A system of 1223 particles, each of which is either an electron or a proton, has a net charge of -5.328×10-17 C. How many protons are in this system (exactly)?

Answers

Answer 1
Answer:

Answer:

Therefore the number of proton in the given system is 450.

Explanation:

Given that, a system has 1223 particles.

Let x number of proton be present in the system.

Then the number of electron is =(1223-x)

The charge of a proton is = 1.602×10⁻¹⁹ C

The charge of an electron = - 1.602×10⁻¹⁹ C

The charge of x protons is =( 1.6×10⁻¹⁹×x) C

The charge of (1223-x) electrons is = - 1.6×10⁻¹⁹ (1223-x) C

According to the problem,

(1.6×10⁻¹⁹×x) +{ - 1.6×10⁻¹⁹ (1223-x)}= -5.328×10⁻¹⁷

⇒1.6×10⁻¹⁹(x-1223+x)=-5.328×10⁻¹⁷

\Rightarrow (2x-1223)=(-5.328* 10^(-17))/(1.6* 10^(-19))

⇒2x-1223= -333

⇒2x= -333+1223

⇒2x=900

\Rightarrow x=(900)/(2)

⇒x=450

Therefore the number of proton is 450.


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The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?

Answers

Answer:

The charge that passes through the starter motor is \Delta Q=158.1 C.

Explanation:

Known Data

  • Avogadro's Number N_(A)=6.02x10^(23)
  • Current, I=170A=170(C)/(s)
  • Charge in an electron, q=1.60x10^(-19)C
  • Time, \Delta t=0.930s
  • Diameter, d=4.60mm=0.0046m
  • Transversal Area, A=((d)/(2))^(2) \pi=((0.0046m)/(2))^(2) \pi=1.66x10^(-5) m^(2)
  • Volume, V=Length*A=(1.2m)(1.66x10^(-5) m^(2))=1.99x10^(-5) m^(3)

First Step: Find the number of the electrons per unit of volume in the wire

We use the formula n=(N_(A))/(V)= (6.02x10^(23) electrons)/(1.99x10^(-5) m^(3)) =3.02x10^(28)el/ m^(3).

Second Step: Find the drag velocity

We can use the following formula v_(d)=(I)/(nqA)=(170C/s)/((3.02x10^(28)m^(-3))(1.60x10^(-19)C)(1.66x10^(-5) m^(2)))  =2.11x10^(-3) m/s

Finally, we use the formula \Delta Q=(nAv_(d)\Delta t)q=(3.02x10^(28) m^(-3))(1.66x10^(-5) m^(2))(2.11x10^(-3) m/s)(0.930s)(1.60x10^(-19)C)=158.1 C.

What does a planet need in order to retain an atmosphere? How does an atmosphere affect the surface of a planet and the ability of life to exist?

Answers

Answer:

Explained

Explanation:

In order to retain atmosphere a planet needs to have gravity. A gravity sufficient enough to create a dense atmosphere around it, so that it can retain heat coming from sun. Mars has shallow atmosphere as its gravity is only 40% of the Earth's gravity. Venus is somewhat similar to Earth but due to green house effect its temperature is very high. Atmosphere has a huge impact on the planets ability to sustain life. Presence of certain kind gases make the atmosphere poisnous for life. The atmosphere should be such that it allows water to remain in liquid form and maintain an optimum temperature suitable for life.

A wire with mass 90.0 g is stretched so that its ends are tied down at points 98.0 cm apart. The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude of 0.300 cm at the antinodes. Part A What is the speed of propagation of transverse waves in the wire

Answers

Answer:

118 m/s

Explanation:

Given :

We know that

f\ =\ (1)/(2l) \sqrt{(T)/(U) }......Eq(1)

Where \sqrt{(T)/(u) } =v

l=length

f=frequency

l= 98.0 cm= 0.98 m

f=60.0 Hz

Now from the Eq(1)

f\ =\ (v)/(2l)

This equation can be written as

v=2fl.............Eq(2)

Putting the value f and l in Eq(2)

v=2*60*0.98

v=117.6 m/s ~ 118 m/s

A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

Answers

To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_(1,2)= Mass of each object

v_(1,2) = Initial Velocity of each object

v_f= Final velocity

Replacing we have that,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

1850*13.8+3100*0 = (1850+3100)v_f

v_f = 5.1575m/s

With the final speed obtained we can determine the acceleration through the linear motion kinematic equations, that is to say

v_f^2-v_i^2 = 2ax

Since there is no initial speed, then

v_f^2 = 2ax

5.1575^2 = 2a (1.91)

a = 6.9633m/s^2

Finally with the acceleration found it is possible to find the friction force from the balance of Forces, like this:

F_f = F_a \n\mu N = m*a \n\mu = (ma)/(N)\n\mu = (ma)/(mg)\n\mu = (a)/(g)\n\mu = (6.9633)/(9.8)\n\mu = 0.7105

Therefore the Kinetic friction coefficient is 0.7105

in the demolition of an old building,a 1,300 kg wrenching ball hits the building at 1.07m/s^2.Calculate the amount of force at which the wrecking ball strikes the building

Answers

Answer:

1391 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1300 × 1.07

We have the final answer as

1391 N

Hope this helps you

Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of the car applies a torque of 456 N·m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

Answers

Answer:

The magnitude of the static frictional force is 1200 N

Explanation:

given information :

radius, r = 0.380 m

applied-torque, τ1 = 456 N

The car has a constant velocity, thus the acceleration is zero

α = 0

Στ = I α

τ1 - τ2 = I α

τ2 = counter-torque

τ1 - τ2 = 0

τ1 = τ2

r x F_(s) = τ1

F_(s) = the static frictional force (N)

F_(s) = τ1 /r

  = 456 N/0.380 m

  = 1200 N