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Artyom0805 [142]
3 years ago
12

How to do this problem

Physics
2 answers:
marusya05 [52]3 years ago
4 0
The answer I'm pretty sure is 3
STALIN [3.7K]3 years ago
3 0
You know the insect's mass.
You know the insect's acceleration.
==> Find the force on the insect. (F=Ma)

==> The force on the car is equal and opposite.
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A college student is working on her physics homework in her dorm room. Her room contains a total of 6.0 x 10^26 gas molecules. A
IceJOKER [234]

Answer:

Temperature, T = 3.62 kelvin

Explanation:

It is given that,

Total number of gas molecules, N=6\times 10^{26}

Her body is converting chemical energy into thermal energy at a rate of 125 W, P = 125 W

Time taken, t = 6 min = 360 s

Energy of a gas molecules is given by :

\Delta E =\dfrac{3}{2}NkT

T=\dfrac{2E}{3Nk}, k is Boltzmann constant

T=\dfrac{2\times P\times t}{3Nk}

T=\dfrac{2\times 125\times 360}{3\times 6\times 10^{26}\times 1.38\times 10^{-23}}

T = 3.62 K

So, the temperature increases by 3.62 kelvin. Hence, this is the required solution.

4 0
3 years ago
A chef fills 50ml container with 43.5 g of cooking oil. what is the density of the oil?
Rama09 [41]

Answer:

0.87 g per ml

Explanation:

D=M/V

43.5/50=0.87

3 0
3 years ago
True or false an experiment in investigating the effects and development variable on the independent variable
Sergeu [11.5K]

true if you are refering to the desing of the experimnt as it does identify the variable

8 0
3 years ago
Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge
Wittaler [7]

Answer:

 F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

        F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

          r₁₂ = y₁ -y₂

          r₁₂ = 0.30 + 0.30

          r₁₂ = 0.60 m

let's calculate

          F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

          F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

            F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

           F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

          Fx = F13x + F12x

          Fx = -1,357 10-2 +0

          Fx = -1.357 10-2 N

          Fy = F13y + F12y

         Fy = 1.0183 10-2 + ​​1 10-1

          Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

         F = Ra (Fx2 + Fy2)

         F = RA [(1.357 10-2) 2 + 0.110183 2]

         F = 0.111015 N

Let's use trigonometry for the angles

         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

6 0
3 years ago
What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a
Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity  = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2

v = r ω

mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}

I = \dfrac{m(2gh - v^2)r^2}{v^2}

I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}

I =mr^2(0.287)

I = 0.287 MR²

3 0
3 years ago
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