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3 years ago

if an object is falling on earth at a terminal velocity of 140 km/hr, does that mean that the acceleration is ZERO?

1 answer:

7 0

If he's falling in a straight line and his speed is not changing, that tells you that his acceleration is zero.

And THAT tells you that the forces on him are balanced, the net force acting on him is zero, and his motion is the same as it would be if there were NO force acting on him.

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Describe how a battery works

Whitepunk [10]

Magic, Nah im just kidding. A battery has two parts, the anode and the cathode. Which anode is positive and cathode is negative, which they are connected to the electrolyte. Once they are connected to a device they once start working from separate ends. Which is the flow of energy

6 0

3 years ago

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

2 years ago

A 1200 kg sports car accelerates from 0 m/s to 30 m/s in 10 s. What is the average power of the engine?

DIA [1.3K]

Answer:

3600N

Explanation:

Given: m = 1200kg, Vo = 0m/s, Vf = 30m/s, Δt = 10s

ΣF = ma

we need to find 'a' first, using the definition of 'a' we get equation:

a = (Vf-Vo)/Δt

a = (30m/s)/10s

a = 3 m/s^2

now substitute into top equation

ΣF = ma

Fengine = (1200kg)(3m/s^2)

Fengine = 3600N

5 0

1 year ago

Read 2 more answers

Need help finding the average speed.

Snowcat [4.5K]

Explanation:

To find the average of these numbers, we just have to add the three numbers together and divide by 3.

2.07 + 0. 74 + 1.33 = 4.14. 4.14 / 3 = 1.38
1.09 + 1.40 + 0.31 = 2.8. 2.8 / 3 ≈ 9.3333333/ 9 1/3
0.95 + 1.61 + 0.56 = 3.12 / 3 = 1.04
0.81 + 1.89 + 1.08 = 3.78 / 3 = 1.26

5 0

3 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So