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3 years ago

How to prove3 equations of motion graphically

1 answer:

6 0

Consider an object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time t = 0 and t = t are u and v respectively. During time t, let s be the total distance travelled by the object.

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Can someone solve this problem and explain to me how you got it

Zarrin [17]

1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

r' = 5r

Therefore, the new force between the charges is:

$F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}$

So, the force has changed by a factor of 25.

The original force between the two charges is

$F=k\frac{q_1 q_2}{r^2}$

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

- The charges are both changed by a factor of 8:

$q_1' = 8q_1$

$q_2' = 8q_2$

Substituting into the equation, we find the new force:

$F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F$

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

Which one i need help

Lostsunrise [7]

Answer:

Explanation:

7 0

3 years ago

An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe —

sattari [20]

Answer:

r = 2,026 10⁹ m and T = 2.027 10⁴ s

Explanation:

For this exercise let's use Newton's second law

F = m a

where the force is electric

F = $k \frac{q_1q_2}{r^2}$

Acceleration is centripetal

a = v² / r

we substitute

$k \frac{q_1q_2}{r^2} = m \frac{v^2}{r}$

r = $k \frac{q_1q_2}{m \ v^2}$ (1)

let's look for the charge in the insulating sphere

ρ = q₂ / V

q₂ = ρ V

the volume of the sphere is

v = 4/3 π r³

we substitute

q₂ = ρ $\frac{4}{3}$ π r³

q₂ = 3 10⁻⁹ $\frac{4}{3}$ π 4³

q₂ = 8.04 10⁻⁷ C

let's calculate the radius with equation 1

r = 9 10⁹ 1.6 10⁻¹⁹ 8.04 10⁻⁷ /(9.1 10⁻³¹ 628 10³)

r = 2,026 10⁹ m

this is the radius of the electron orbit around the charged sphere.

Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio

v = x / t

the distance traveled in a circle is

x = 2π r

In this case, time is the period

v = 2π r /T

T = 2π r /v

let's calculate

T = 2π 2,026 10⁹/628 103

T = 2.027 10⁴ s

4 0

3 years ago

2. In the same tournament, a player is positioned 35 m (40° W of S] of the net. He shoots the puck

andrezito [222]

Answer:

The displacement of the net from player 2 in component form = (-47.498î - 26.812j)

The displacement of the net from player 2 in statement form is 54.54 m and 29.44° (S of W) or 60.56° (W of S)

Explanation:

The sketch of the bearings described in the question is presented in the attached image to this solution.

Method 1

Using component method

Taking the player 1's position as the origin,

The displacement of the player 2 from the origin is (25î) m

The displacement of the net from the origin is 35[(sin θ)î + (cos θ)j]

But θ is the angle of the net's displacement reading from the positive x-axis in the anticlockwise direction. θ = 230°

Displacement of the net from the origin = 35[(cos 230°) + (cos 230°)]

= 35[-0.6428î - 0.7660j]

= (-22.498î - 26.812j) m

In component form, taking note of the directions of the respective displacements calculated (check the attached image)

(The displacement of the net from player 1) = (The displacement of player 2 from player 1) + (The displacement of the net from player 2)

Since we have agreed that player 1 is the origin

(The displacement of the net from origin) = (The displacement of player 2 from origin) + (The displacement of the net from player 2)

(-22.498î - 26.812j) = (25î) + (The displacement of the net from player 2)

The displacement of the net from player 2 = (-22.498î - 26.812j) - (25î) = (-47.498î - 26.812j)

The magnitude of this displacement = √[(-47.498)² + (-26.812)²]

= √(2,256.060004 + 718.883344) = 54.54 m

Direction = tan⁻¹ (-26.812/-47.498) = 209.44° (the signs on the components show that the direction is the third quadrant from the positive x-axis in the anti-clockwise direction)

Hence, the displacement of the net from player 2 is 54.54 m and 29.44° (S of W)

Method 2

Using trignometry,

We will use cosine and sine rule to obtain the required magnitude and direction of the displacement of the net from player 2

Cosine rule

Magnitude = √[35² + 25² - (2×25×35×cos 130°)] = √2,974.8783169514 = 54.54 m

Sine rule

(Sin θ)/35 = (Sin 130°)/54.54

Sin θ = (35 × Sin 130°)/54.54 = 0.4916

θ = Sin⁻¹ (0.4916) = 29.44°

This answer matches the answers from method 1.

Hope this Helps!!!