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3 years ago

A 50-kilogram object in outer space is attracted to a nearby planet with a

Physics

1 answer:

professor190 [17]3 years ago

8 0

Answer: 8m/s^2

Explanation:

By Newton's second law we have that.

F = m*a

Where F is force, m is mass and a is acceleration.

We know that m = 50kg and F = 400N

400N = 50kg*a

a = (400/50) m/s^2 = 8m/s^2

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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

How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

4 years ago

Which is a common disease involving the sores on the esophagus, stomach, and/or small intestine?

Elanso [62]

Answer: Peptic Ulcer

7 0

3 years ago

A shock wave first forms when an object ___ the speed of sound.

Anna11 [10]

a shock wave first forms when an object Reaches the speed of sound When an object reaches the speed of sound, it will create some sort of disturbance in the air that created discontinuous change in pressure, medium density and the temperature around it, which will eventually became a shock wave.

5 0

3 years ago

Read 2 more answers

A 0.150-kg lump of clay is dropped from a height of 1.15 m onto the floor. it sticks to the floor and does not bounce.

Serggg [28]

Missing question: "What is the magnitude of the impulse I imparted to the clay by the floor during the impact?"

Solution:
The impulse is equal to the variation of momentum of the object:
 $I=\Delta p=m \Delta v$
where m is the mass of the object and $\Delta v$ is the variation of velocity of the object during the impact.

By using energy conservation, we can calculate the velocity of the object before the impact. In fact, the initial potential energy of the object is all converted into kinetic energy before the impact:
$mgh= \frac{1}{2}mv^2$
from which we find v:
$v= \sqrt{2gh}= \sqrt{2 \cdot 9.81 m/s^2 \cdot 1.15 m}=4.75 m/s$

This is the velocity of the object just before the impact. After the impact, the object comes to rest: this means that the variation of velocity of the object is equal to 4.75 m/s. Therefore, the impulse is

$I=m \Delta v=(0.150 kg)(4.75 m/s)=0.71 Kg m s^{-1}$

6 0

3 years ago