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sergeinik [125]

3 years ago

8

A 4kg mass traveling eastwards at 4m.s per second collides with a 3kg mass traveling westward as 8m.s per second..calculate the

total momentum of mass before they colide

1 answer:

quester [9]3 years ago

7 0

The following instructions are my idea.

P=m*v

1. notice that the to masses move in opposite directions -> one has positive and one has negative speed

2. P1=4*(-4)=-16[kg*m/s]
P2=3*8=24[kg*m/s]
3. Add the momentums:

P1+P2=-16+24=8[kg*m/s]

The total momentum of mass equals 8 kg*m/s

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A constant force of 2.5 N to the right acts on a 4.5 kg mass for 0.90 s.

Alborosie

Answer:

(a) $v_f=0.5\frac{m}{s}$

(b) $v_f=-11\frac{m}{s}$

Explanation:

(a) Since a constant external force is applied to the body, it is under an uniformly accelerated motion. Using the following kinematic equation, we calculate the final velocity of the mass if it is initially at rest( $v_0=0$ ):

$v_f=v_0+at\\v_f=at(1)$

According to Newton's second law:

$F=ma\\a=\frac{F}{m}(2)$

Replacing (2) in (1):

$v_f=\frac{F}{m}t\\v_f=\frac{2.5N}{4.5kg}(0.9s)\\v_f=0.5\frac{m}{s}$

(b) In this case we have $v_0=-11.5\frac{m}{s}$ . So, we use the final velocity equation:

$v_f=v_0+at\\v_f=v_0+\frac{F}{m}t\\v_f=-11.5\frac{m}{s}+\frac{2.5N}{4.5kg}(0.9s)\\v_f=-11\frac{m}{s}$

8 0

3 years ago

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

Viktor [21]

Answer:

A

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

B

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

4 0

3 years ago

Anna learned that every magnet has a north and a south end called poles. One day, while she is playing with magnets she notices

My name is Ann [436]

Just from that one observation, choice-B
would be a reasonable conclusion.

4 0

3 years ago

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What physical property makes the red light different from blue light, or radio waves different from microwaves?

inna [77]

Only their wavelength does.

Blue light waves have only roughly half the wavelength of red light waves, and the so-called "microwaves" are the radio waves with the shortest wavelengths.

4 0

3 years ago

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If carbon has an atomic number of 6, how many protons and neutrons are found in the carbon-14 atom? A.

Alex Ar [27]

The correct answer is B. 6 protons and 8 neutrons

Carbon-14 has same atomic number of 6. It has a nucleon number of 14
Atomic number = proton number = 6
Neutron number = nucleon number - atomic number = 14 - 6 = 8

Hope it helped!

5 0

3 years ago