What’s 2.3 miles into kilometers

There are two forces on the 2 kg box in the overhead view of the following figure, but only one is shown. For F1=20N, a= 12 m/s2

maw [93]

Answer:

second force = 32.784

Magnitude = $\sqrt{32.784$

θ = -90°

Explanation:

a)

Fnet = ma

F1 + F2 = ma

20N + F2 = 2(12 × cos30° + 12 ×sin30°)

F2 = 2 × 12 ( sin 30° + cos 30°)

= 24 × ( 1 + √3 )÷ 2

=12 (1 +√3 )

= 32.784

b) $\sqrt{12(1 +\sqrt{3}}$

= $\sqrt{12 ( 1+ 1.732)}$

= $\sqrt{12 (2.732)}$

= $\sqrt{32.784}$

c)

θ = 30° + 180°

θ = 210°

210° - 300°

θ = -90°

8 0

2 years ago

A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in

Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

$\boxed{u \ or \ v_i = initial \ velocity}$
$\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}$
$\boxed{a = acceleration \ (constant)}$
$\boxed{d = distance \ travelled}$

The formula we follow for this problem are as follows:

$\boxed{ \ v^2 = u^2 + 2ad \ }$

a = acceleration (in m/s²)
u = initial velocity
v = final velocity
d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

$\boxed{ \ 0 = v^2 + 2ad \ }$

$\boxed{ \ v^2 = -2ad \ }$

Both sides are divided by -2d, we get $\boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }$

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

$\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }$

Here d' is the stopping distance that we want to look for.

$\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }$

We crossed out 2 in above and below.

$\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }$

We multiply both sides by d.

$\boxed{ \ v^2 d' = 49.0v^2 d \ }$

We crossed out v^2 on both sides.

$\boxed{\boxed{ \ d' = 49.0d \ }}$

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>

Determine the acceleration of the stuffed bear brainly.com/question/6268248
Particle's speed and direction of motion brainly.com/question/2814900
About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0

2 years ago

Read 2 more answers

A small cylinder rests on a circular turntable that is rotating clockwise at a constant speed. Which set of vectors gives the di

I am Lyosha [343]

The question is missing the diagram. Also, the choices must have pictorial representation. So, I have attached the missing diagram and the pictorial representation of the vectors.

Answer:

The correct representation is attached below. Force and acceleration will be towards the center of rotation while the velocity will be along the tangent to the circular motion. <u>Option (D).</u>

Explanation:

From the figure, we can conclude the following points:

1. The cylinder is under a uniform circular motion as the circular table is moving at constant speed.

2. For a circular motion, velocity acts along the tangent to the circular path.

3. For a circular motion, centripetal force acts on the body that causes it move around a circular path.

4 The direction of the centripetal force is radially inward towards the center of rotation.

5. The centripetal force causes a centripetal acceleration acting on the body.

6. From Newton's second law, the net acceleration of a body is in the same direction as that of the net force acting on it. So, centripetal acceleration also acts in the radially inward direction.

Therefore, from the above conclusions, it is clear that velocity will act in the horizontal direction at the given instance of time and force and acceleration will act vertically down for the given instance.

This is shown in the picture below. The option (D).

4 0

2 years ago

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined

MrRa [10]

Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

$F_{B}=-5755N$

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

$\frac{mv^{2} }{R}=W+F_{B}\\ F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N$

The minus sign for downward direction

6 0

2 years ago

La altura de un tornillo de banco respecto a la superficie es de 80 cm expresar dicha medida en pies..

Andrej [43]

Answer:

this measurement if feet is: 2.624672 ft

Explanation:

Notice that 80 cm can be expressed as 0.8 meters, and In order to convert from meters to feet, one needs to multiply the meter measurement times 3.28084. Therefore:

0.80 m can be written in feet as: 0.80 * 3.28084 feet = 2.624672 feet

3 0

2 years ago

What’s 2.3 miles into kilometers￼

What’s 2.3 miles into kilometers