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2 years ago

What is the momentum of a 0.000015 kg mosquito flying straight at you with a velocity of 5.20 m/s?

Physics

1 answer:

Setler [38]2 years ago

3 0

Answer:

Momentum, $p=7.8\times 10^{-5}\ N-m$

Explanation:

Given that,

Mass of a mosquito, m = 0.000015 kg

Velocity of the mosquito, v = 5.2 m/s

We need to find the momentum of the mosquito. The momentum of an object is given by :

p = mv

Put all the values in the above formula.

$p=0.000015\ kg\times 5.2\ m/s\\\\p=0.000078\ N-m\\\\\text{or}\\\\p=7.8\times 10^{-5}\ N-m$

So, the momentum of the mosquito is $7.8\times 10^{-5}\ N-m$ .

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A projectile is fired directly upward from the ground with an initial velocity of 112 f/sec its distance s above the ground afte

Travka [436]

Answer:

3.48 seconds

Explanation:

At maximum height Vf=0 m/s

Vf= Vi - g*t

⇒g*t= Vi

⇒t= Vi/g

⇒t= 112/32.17 sec

⇒ t= 3.48 s

so the projectile will achieve its maximum height in 3.48 seconds.

7 0

2 years ago

I'm not really sure how to go about creating the equation, can anyone help me?

AlexFokin [52]

The displacement vector (SI units) is
$\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}$

The speed is a scalar quantity. Its magnitude is
$v= \sqrt{A^{2}t^{2}+A^{2}(t^{3}-6t^{2})^{2}} \\ v=A \sqrt{t^{2}+t^{6}-12t^{5}+36t^{4}} \\ v=At \sqrt{t^{4}-12t^{3}+36t^{2}+1}$

Answer: At√(t⁴ - 12t³ + 36t² + 1)

3 0

3 years ago

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/

Hunter-Best [27]

Answer:

$n_{T} = 31.68\,rev$

Explanation:

The angular acceleration is:

$\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}$

$\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}$

And the angular deceleration is:

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}$

$\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}$

The total number of revolutions is:

$n_{T} = n_{1} + n_{2}$

$n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}$