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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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Please help on this one?

navik [9.2K]

the upside down image means an inverted image and for an inverted image, magnification is negative

so the answer is -m

4 0

4 years ago

Is the fuel filter considered to be part of the cars engine?

nikdorinn [45]

Yes it is concidered to be apart of the cars engine.

3 0

4 years ago

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

#1

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

$F_{net} = F_{app} - F_f$

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

#2

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

$F_{net} = F_{app} - F_f$

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

#3

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

$F_{net} = F_{app} - F_f$

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

4 years ago

The tsunami described in the passage produced a very erratic pattern of damage, with some areas seeing very large waves and near

Lunna [17]

Answer:B. The superposition of waves from the primary source and reflected waves produced regions of constructive and destructive interference.

Explanation: Tsunami is described as several waves originating from a water body mainly an ocean caused by Large scale earth movements taking place under the sea.

Superimposition of wave is the movement of one wave on another,it can be constructive or destructive.

Constructive interference is a wave interference that take place causing the Crest of one wave to align with that of another wave leading to high amplitude.

Destructive interference is a wave interference that take place causing the Trouph of one wave to align with the crest of another wave leading to low amplitude wave.

5 0

3 years ago

A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the t

Alexandra [31]

Answer:

the coin does not slide off

Explanation:

mass (m) = 5 g = 0.005 kg

distance (r) = 15 cm = 0.15 m

static coefficient of friction (μs) = 0.8

kinetic coefficient of friction (μk) = 0.5

speed (f) = 60 rpm

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the angular speed of the table

ω = 2πf

ω = 2 x π x 60 x $\frac{1}{60}$

ω = 6.3 s^{-1]

Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding

static force (Fs) = ma

static force (Fs) = μs x Fn = μs x m x g

Fs = 0.8 x 0.005 x 9.8 = 0.0392 N

Fs = ma

0.0392 = 0.005 x a

a = 7.84 m/s^{2}

$(Vmax)^{2}$ = a x r

$(Vmax)^{2}$ = 7.84 x 0.15

Vmax = 1.08 m/s

ωmax = $\frac{Vmax}{r}$

ωmax = $\frac{1.08}{0.15}$ = 7.2 s^{-1}

now that we have the maximum angular acceleration of the table, we can calculate its maximum speed in rpm

Fmax = $\frac{ωmax}{2π}$

Fmax = $\frac{7.2}{2 x π}$ = 68.7 rpm

since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.

4 0

4 years ago