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1 year ago

7

The label on the box of cleanser states that it contains ch3cooh. what elements are in this compound.

1 answer:

Alex17521 [72]1 year ago

6 0

It contains Carbon, Hydrogen and Oxygen

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One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned

liq [111]

Answer:

$T = 3183 N$

Explanation:

When the screw is turned by two turns then change in the length of the wire is given as

$\Delta L = 2(2\pi r)$

$\Delta L = 4\pi r$

$\Delta L = 4(\pi)(1.0 mm)$

$\Delta L = 12.56 mm$

now we know by the formula of Young's modulus

$Y = \frac{T/A}{\Delta L/L}$

so we have

$T = \frac{AY \Delta L}{L}$

$T = \frac{\pi (0.55 \times 10^{-3})^2(2.0 \times 10^{11})(12.56 \times 10^{-3})}{0.75}$

$T = 3183 N$

3 0

2 years ago

Rotation of the lever OA is controlled by the motion of the contacting circular disk of radius r = 300 mm whose center is given

sergiy2304 [10]

Answer:

The angular velocity is

5.64rad/s

Explanation:

This problem bothers on curvilinear motion

The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s

We know that the velocity v is given as

v= ωr

Where ω is the angular velocity

r is 300mm to meter = 0.3m

the radius of the circle

described by the level

v=1.64m/s

Making ω subject of the formula and solving we have

ω=v/r

ω=1.64/0.3

ω=5.46 rad/s

3 0

1 year ago

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,

seropon [69]

Answer:

The angular frequencies of all the 3 pendulums shall be same.

Explanation:

The time period of a simple pendulum with the approximation $sin(\theta )\approx \theta$ is given by:

$T=2\pi\sqrt{\frac{l}{g}}$

The angular frequency $\omega$ is given by

$\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}$

As we can see that the angular frequency is independent on the initial angle (valid strictly for small angle approximations) we conclude that the angular frequencies of the 3 pendulums are the same.

6 0

2 years ago

Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long peri

maksim [4K]

Answer:

The activity is 811.77 Bq

Solution:

As per the question:

Half life of Thorium, $t_{\frac{1}{2}} = 1.405\times 10^{10}\ yrs$

Mass of Thorium, m = 200 mg = 0.2 g

M = 232 g/mol

Now,

No. of nuclei of Thorium in 200 mg of Thorium:

$N = \frac{N_{o}m}{M}$

where

$N_{o }$ = Avagadro's number

Thus

$N = \frac{6.02\times 10^{23}\times 0.2}{232} = 5.19\times 10^{20}$

Also,

Activity is given by:

$\frac{0.693}{t_{\frac{1}{2}}}\times N$

$A= \frac{0.693}{1.405\times 10^{10}}\times 5.19\times 10^{20} = 2.56\times 10^{10}\ \yr$

$A = \frac{2.56\times 10^{10}}{365\times 24\times 60\times 60} = 811.77\ Bq$

5 0

2 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

ANTONII [103]

1) 9.18 s

In the first part of the motion, the rocket accelerates at a rate of

$a_1=13.5 m/s^2$

For a time period of

$t_1=3.50 s$

So we can calculate the velocity of the rocket after this time period by using the SUVAT equation:

$v_1=u+a_1t_1$

where u = 0 is the initial velocity of the rocket. Substituting a1 and t1,

$v_1=(13.5)(3.50)=47.3 m/s$

In the second part of the motion, the rocket decelerates with a constant acceleration of

$a_2 = -5.15 m/s^2$

Until it comes to a stop, to reach a final velocity of

$v_2 = 0$

So we can use again the same equation

$v_2 = v_1 + a_2 t_2$

where $v_1 = 47.3 m/s$ . Solving for t2, we find after how much time the rocket comes to a stop:

$t_2 = -\frac{v_1}{a_2}=-\frac{47.3}{5.15}=9.18 s$

2) 299.9 m

We have to calculate the distance travelled by the rocket in each part of the motion.

The distance travelled in the first part is given by:

$d_1 = ut_1 + \frac{1}{2}a_1 t_1^2$

Using the numbers found in part a),

$d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m$

The distance travelled in the second part of the motion is

$d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2$

Using the numbers found in part a),

$d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m$

So, the total distance travelled by the rocket is

d = 82.7 m + 217.2 m = 299.9 m

6 0

2 years ago