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DENIUS [597]
3 years ago
12

What force would be required to accelerate a 1,100kg car to 0.5 m/s2

Physics
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

the force required to accelerate a 1,100kg car is 550N

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What factors that affect magnetic force
Reil [10]

Answer:

Strong electrical currents in close proximity to the magnet.

Other magnets in close proximity to the magnet.

Neo magnets will corrode in high humidity environments unless they have a protective coating.

Explanation: Heat radiation

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A bike travels at a constant speed of 3.1 m/s for 6 s. how far does it go?
IgorC [24]
Ah for this problem you are thinking quite a bit hard on. The problem is actually simpler than it looks. The problem states that a bike travels at a constant speed of 3.1 m/s for 6 s and asks how far will it go?. To figure this out you simply need to take 3.1 times 6 s because every second the bike travels 3.1 m. So the answer to this problem would be 18.6 m
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4 years ago
The strength of the force of gravity depends on a. the masses of the objects and their speeds. b. the masses of the objects and
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3 years ago
A 5 m3 tank containing 5kg of an unknown ideal gas at 500 kPa is connected through a valve to another tank containing 10 kg of t
Ivan

Answer:

a) V_{T} = 9\,m^{2}, b) m_{T} = 15\,kg, c) P_{T} = 416.667\,kPa

Explanation:

a) The equation of state for ideal gas is:

P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T

Given the existence of an isothermal process, the following relation is derived:

\frac{P_{1}\cdot V_{1}}{m_{1}} = \frac{P_{2}\cdot V_{2}}{m_{2}}

The volume of the other tank is:

V_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}

V_{2} = \left(\frac{10\,kg}{5\,kg} \right)\cdot \left(\frac{200\,kPa}{500\,kPa}\right)\cdot (5\,m^{3})

V_{2} = 4\,m^{3}

The total volume is:

V_{T} = V_{1} + V_{2}

V_{T} = 5\,m^{3} + 4\,m^{3}

V_{T} = 9\,m^{2}

b) The total mass is:

m_{T} = m_{1} + m_{2}

m_{T} = 5\,kg + 10\,kg

m_{T} = 15\,kg

c) The pressure of the gas in the two tanks is:

P_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{V_{1}}{V_{2}}\right)\cdot P_{1}

P_{T} = \left(\frac{15\,kg}{5\,kg}\right)\cdot \left(\frac{5\,m^{2}}{9\,m^{2}} \right)\cdot (500\,kPa)

P_{T} = 416.667\,kPa

3 0
4 years ago
A magnet attracts more iron dusts at its end​
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Its a MAGNETIC FIELD that can let a magnet attracts more iron dusts

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