the land surface. A region may have more than one aquifer beneath it and even most deserts are above aquifers. The source region for an aquifer beneath a desert is likely to be far from where the aquifer is located; for example, it may be in a mountain area.

The amount of water that is available to enter groundwater in a region is influenced by the local climate, the slope of the land, the type of rock found at the surface, the vegetation cover, land use in the area, and water retention, which is the amount of water that remains in the ground. More water goes into the ground where there is a lot of rain, flat land, porous rock, exposed soil, and where water is not already filling the soil and rock.

The residence time of water in a groundwater aquifer can be from minutes to thousands of years. Groundwater is often called “fossil water” because it has remained in the ground for so long, often since the end of the ice ages.

8 0
2 years ago
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A heat pump with a COP of 3.15 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the te
Jet001 [13]

Answer: 1026s, 17.1m

Explanation:

Given

COP of heat pump = 3.15

Mass of air, m = 1500kg

Initial temperature, T1 = 7°C

Final temperature, T2 = 22°C

Power of the heat pump, W = 5kW

The amount of heat needed to increase temperature in the house,

Q = mcΔT

Q = 1500 * 0.718 * (22 - 7)

Q = 1077 * 15

Q = 16155

Rate at which heat is supplied to the house is

Q' = COP * W

Q' = 3.15 * 5

Q' = 15.75

Time required to raise the temperature is

Δt = Q/Q'

Δt = 16155 / 15.75

Δt = 1025.7 s

Δt ~ 1026 s

Δt ~ 17.1 min

5 0
3 years ago
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Describe a life cycle of a star
vagabundo [1.1K]
Planetary Nebula are the outer layers of a star that are lost when the star changes from a red giant to a white dwarf. A star is a luminous globe of gas producing its own heat and light by nuclear reactions (nuclear fusion). They are born from nebulae and consist mostly of hydrogen and helium gas. Is this what you needed?
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3 years ago
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What is Newton's third law motion answer part a and b
Olenka [21]

Answer:

hope it helps

Explanation:

Newtons third law is that objects exert equal and opposite forces on each other.

'every action has an equal and opposite reaction'.

3 0
2 years ago
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ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit
horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

F = \dfrac{d^4G\delta}{8D^3N}

\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm

F = \dfrac{d^4G}{8D^3}\times 0.004

F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004

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stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

\tau = \dfrac{8FDk_s}{\pi d^3}

\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}

              = 442.6 Mpa

The tensile strength of the steel material of  ASTM A229 is equal to 1300 Mpa

now,

\tau_s \leq 0.45 S_u

\tau_s \leq 0.45 \times 1300

\tau_s \leq 585\ Mpa

since corrected stress is less than the \tau_s

hence, spring will return to its original shape.

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3 years ago
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