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Dahasolnce [82]

3 years ago

7

During the French and Indian war France and Great Britain fought for control of north American territory what impacted the end o

f the war have on the American colonies

1 answer:

Marta_Voda [28]3 years ago

4 0

Answer:

during American War, George Washington was voted as a military chef. So USA and France decided to be allies (France lost 7years war so they wanted to revenge to Great Britain).

after revolution

In 1778, Bill of Rights was made after that, every Americans got liberty, New laws with president and their own country (USA).

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A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

a)

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

b)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

c)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

3 years ago

How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the

KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0

3 years ago

In the compound fe203 irons oxidization number is + 3 + Oxygen's oxidation number is

PtichkaEL [24]

Answer:

-2

Explanation:

Since its a compound, it's charge is 0.

So, there are 2 Fe and 3 Oxygen,

let x = oxidation number of oxygen

2(+3) + 3x =0

x =-2

Side note : even if the oxidation number is positive, make sure to put the positive sign like how you would in numbers

8 0

3 years ago

Water at a gauge pressure of 3.8 atm at street level flows into an office building at a speed of 0.68 m/s through a pipe 5.4 cm

Nataliya [291]

kxubj ysufuxtfuvhxub

7 0

3 years ago

Suppose that you take a 10 kg mass on the surface of the earth and then place it on the moon. What will

sergey [27]

Answer:

10 kg

Explanation:

The mass of an object does not change even if the amount of gravtiy changes.

5 0

3 years ago