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3 years ago

Help quick p;leaseeeeeeeeeeee its about direction and magnitude

Physics

1 answer:

KonstantinChe [14]3 years ago

7 0

Explanation:

The net force of each square is the combination of the forces in each direction. The direction is the... direction the square would go in due to the net force. The magnitude of the net force is how large it is. So if you had a force pushing 2N to the left and 1N to the right, then the net force would be 1N to the left; because the two oppose eachother. If they were going in the same direction, then they'd add to each other. And perpendicular net forces (like one pushing up and another pushing left) can create net forces in diagonal directions.

I'm not going to do all of these for you because they're basically all the same thing and it's good practice for you anyway. But I'll do the first three just so you can get the idea:

1. The net force's magnitude is 4N and it's direction is to the right.

2. The net force's magnitude is 4N and it's direction is to the left.

3. The net force's magnitude is 0N and it has no direction because they are equal forces acting in opposite directions.

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A boy is running with a kinetic energy of 82"" j if the boy has a mass of 80 kg what is his speed

weqwewe [10]

Explanation:

$v = \sqrt \ \frac{ke}{ \frac{1}{2} m}$

$v = \sqrt \ \frac{82}{ \frac{1}{2} (80)}$

v = 1.4317821063276 m/s

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3 years ago

Why do boys like boys

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3 years ago

Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initia

Sedaia [141]

Answer:

$a_2\ =\ -33.65\ m/s^2$

Explanation:

Given,

For the first rocket,

Initial velocity of the first rocket A = $u_1\ =\ 4600\ m/s.$

Acceleration of the first rocket = $a_1\ =\ -18\ m/s^2$

For the second rocket,

Initial velocity of the second rocket B = $u_2\ =\ 8200 m/s.$
Displacement of both the rockets A and B = s = 0 m

Fro the first rocket,

Let 't' be the time taken by the first rocket A for whole the displacement

$\therefore s\ =\ u_1t\ +\ \dfrac{1}{2}a_1t^2\\\Rightarrow 0\ =\ 4600t\ -\ 0.5\times 18t^2\\\Rightarrow t\ =\ \dfrac{4600}{0.5\times 18}\\\Rightarrow t\ =\ 511.11 sec$

Let $a_2$ be the acceleration of the second rocket B for the same time interval

from the kinematics,

$\therefore s\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s\ =\ u_2t\ +\ \dfrac{1}{2}a_2t^2\\\Rightarrow a_2\ =\ \dfrac{2s\ -\ 2u_2t}{t^2}\\\Rightarrow a_2\ =\ \dfrac{0\ -\ 2u_2t}{t^2}\\$

$\Rightarrow a_2\ =\ -\dfrac{2u_2}{t}\\\Rightarrow a_2\ =\ -\dfrac{2\times 8600}{511.11}\\\Rightarrow a_2\ =\ -33.65\ m/s^2$

Hence the acceleration of the second rocket B is -33.65\ m/s^2.

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3 years ago

Which organism contains organ systems?

elena55 [62]

Pig because it is made up of many organs system like cardiovascular and muscular system

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4 years ago

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and was approaching at

GuDViN [60]

Answer:

a) v = 11.24 m / s , θ = 17.76º b) Kf / K₀ = 0.4380

Explanation:

a) This is an exercise in collisions, therefore the conservation of the moment must be used

Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved

Recall that moment is a vector quantity so it must be kept on each axis

X axis

initial moment. Before the crash

p₀ₓ = m₁ v₁

where v₁ = -25.00 me / s

the negative sign is because it is moving west and m₁ = 900 kg

final moment. After the crash

$p_{x f}$ = (m₁ + m₂) vx

p₀ₓ = p_{x f}

m₁ v₁ = (m₁ + m₂) vₓ

vₓ = m1 / (m₁ + m₂) v₁

let's calculate

vₓ = - 900 / (900 + 1200) 25

vₓ = - 10.7 m / s

Axis y

initial moment

$p_{oy}$ = m₂ v₂

where v₂ = - 6.00 m / s

the sign indicates that it is moving to the South

final moment

p_{fy}= (m₁ + m₂) $v_{y}$

p_{oy} = p_{fy}

m₂ v₂ = (m₁ + m₂) v_{y}

v_{y} = m₂ / (m₁ + m₂) v₂

we calculate

$v_{y}$ = 1200 / (900+ 1200) 6

$v_{y}$ = - 3,428 m / s

for the velocity module we use the Pythagorean theorem

v = √ (vₓ² + v_{y}²)

v = RA (10.7²2 + 3,428²2)

v = 11.24 m / s

now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)

tan θ = $v_{y}$ / Vₓ

θ = tan-1 v_{y} / vₓ)

θ = tan -1 (3,428 / 10.7)

θ = 17.76º

This angle is from the west to the south, that is, in the third quadrant.

b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship

Kf = 1/2 (m1 + m2) v2

K₀ = ½ m₁ v₁² + ½ m₂ v₂²

Kf = ½ (900 + 1200) 11.24 2

Kf = 1.3265 105 J

K₀ = ½ 900 25² + ½ 1200 6²

K₀ = 2,8125 10⁵ + 2,16 10₅4

K₀ = 3.0285 105J

the wasted energy is

Kf / K₀ = 1.3265 105 / 3.0285 105

Kf / K₀ = 0.4380

this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy

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3 years ago