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

⇒2x-1223= -333

⇒2x= -333+1223

⇒2x=900

⇒x=450

**Therefore the number of proton is 450.**

Twopucksofequalmasscollideonafrictionlesssurface,asillustratedinthefigure.Immediatelyafterthe collision, the speed of the black puck is 1.5 m/s. What is the speed of the white puck immediately after the collision?

Parallel Plates Consider a very large conducting plate at potential V0 suspended a distance d above a very large grounded plane. Find the potential between the plates. The plates are large enough so that they may be considered to be infinite. This means that one can neglect fringing fields.

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

A car, starting from rest, accelerates in a straight-line path at a constant rate of 2.0 m/s2. How far will the car travel in 12 seconds?

A vibrating tuning fork makes 500 vibrations in one second. What is the wavelength of the sound produced if the air temperature is 20°C? Group of answer choices0.500 m0.686 m0.343 m1.46 m1.87 m

Parallel Plates Consider a very large conducting plate at potential V0 suspended a distance d above a very large grounded plane. Find the potential between the plates. The plates are large enough so that they may be considered to be infinite. This means that one can neglect fringing fields.

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

A car, starting from rest, accelerates in a straight-line path at a constant rate of 2.0 m/s2. How far will the car travel in 12 seconds?

A vibrating tuning fork makes 500 vibrations in one second. What is the wavelength of the sound produced if the air temperature is 20°C? Group of answer choices0.500 m0.686 m0.343 m1.46 m1.87 m

**Answer:**

The charge that passes through the starter motor is .

**Explanation:**

__Known Data__

- Avogadro's Number
- Current,
- Charge in an electron,
- Time,
- Diameter,
- Transversal Area,
- Volume,

__First Step: Find the number of the electrons per unit of volume in the wire__

We use the formula .

__Second Step: Find the drag velocity__

We can use the following formula

Finally, we use the formula .

**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.

**Answer:**

118 m/s

**Explanation:**

**Given :**

We know that

......Eq(1)

Where =v

l=length

f=frequency

l= 98.0 cm= 0.98 m

f=60.0 Hz

Now from the Eq(1)

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

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

Where,

= Mass of each object

= Initial Velocity of each object

= Final velocity

Replacing we have that,

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

Since there is no initial speed, then

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

Therefore the Kinetic friction coefficient is 0.7105

**Answer:**

**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

Hope this helps you

**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 = τ1

= the static frictional force (N)

= τ1 /r

= 456 N/0.380 m

= **1200 N**