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3 years ago

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what occurs in a chemical reaction? A. Reactants are formed without chemical bonds being broken B. Reactants are formed from pro

ducts by the breaking and forming of bonds C. Products are formed without chemical bonds being broken D. Products are formed from reactants by the breaking and forming of new bonds

2 answers:

Ann [662]3 years ago

3 0

The answer is D. Products are formed from reactants by the breaking and forming of new bonds.

tekilochka [14]3 years ago

3 0

The correct answer would be D also

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In _____, marketers determine price based on what consumers are willing to pay and then subtract desired margins to yield target

Alona [7]

I would have to say "a. cost-based pricing"

3 0

3 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

4.72 A full-wave bridge-rectifier circuit with a 1-k load operates from a 120-V (rms) 60-Hz household supply through a 12-to-1 s

melisa1 [442]

Answer:

a) 12.74 V

b) Two pairs of diode will work only half of the cycle

c) 8.11 V

d) 8.11 mA

Explanation:

The voltage after the transformer is relationated with the transformer relationshinp:

$V_o=Vrms*\frac{1}{12}\\V_o=10Vrms$

the peak voltage before the bridge rectifier is given by:

$V_{op}=Vo*\sqrt{2}\\V_{op}=14.14V$

The diodes drop 0.7v, when we use a bridge rectifier only two diodes are working when the signal is positive and the other two when it's negative, so the peak voltage of the load is:

$V_l=V_{op}-2(0.7)\\V_l=12.74V$

As we said before only two diodes will work at a time, because the signal is half positive and half negative,so two of them will work only half of the cycle.

The averague voltage on a full wave rectifier is given by:

$V_{avg}=2*\frac{V_l}{\pi}\\V_{avg}=8.11V$

Using Ohm's law:

$I_{avg}=\frac{V_{avg}}{R}\\\\I_{avg}=8.11mA$

7 0

3 years ago

A concave (convergent) mirror has a focus of 3.4 cm. A tree that is 0.5cm tall is situated 4.5 cm away from the mirror. Calculat

Elanso [62]

The image position is 13.9 cm from the mirror

Explanation:

To solve this problem, we can use the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem, we have:

f = 3.4 cm is the focal length of the mirror (positive for a concave mirror)

p = 4.5 cm is the position of the object (the tree)

Solving the equation for q, we find the position of the image:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{3.4}-\frac{1}{4.5}=0.0719 cm^{-1}\\q=\frac{1}{0.0719}=13.9 cm$

Learn more about mirrors:

brainly.com/question/8737441

#LearnwithBrainly

3 0

3 years ago

(this is somehow part of my science unit, dont ask why)

nirvana33 [79]

Answer:

Height of coffee mug = 5 cm

Explanation:

Changing all the data given in to unit cm

50 mm = 5 cm

20 cm = 20 cm

8 m = 800 cm

0.1 km = 100 m = 10000 cm

Since we need to find the height of coffee mug, it's height will not be more than 10 cm.

So 5 cm is the correct choice.

6 0

4 years ago