Let's break down and solve the problems presented in the image:

Problem 1

A vehicle moves at 60 km/h. Transform this speed to m/s.

We are given the speed in kilometers per hour (km/h), and we need to convert it to meters per second (m/s). We can use the following conversion factor:

$1 \text{ km} = 1000 \text{ m}, \quad 1 \text{ hour} = 3600 \text{ seconds}$

The speed in m/s is calculated as:

$\text{Speed in m/s} = 60 \, \frac{\text{km}}{\text{hr}} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{hr}}{3600 \, \text{s}} = 16.67 \, \text{m/s}$

Thus, the speed is approximately 16.6 m/s.

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Problem 2

A vehicle starts from rest and accelerates to reach a speed of 50 km/h in 15 seconds. Find:

a) The acceleration.

We can use the following kinematic equation to find the acceleration $a$:

$v = u + at$

Where:

• $v = 50 \, \text{km/h} = 50 \times \frac{1000}{3600} \, \text{m/s} = 13.89 \, \text{m/s}$
• $u = 0 \, \text{m/s}$ (since the vehicle starts from rest)
• $t = 15 \, \text{s}$

Substituting the values into the equation:

$13.89 = 0 + a(15)$

Solving for $a$:

$a = \frac{13.89}{15} = 0.926 \, \text{m/s}^2$

So, the acceleration is 0.93 m/s².

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b) The distance traveled.

We can use the following kinematic equation to calculate the distance:

$s = ut + \frac{1}{2}at^2$

Since $u = 0$:

$s = \frac{1}{2} \times 0.93 \times (15)^2 = 0.465 \times 225 = 104.625 \, \text{m}$

Thus, the distance traveled is approximately 104.6 meters.

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Problem 3

The following object is in equilibrium under the action of three forces. Find the unknown forces and the mass of the object.

This problem requires a free-body diagram analysis, which involves decomposing the forces in the horizontal (x) and vertical (y) directions.

• Force 1 ($F_1 = 20 \, \text{N}$) is applied at an angle of $60^\circ$ from the horizontal.
• Force 2 ($F_2 = 40 \, \text{N}$) is applied at an angle of $140^\circ$ from the horizontal.
• Force 3 ($F_3$) is unknown, and it acts along the horizontal axis.

We assume that the object is at equilibrium, so the sum of forces in both the x and y directions must be zero.

For the x-axis:

$F_3 - F_1 \cdot \cos(60^\circ) - F_2 \cdot \cos(140^\circ) = 0$

For the y-axis:

$F_1 \cdot \sin(60^\circ) + F_2 \cdot \sin(140^\circ) = 0$

Using trigonometric values:

• $\cos(60^\circ) = 0.5$
• $\sin(60^\circ) = 0.866$
• $\cos(140^\circ) = -0.766$
• $\sin(140^\circ) = 0.642$

By substituting the values into the equations, you can solve for $F_3$ and the mass $m$ using Newton’s second law.

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If you need further steps to solve the last problem or need additional clarification, feel free to ask!