All categories

3 years ago

What do the arrows in this photograph represent

Physics

1 answer:

babunello [35]3 years ago

7 0

Answer: D. An action-reaction force pair

Explanation: When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body. There are two forces resulting from this interaction - a force on the chair and a force on your body. Another example would be a person pushing against a wall (action force), and the wall exerts an equal and opposite force against the person.

You might be interested in

A common cylindrical copper wire used in a lab is 841 m long. Find the radius (in mm) of a wire necessary to have 0.5 Ohms of re

slava [35]

The relationship between the resistance R of a wire and its resistivity $\rho$ is given by
$R= \frac{\rho L}{A}$
where L is the length of the wire and A is its cross sectional area.

In the problem, we have $R=0.5 \Omega$ , $\rho = 1.68 \cdot 10^{-8} \Omega m$ and $L=841 m$ . So we can solve the find the area A:
$A= \frac{\rho L}{R}=2.83 \cdot 10^{-5} m^2$

For a cylindrical wire, the cross sectional area is given by
$A= \pi r^2$
where r is the radius. We know the value of the area A, so now we can find the radius of the wire:
$r= \sqrt{ \frac{A}{\pi} }= \sqrt{ \frac{2.83 \cdot 10^{-5}m^2}{\pi} }=0.003 m=3.0 mm$

3 0

3 years ago

Period vs string length

Galina-37 [17]

Trial 3 is the correct answer

8 0

3 years ago

There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center)

Aneli [31]

Answer:

q₁ =± 1.30 10⁻⁶ C and q₂ = ± 1.28 10⁻⁶ C

Explanation:

We will solve this problem with Coulomb's law

F = K q₁q₂ / r²

Where the Coulomb constant is value 8.99 10⁹ N m² / C²

Let's apply this equation to our problem

Case 1

F1 = k q₁ q₂ / r₁²

Where r₁ = 0.440 m and F1 = 0.0765 N

Case 2

The charges are the same

F2 = k q q / r₂²

With r₂ = r₁ = 0.440 m, the spheres are fixed and the force is F2 = 0.100 N

When the spheres are joined with the wire, the charge is distributed, distributed and matched in the two spheres

q₁ + q₂ = 2 q

Let's replace

F2 = k ½ (q₁ + q₂) / r²

Let's write the two equations and solve the system of equations

F1 = k q₁ q₂ / r²

F2 = ½ k (q₁ + q₂) / r²

F1 r² / k = (q₁ q₂)

F2 r² / k = (q₁ + q₂)/2

q₁ = 2F2 r² / k - q₂

We substitute in the other equation

F1 r² / k = (2F2 r² / k - q₂) q₂

0 = -F1 r² / k + (2F2 r² / k) q₂ - q₂²

Let's solve the second degree equation

F1 r² / K = 0.0765 0.440² / 8.99 10⁹

F1 r² / K = 1.65 10⁻¹²

(2F2 r² / k) =2 0.10 0.44² / 8.99 10⁹

(2F2 r2 / k) = 4.30 10⁻¹²

q₂² - 4.30 10⁻¹² q₂ + 1.65 10⁻¹² = 0

q₂ = ½ {4.30 10⁻¹² ± √ [(4.30 10⁻¹²)² - 4 1.65 10⁻¹²]}

q₂ = ½ {4.30 10⁻¹² ± 2,569 10⁻⁶}

q₂ = ± 1.2845 10⁻⁶ C

Now we calculate q1

F1 = k q₁ q₂ / r²

q₁ = F1 r² / (k q₂)

q₁ = 0.0765 0.440² / (8.99 10⁹ 1.2845 10⁻⁶)

q₁ = 1.30 10⁻⁶ C

3 0

3 years ago

Why the rusting of iron is slow in the deserts than in the coastal areas?

gulaghasi [49]

Answer:

Rusting of iron is slow in desert area because rusting happens when iron articles react with atmospheric oxygen and moisture and produced a substance called rust. in desert areas there is less moisture in air as compared to coastal area therefore rusting is slow.

6 0

3 years ago

When a wind turbine is operating, 15000 kg of air, moving at a speed of 12m/s pass through it in one second driving the turbine

ICE Princess25 [194]

Answer:

15k

Explanation:

5 0

2 years ago