Which safety precaution is of primary interest to ALL laboratory activities?

Two workers pull horizontally on a heavy box. but one pulls twice as hard as the other. The larger pull is directed at 21.0° wes

pantera1 [17]

Answer:

The magnitude of F1 is

$|F1|=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

Explanation:

<u>Net Force </u>

Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.

The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.

Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

$\displaystyle -2F\ sin21^o+F\ cos\alpha =0$

The sum of the vertical components of F1 and F2 must equal the total force Ft

$\displaystyle -2F\ cos21^o+F\ sin\alpha =460$

Solving for $\alpha$ in the first equation

$\displaystyle cos\alpha =\frac{2F\ sin21^o}{F}=2sin21^o$

$\displaystyle cos\alpha =0.717=>\alpha =44.214^o$

$\displaystyle F(2cos21^o+sin\alpha)=460$

$\displaystyle F=\frac{460}{2cos21^o+sin\alpha}$

$\displaystyle F=\frac{460}{2cos21^o+sin44.214^o}$

$\displaystyle F=179.37\ N$

The magnitude of F1 is

$|F1|=2*F=358.74 \ N$

The magnitude of F2 is

$|F2|=179.37\ N$

And the direction of F2 is

$\alpha = 44.214^o$

4 0

3 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

1 year ago

What are some non examples of temperature ?

Alja [10]

non examples of temperature are dixionanon , fairinheat, cabrowskin, and lastly ancomthere

6 0

3 years ago

Emotions refer to the act or process of knowing. True or False

vaieri [72.5K]

Answer:

True

Explanation:

Because they are central to the human experience and are related to everything we do .

4 0

3 years ago

Number of complete 90.9 MHz radio waves over a 1.50 km distance

zimovet [89]

You could answer this right away IF you knew the length of each wave, right ?

Well, Wavelength = (speed) / (frequency).

Speed = 3 x 10⁸ m/s (the speed of light)
and
Frequency = 90.9 x 10⁶ Hertz.

So the length of each wave is 3 x 10⁸ / 90.9 x 10⁶ meters.

To answer the question, see how many pieces you have to cut
that 1.5 km into, in order for each piece to be 1 wavelength.
It'll be

(1,500 meters) divided by (3 x 10⁸ meters/sec) / (90.9 x 10⁶ Hz)

To divide by a fraction, flip the fraction and then multiply:

(1500 meters) times (90.9 x 10⁶ Hz)/(3 x 10⁸ meters/sec)

= 454.5

5 0

3 years ago