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3 years ago

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I am thinking of an unreactive nonmetal and Aisha suggested I use bromine. Is this a good choice? Explain your reasoning. If not

which nonmetal would you suggest?

1 answer:

galina1969 [7]3 years ago

3 0

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A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?

Yuri [45]

W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.

5 0

3 years ago

Read 2 more answers

Tyler's favourite number is 8642, and he tries to fit it into everything he does. If he wishes to do that much work by climbing

Brums [2.3K]

Answer:

2014

Explanation:

none

6 0

3 years ago

Please help!!!!!!!!!

IgorC [24]

In an Internal Combustion Engine, the fuel is singed in the chamber or vessel. Example: Diesel or Petrol motor utilized as a part of Cars.
The internal engine has its vitality touched off in the barrel, as 99.9% of motors today. In an External Combustion Engine, the inner working fuel is not consumed. Here the liquid is being warmed from an outer source. The fuel is warmed and extended through the interior instrument of the motor bringing about work. Eg. Steam Turbine, Steam motor Trains. An outer burning case is a steam motor where the warming procedure is done in a kettle outside the motor.

5 0

3 years ago

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both fila

Lemur [1.5K]

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

or

$\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}$

substituting the values in the above question we get

$\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}$

or

$\frac{A_1}{A_2} }$ =0.3659

6 0

3 years ago

A lawn roller is rolled across a lawn by a force of 107 N along the direction of the handle, which is 13.5 ◦ above the horizonta

-BARSIC- [3]

Answer:

22.02 m

Explanation:

given,

Force, F = 107 N

angle made with horizontal = 13.5◦

Power develop by the lawn roller = 69.4 W

time = 33 s

distance = ?

Force along horizontal= F cos θ

= 107 cos 13.5°= 104 N

Power = $\dfrac{work\ done}{time}$

69.4 = $\dfrac{W}{33}$

W = 2290.2 J

Work done= Force x displacement

displacement= $\dfrac{2290.2}{104}$

= 22.02 m

6 0

3 years ago