Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal speeds of 4.0 m/s at an instant when the distance separating the two is equal to 25 cm. How far apart will they be when closest to one another?

Answers

Answer 1

Answer:

Answer:

r₁ = 20.5 cm

Explanation:

In this exercise we can use the conservation of energy

the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign

starting point.

Em₀ = U_g + U_e + K = $-G (m_1m_2)/(r) +k (q_1q_2)/(r) - 2 ( (1)/(2) m v^2)$

the two in the kinetic energy is because they are two particles

final point. When it is detained

Em_f = U_g + U_e = $-G (m_1m_2)/(r_1) + k (q_1q_2)/(r_1)$

the energy is conserved

Em₀ = em_f

the charges and masses of the two particles are equal

$-G (m^2)/(r) + k (q^2)/(r) + m v^2 = - G (m^2)/(r_1) + k (q^2)/(r_1)$

sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

-5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁ + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

3.24 10⁻⁵ - 7.2 10⁻⁵ = 8.1 10⁻⁶ / r₁

-3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

r₁ = 2.045 10⁻¹ m

r₁ = 20.5 cm

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A novice scuba diver practicing in a swimming pool takes enough air from his tank to fully expand his lungs before abandoning the tank at depth L and swimming to the surface. He ignores instructions and fails to exhale during his ascent. When he reaches the surface, the difference between the external pressure on him and the air pressure in his lungs is 9.3 kPa. From what depth does he start?

Answers

Answer:

11.625 m.

Explanation:

Difference of pressure will be due to hydro-static pressure due to column of water of height L.

Pressure of water column = L d g , where L is depth ,

d is density of water = 1000kg /m³

g = 9.8 ms²

Pressure difference = 9.3 kPa = 9300 Pa

So Ldg = 9300

L X 1000 X 0.8 =9300

800 L = 9300

L = 11.625 m.

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answers

Answer:

425.1 W

Explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W

How do impacts by comets and asteroids influence Earth’s geology, its atmosphere, and the evolution of life?

Answers

Answer:

Explanation:

A comet is a celestial body made up of ice and dust and assumed to have a tail.

As per considering ancient history, they are even termed as death-dealers and have doomed many planets.

If we talk about Earth, comets and asteroids have played a massive role in changing Earth's geology, atmosphere and evolution. Its believed that dinosaurs population is wiped out from Earth because of comets that falls during that time. Many new diseases are even brought by them.

Large comets can result into global environmental damage and can even lead to mass extension.The dust from the impact and the heat creates many harmful oxides resulting into acid rain and can kill thousand of organism.

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.

Answers

Answer:

av=0.333m/s, U=3.3466J

$v_(A2)=-1.333m/s,\n v_(B2)=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\n\nP=mv\n\n\therefore m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_(B2)-v_(A2)$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_(B2)=v_(A2)=v_2\n\n2* 2+10* 0=2v_2+10v_2\n\nv_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_(el,1)=K_2+U_(el,2)\n\n#Springs \ in \ equilibrium \ before \ collision\n\nU_(el,2)=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\n\nU_(el,2)=0.5* 2* 2^2-0.5(2+10)(0.333)^2\n\nU_(el,2)=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

$2*2+10*0=2v_A_2+10v_B_2\n\n2=v_A_2+5v_B_2\n\n#Eqtn 2:\nv_A_1-v_B_1=v_(B2)-v_(A2)\n\n2-0=v_(B2)-v_(A2)\n\n2=v_(B2)-v_(A2)\n\n#Solve \ to \ eliminate \ v_(A2)\n\n6v_(B2)=2.0\n\nv_(B2)==0.667m/s\n\n#Substitute \ to \ get \ v_(A2)\n\nv_(A2)=(4)/(6)-2=1.333m/s$

The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?

Answers

Final answer:

The problem discusses the change in acceleration when a passenger is added to a car. It requires understanding of Newton's second law of motion, force equals mass times acceleration, and then recalculating the acceleration with the passenger added to the total mass.

Explanation:

This problem pertains to Newton's second law of motion, stating that the force applied on an object equals its mass times its acceleration (F = ma). Given that the initial acceleration of the Lamborghini Huracan with a driver is 0.80g or 0.80*9.80 m/s², we can calculate the force applied by the car. By multiplying the car's mass (1510 kg) with its acceleration, we will find the force.

Οnce we have the force, we can calculate the new acceleration if the 80 kg passenger rode along. Given that the force is constant, we determine the car's new acceleration by dividing this force with the new total mass (car mass + passenger's mass). So the question ultimately requires an application of the concepts of force, mass, and acceleration.

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Final answer:

The new acceleration of the Lamborghini Huracan with an added passenger can be calculated by finding the initial force using the car's mass and acceleration, and then using this force with the increased mass (original mass + passenger's mass) to find the new acceleration. The new acceleration will be less than the initial acceleration due to the increased mass.

Explanation:

To determine the new acceleration of the Lamborghini Huracan with an added passenger, we first calculate the initial force acting on the car. This can be done by using Newton's second law which states that Force = mass * acceleration. Initially, the acceleration is 0.80g (where g is acceleration due to gravity = 9.81 m/s²), and the mass is 1510 kg (including the driver). Therefore, the initial force = 1510 kg * 0.8 * 9.81 m/s².

However, when an 80-kg passenger rides along, the total mass becomes 1510 kg + 80 kg = 1590 kg. To find the new acceleration, we keep the force constant (as it is not affected by the introduction of the passenger) and rearrange the formula F = m*a as a = F/m. Use the increased mass to find the new acceleration. Please note that the new acceleration will be less than the initial acceleration due to increased mass.

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A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spectrometer. What is the mass of this ion?