All categories

3 years ago

How much force is required to accelerate a 0.10 gram mosquito at 20m/?

Physics

2 answers:

EastWind [94]3 years ago

8 0

Answer: 0.002 newtons

===================================================

Work Shown:

I'm assuming you meant to write "at 20 m/s^2" at the end. If so, then,

0.10 grams = 0.10/1000 = 0.0001 kg

m = 0.0001 kg = mass

a = 20 m/s^2 = acceleration

F = ma .... newton's second law

F = 0.0001*20

F = 0.002

The force required is 0.002 newtons

uysha [10]3 years ago

6 0

Answer:Force is 2

Explanation:

F=ma

F= 0.10x20

F=2

You might be interested in

A(n) 83 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisher

Anna71 [15]

here we can say that there is no external force on fisherman and dock

so here we will use momentum conservation theory

As per momentum conservation

initial momentum of fisherman + boat = final momentum of fisherman + boat

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

now we will have

$83(3.1) + 139(0) = 83 v + 139 v$

$257.3 = 222v$

$v = 1.16 m/s$

so the speed of boat and fisherman will be 1.16 m/s

3 0

3 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago

PLSSSSS HELP!!!!!!! ASAP!!!!! MARK BRAINLIEST!!!!

Ymorist [56]

Answer: I would say the answer is B.

Explanation: It looks and sounds the most correct.

5 0

3 years ago

Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.

lapo4ka [179]

Answer:

This is how I do it:

5.24 rounded to one significant figure is 5
43.261 rounded to one significant figure is 40.
7.3458 rounded to one significant figure is 7.
5 + 40 + 7 is 52 g

Hope this helps you.

Explanation:

7 0

3 years ago

A 20-kg child is tossed up into the air by her parent. The child is 2 meters off the ground traveling 5 m/s.

nignag [31]

The kinetic energy (KE) is 250 J and the gravitational potential energy (GPE) is 392 J

4 0

3 years ago