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3 years ago

Defination of thrust

Physics

2 answers:

Phoenix [80]3 years ago

3 0

push (something or someone) suddenly or violently in the specified direction

sergeinik [125]3 years ago

3 0

The four definitions of thrust:

1. Push (something or someone) suddenly or violently in the specified direction.

2. A sudden or violent lunge with a pointed weapon or a bodily part.

3. The propulsive force of a jet or rocket engine.

4. A reverse fault of low angle, with older strata displaced horizontally over younger.

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My name is Ann [436]

The crate is moving at constant velocity when the forces acting on it are

balanced.

The value of the force <em>F</em> required to pull the crate with constant velocity is; $\underline{F = \sqrt{2} \cdot \mu \cdot m\cdot g}$

Reasons:

Mass of the crate = m

Cross section of through = Right angled

Orientation (inclination) of the through to horizontal = 45°

Coefficient of kinetic friction = μ

Required:

The value of the required to pull the crate along the through at constant

velocity.

Solution:

When the through is moving at constant velocity, we have;

Friction force acting on crate = Force pulling the crate

Friction force = Normal reaction × Coefficient of kinetic friction

Normal reaction on an inclined plane = $\mathbf{F_N}$

Each side of the through gives a normal reaction.

The vertical component of the normal reaction on each side of the through

is therefore;

$F_N$ ·j = $\mathbf{F_N}$ × sin(θ)

The sum of the vertical component = $F_N$ ·j + $F_N$ ·j = 2· $F_N$ ·j = 2· $F_N$ ×sin(θ)

The sum of the vertical component of the normal reactions = The weight of the crate

Therefore;

2· $F_N$ ×sin(θ) = m·g

θ = 45°

Therefore;

2· $F_N$ ×sin(45°) = m·g

$\displaystyle sin(45^{\circ}) = \mathbf{\frac{\sqrt{2} }{2}}$

Therefore;

$\displaystyle 2 \cdot F_N \cdot sin(45^{\circ}) = 2 \cdot F_N \times \frac{\sqrt{2} }{2} = \sqrt{2} \cdot F_N$

$\displaystyle F_N = \mathbf{ \frac{m \cdot g}{\sqrt{2} }}$

Which gives;

$\displaystyle Force \ required, \ F = Sum \of \ friction \ forces \ = 2 \times F_N \times \mu = \mathbf{ 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu}$

$\displaystyle Force \ required, \ F = 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu = \mathbf{ \sqrt{2} \cdot \mu \cdot m \cdot g}$

Force required to pull the crate at constant velocity, <u>F = √2·μ·m·g</u>

Learn more about force of friction here:

brainly.com/question/6561298

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2 years ago

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Yes ples omggggggggg

Alexxx [7]

Answer:

The answer is A.

Explanation:

They both work together to circulate oxygen and blood throughout the body.

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3 years ago

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Two spheres having masses M and 2M and radii R and 3R, respectively, are released from rest when the distance between their cent

Andrei [34K]

Answer:

$v_2 = \sqrt{\frac{GM}{3R}}$

$v_1 = 2\sqrt{\frac{GM}{3R}}$

Explanation:

As we know by energy conservation that change in gravitational potential energy of the system = change in kinetic energy of the two ball

So here we can say

$-\frac{GM(2M)}{12R} + 0 = -\frac{GM(2M)}{4R} + \frac{1}{2}Mv_1^2 + \frac{1}{2}(2M)v_2^2$

Also since there is no external force on the system of two masses so here total momentum of the two balls will remains conserved

$0 = Mv_1 + 2Mv_2$

$v_1 = -2v_2$

now we have

$\frac{GM^2}{2R} - \frac{GM^2}{6R} = \frac{1}{2}M(-2v_2)^2 + \frac{1}{2}(2M)v_2^2$

$\frac{GM^2}{3R} = Mv_2^2$

$v_2 = \sqrt{\frac{GM}{3R}}$

$v_1 = 2\sqrt{\frac{GM}{3R}}$

4 0

4 years ago