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3 years ago

a book is sitting still on your desk. are the forces acting on the book balanced or unbalanced? explain.

2 answers:

8 0

<span>They are balanced. If the forces were not balanced the book would move*. In this example, the downward force of gravity on the book is counterbalanced by the upthrust of the desk. </span>

fgiga [73]3 years ago

4 0

The answer to this is balanced. Remember the downward force of gravity on the book is counterbalanced by the upthrust of the desk. Note that moving objects can also have balanced forces acting on them. if an object is moving at a constant speed, the force acting on it are balanced. It they are unbalanced the object would either accelerate or decelerate.

Hope this helped. <span />

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A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16

erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle $\phi$ = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

$V =\pi D( \frac{N}{60} )$

$V = 3.14 (0.034) \frac{(1650)}{60}$

$V = 2.93 \frac{m}{s}$

Form factor for the pinion gear is

Y = 0.303

Now

$K_{v} = \frac{6.1 +0.303}{6.1} = 1.049$

Force on gear tooth

$F = \frac{P}{V}$

$F = \frac{1200}{2.93}$

F = 408.73 N

Now the bending stress is given by the formula

$\sigma = \frac{K_{v} F}{m b y}$

$\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}$

$\sigma$ = 35.38 M pa

This is the value of bending stress on the pinion

8 0

3 years ago

2. A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotatin

ycow [4]

Answer:

$\frac{0.065}{r}$

Explanation:

The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

$v=\sqrt{\mu gr}....................(1)$

where $\mu$ is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.

Given;

v = 0.8m/s

g = $9.81m/s^2$

r = ?

$\mu=?$

In order to solve for $\mu$ , we can simply make it the subject of formula from equation (1) as follows;

$v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)$

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.

Therefore;

$\mu=\frac{0.8^2}{9.81r}$

$\mu=\frac{0.64}{9.81r}\\\\\mu=\frac{0.065}{r}$

8 0

3 years ago

Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,

adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

x_cm = 1 /M ∑ $x_{i}$ m_{i}

y_cm = 1 /M ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

x4=?

y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

M = 2.00 + 2.20 + 3.40

M = 7.6 kg

0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

x4 = -6.8 / 6

x4 = -1,133 m

Axis y

0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

y4 = -11/6

y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0

3 years ago

An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance o

Minchanka [31]

Answer:

B. 0.16 m

Explanation:

The vertical distance by which the player will miss the target is equal to the vertical distance covered by the dart during its motion.

Since the dart is thrown horizontally, the initial vertical velocity is zero:

$v_y = 0$

While the horizontal velocity is

$v_x = 15 m/s$

The horizontal distance covered is

$d_x = 2.7 m$

Since the dart moves by uniform motion along the horizontal direction, the time it takes for covering this distance is

$t=\frac{d_x}{v_x}=\frac{2.7 m}{15 m/s}=0.18 s$

along the vertical direction, the motion is a uniformly accelerated motion with constant downward acceleration g=9.8 m/s^2, so the vertical distance covered is given by

$d_y = \frac{1}{2}gt^2=\frac{1}{2}(9.8 m/s^)(0.18 s)^2=0.16 m$

8 0

3 years ago

1 point

Novay_Z [31]

Answer:

Explanation:

Not sure what your options are but anything that says something like

"at the block surface in contact with the ramp along the line from V to Z" is probably a good shot.

4 0

3 years ago