List of places in our bodies where HIV is present

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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

8 0

2 years ago

Read 2 more answers

Which sentence uses the correct adverb to make a comparison?

Arturiano [62]

Option C would be the right one

8 0

3 years ago

When you drop a 0.43 kg apple, Earth ex erts a force on it that accelerates it at 9.8m / (s ^ 2) toward the earth's surface. Acc

Sever21 [200]

Answer:

The answer to the question is highlighted in the box

8 0

1 year ago

Will mark the brainliest

AysviL [449]

Answer:

The impulse transferred to the nail is 0.01 kg*m/s.

Explanation:

The impulse (J) transferred to the nail can be found using the following equation:

$J = \Delta p = p_{f} - p_{i}$

Where:

$p_{f}$ : is the final momentum

$p_{i}$ : is the initial momentum

The initial momentum is given by:

$p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}$

Where 1 is for the hammer and 2 is for the nail.

Since the hammer is moving down (in the negative direction):

$v_{1_{i}} = -10 m/s$

And because the nail is not moving:

$v_{2_{i}}= 0$

$p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s$

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

$p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Since the hammer and the nail are moving in the negative direction:

$v_{1_{f}}$ = $v_{2_{f}}$ = -9.7 m/s

$p_{f} = -9.7 m/s(7 \cdot 10^{-3} kg + 0.25 kg) = -2.49 kg*m/s$

Finally, the impulse is:

$J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s$

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!

7 0

3 years ago

A series-parallel circuit consists of two parallel circuits connected in series across a 45-V source. One parallel branch consis

musickatia [10]

Answer:

The answer to the question is

The current through R4 = 0.5865 mA

Explanation:

To solve this we list out the known thus

Voltage source = 45-V

Firsrt parallel circiuit has resistances of R1 = 17 kΩ and R2 = 23 kΩ

Second parallel circiuit has resistances of R3 = 45 kΩ and R4 = 55 kΩ

we first find the current flowing in the circuit by finding thr sum of the total resistance in eah parallel circuit

Firsrt parallel circiuit, for circuit in parallel, sum of resistance 1/RT1 = 1/ R1 + 1/R2 = 1/17+1/23 = 0.1023017 therefore RT1 = 1/0.1023017 = 9.775 kΩ

Similarly we have for the second parallel circuit 1/RT2 = 1/R3 + 1/R4

= 1/45 + 1/55 = 0.0404 Hence RT2 = 1/0.0404 = 24.75 kΩ

This means that the 9.775 kΩ and the 24.75 kΩ are in series hence total resistance of the circuit = sum of all resistances in series

= 9.775 kΩ + 24.75 kΩ = 34.525 kΩ

However current, I is given by V/R = 45-V/34.525 kΩ = 1.303 × 10⁻³ A or 1.303 mA

From the current divider rule, I4 = I × (R4/(R3+R4)

That is the currrent flowing throuhgh R4 = 1.303 × (45/(45+55)) = 0.5865 mA

3 0

4 years ago