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3 years ago

Carbon and hydrogen are examples of pure substances or

Physics

1 answer:

bogdanovich [222]3 years ago

6 0

Carbon and hydrogen are also examples of elements

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Which of the following has the greatest kinetic energy? A. a mass of 4m at velocity v. B. a mass of 3m at velocity 2v. C. a mass

Artist 52 [7]

Answer:

Option C

Explanation:

Kinetic energy is the energy that the body possesses by virtue of its motion.

The formula for Kinetic energy is given by $\frac{1}{2} mv^2$

Using this formula let us find kinetic energy for the bodies given and find out which is the greatest

A) KE = $\frac{1}{2} (4m)(v^2) = 2mv^2$

B) KE = $\frac{1}{2} (3m)(2v)^2 = 6mv^2$

C) KE = $\frac{1}{2} (2m)(3v)^2 = 9mv^2$

D) KE = $\frac{1}{2} (3)(4v)^2 = 8mv^2$

Comparing these we find that 9mv^2 is the highest.

Hence option C is the answer.

4 0

3 years ago

What is the half-life for the radioactive decay of particle that

pogonyaev

Answer:

ytyug

Explanation:

4 0

3 years ago

Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$

$(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}$

Solving the quadratic equation:

$q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C$

for this values q_1 wil be:

$q_1_o = -1.325 *10^{-6}C | 4.17*10^{-6}C$

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0

3 years ago

A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is

goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, $v_i$ = 1.4 m/s

final velocity of the child, $v_f$ = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, $f_k$ = 41 N

The work done by the child is calculated as;

$\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech} + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96 \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J$

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

5 0

3 years ago

] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I

Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

$\mathtt{=( 0.42\times 9000\times 1055.06) J}$

= 3988126.8 J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec $=\dfrac{750}{3.99}$

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned $= \dfrac{1.1}{100} \times 187.97 \ lb/s$

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0

3 years ago