During what era did the French and Indian War occur?

2 answers:

7 0

<span>French and Indian War began in 1754 and ended with the Treaty of Paris in 1763. The war provided Great Britain enormous territorial gains in North America, but disputes over subsequent frontier policy and paying the war’s expenses led to colonial discontent, and ultimately to the American Revolution.</span>

frutty [35]3 years ago

5 0

The French and Indian war started in 1754 and it ended in 1763 with the Treaty Of Paris

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Which simple experiment could be conducted to show an example of how lightning is created? A) Rub a balloon so it picks up extra

alexandr1967 [171]

I think the answer is d

7 0

2 years ago

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.700 m2 .

Darina [25.2K]

Answer:

The value is $E = 8.9 *10^{-5} \ J$

Explanation:

From the question we are told that

The area is $A = 0.700 \ m^2$

The root mean square value is $E_{rms} = 0.0400 \ V/m$

The time taken is $t = 30.0 \ s$

Generally the energy is mathematically represented as

$E = c * \sepsilon_o * A * t * E_{rms}^2$

=> $E = 3.0*10^{8} * 8.85*10^{-12} * 0.700 * 30 * (0.04)^2$

=> $E = 8.9 *10^{-5} \ J$

3 0

2 years ago

At 600.0 k the rate constant is 6.1× 10–8 s–1. what is the value of the rate constant at 785.0 k?

photoshop1234 [79]

Missing details. Complete text is:"The following reaction has an activation energy of 262 kJ/mol:
C4H8(g) --> 2C2h4(g)
At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 785.0 K?"
To solve the exercise, we can use Arrhenius equation:
$\ln( \frac{K_2}{K_1} ) = \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2} )$
where K are the reaction rates, Ea is the activation energy, R=8.314 J/mol*K and T are the temperatures. Using T1=600 K and T2=785 K, and Ea=262 kJ/mol = 262000 J/mol, on the right side of the equation we have
$\frac{Ea}{R}( \frac{1}{T_1}- \frac{1}{T_2} )=12.38$
And so
$\ln( \frac{K_2}{K_1})=12.38$
And using $K_1=6.1\cdot 10^{-8} s^{-1}$ , we find K2:
$K_2=K_1 e^{12.38}=0.0145 s^{-1}$

5 0

3 years ago

The amount of radiant power produced by the sun is approximately 3.9 ✕ 1026 W. Assuming the sun to be a perfect blackbody sphere

fredd [130]

Answer:

T=5797.8 K

Explanation:

Given that

Power P = 3.9 x 10²⁶ W

Radius ,r= 6.96 x 10⁸ m

We know that ,From Plank's law

P = σ A T⁴