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3 years ago

Find the magnitude of the sum

Physics

1 answer:

Hunter-Best [27]3 years ago

7 0

Answer:

8.57 m

Explanation:

To solve the problem, we have to decompose the two vectors along the two directions first:

Vector A:

- x component: Ax = +9.66 m

- y compoment: Ay = 0 (the vector lies along the x-axis)

Vector B:

- x component: $B_x = -(12.0 m) cos 45^{\circ}=-8.49 m$

- y component: $B_y = (12.0 m) sin 45^{\circ}=8.49 m$

So now we can find the sum of the two vectors by adding the components along each axis:

$R_x = A_x + B_x = 9.66 m - 8.49 m = 1.17 m$

$R_y = A_y + B_y = 0 + 8.49 m = 8.49 m$

And the magnitude of the sum is given by Pythagorean theorem:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(1.17 m)^2+(8.49 m)^2}=8.57 m$

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in the diagram, q1,q2, and q3 are in a straight line. each of these particles has a charge of -2.35x10^-6 C. particles q1 and q2

julia-pushkina [17]

The magnitude of the net force that is acting on particle q₃ is equal to 6.2 Newton.

Given the following data:

Charge = $-2.35 \times 10^{-6}$ C.

Distance = 0.100 m.

Scientific data:

Coulomb's constant = $8.988\times 10^9 \;Nm^2/C^2$

<h3>How to calculate the net force.</h3>

In this scenario, the magnitude of the net force that is acting on particle q₃ is given by:

F₃ = F₁₃ + F₂₃

Mathematically, the electrostatic force between two (2) charges is given by this formula:

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

Where:

q represent the charge.

r is the distance between two charges.

k is Coulomb's constant.

Note: d₁₃ = 2d₂₃ = 2(0.100) = 0.200 meter.

For electrostatic force (F₁₃);

$F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.200^2}\\\\F_{13} = \frac{0.0496}{0.04}$

F₁₃ = 1.24 Newton.

For electrostatic force (F₂₃);

$F_{13} = 8.988\times 10^9 \times \frac{(-2.35 \times 10^{-6} \times [-2.35 \times 10^{-6}])}{0.100^2}\\\\F_{13} = \frac{0.0496}{0.01}$

F₂₃ = 4.96 Newton.

Therefore, the magnitude of the net force that is acting on particle q₃ is given by:

F₃ = 1.24 + 4.96

F₃ = 6.2 Newton.

Read more on charges here: brainly.com/question/14372859

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2 years ago

Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring c

klemol [59]

Answer:

$v_0=17.3m/s$

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

$E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2$

Since we only care about the velocity $v_0$ , we can keep only the second and third parts of the equation and solve:

$mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s$

So, the speed of the ball just after the launch is 17.3m/s.

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3 years ago

How long would it take to travel one light year?

Llana [10]

It depends on what speed you are going at. Assuming you are in our fastest spacecraft as of now which has a speed of 17,500 mph
1 light second = 186,282 mph
There are 31,536,000 in one year so 17,500x=31,534,000 and your answer would be in years. It would take you roughly 1802 years.
(Sorry about mph I'm American.)

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3 years ago

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Rudik [331]

It is true my dude.

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3 years ago

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Suppose you have a car with a battery that applies 12.5 V to the starter.

Marat540 [252]

Answer:

R = 0.1 ohms

Explanation:

It is given that,

Voltage of the battery, V = 12.5 V

Current flowing in the car's starter, I = 125 A

We need to find the effective resistance of a car's starter. It can be calculated using Ohm's law. Let R is the resistance.

$V=IR\\\\R=\dfrac{V}{I}\\\\R=\dfrac{12.5}{125}\\\\R=0.1\ \Omega$

So, the resistance of the car's starter is 0.1 ohms.

7 0

3 years ago