I need help quickly pls

If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial accelerat

katrin [286]

To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

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

Here,

k = Coulomb's constant

q = Charge of proton and electron

r = Distance

Replacing we have that,

$F = (9*10^9)(\frac{(1.602*10^{-19})^2}{2.5*10^{-10}})$

$F = 3.6956*10^{-9}N$

The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.

The acceleration of the electron is given as

$a_e = \frac{F}{m_e}$

$a_e = \frac{3.6956*10^{-9}}{9.11*10^{-31}}$

$a_e = 4.0566*10^{21}m/s^2$

The acceleration of the proton is given as,

$a_p = \frac{F}{m_p}$

$a_p = \frac{3.6956*10^{-9}}{1.672*10^{-27}}$

$a_p = 2.21*10^{18}m/s^2$

3 0

3 years ago

A box of mass 10 kg is pulled from the hold of a ship with an acceleration of 1 m/s^2 by a vertical rope attached to it .find th

wlad13 [49]

F=ma
Mass times acceleration
We have g (10ms^_2) and a (1 given)
So total would be
10 kg times (10+1) =
110 N

5 0

3 years ago

The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

love history [14]

Answer:

Part a)

$EMF = 14 \times 10^{-3} V$

Part b)

$EMF = 15.67 \times 10^{-3} V$

Explanation:

As we know that magnetic flux through the loop is given as

$\phi = B.A$

now we have

$\phi = B\pi r^2$

now rate of change in flux is given as

$\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}$

now we know that

$A = \pi r^2$

$0.285 = \pi r^2$

$r = 0.30 m$

Now plug in all data

$EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)$

$EMF = 14 \times 10^{-3} V$

Part b)

Now the radius of the loop after t = 1 s

$r_1 = r_0 + \frac{dr}{dt}$

$r_1 = 0.30 + 0.037$

$r_1 = 0.337 m$

Now plug in data in above equation

$EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)$

$EMF = 15.67 \times 10^{-3} V$

5 0

3 years ago

Nepal has high potential for producing hydroelectricity however it is difficult too.

Stels [109]

Answer:

I'm not a scholar of hydroelectric power in Nepal, so consider my answers carefully, below.

Explanation:

High Potential: Hydroelectric power comes from the potential energy stored in a mass that is above Earth's surface. As the word "hydro" implies, the mass in this case is water. Water from snow and glacier melt, and from normal precipitation (rain) in mountainous regions eventually cascades down the mountains in fast-flowing rivers or waterfalls. Often, there are lakes or man-made reservoirs to collect and store the water before it flows down. Mt. Everest is 8848 meters tall (about 29,000 feet). If a lake forms at just 2,000 meters, one can calculate the amount of energy in each kilogram of water stored in the lake that represents the potential energy available at that altitude. 1 kg of water at 2,000 meters has potential energy, PE, according to the equation: PE = mgh, where m is the mass in kg, g is Earth's acceleration due to gravity (9.8 m/sec^2), and h is height, in meters.

PE = mGH

PE = (1 kg)*(9.8 m/sec^2)*(2,000 meters) = 19,000 kg*m/sec^2

1 kg*m/sec^2 is the SI unit for 1 Joule, a measure of energy.

This potential energy can be converted into electrical energy by releasing the water so that it can flow down to a water-powered turbine that spins magnets and coils of wire that produce electricity. The 19,000 Joules of water potential energy can be converted to electrical power, less any inefficiency in the system, such as friction.

Nepal has the natural advantage in that it has many high mountain ranges with water flows that can be used for generating electrical power. The result is low operating costs (the fuel is the flowing water) and no greenhouse gas emissions

The difficulty in developing hydroelectric power in Nepal is due to the same factor that gives it an advantage: it is difficult constructing large hydroelectric plants in such rough terrain, and the power lines that are needed to transport the power to its destination are expensive and difficult to maintain and repair.

8 0

2 years ago

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball initially moving in th

inn [45]

Final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

<h3>What is conservation of momentum?</h3>

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. By the law of conservation of momentum,

$m_1u_1 + m_2u_2 = (m_1+m_2)v$

Here, (m) is the mass, (u) is initial velocity before collision, v is final velocity after collision and (subscript 1, and 2) are used for body 1 and 2 respectively. Rewrite the formula for final velocity as,

$v=\dfrac{m_1u_1 + m_2u_2}{(m_1+m_2)}$

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s, has a head-on collision with a 0. 090-kg ball, initially moving in the same direction at a speed of 3.44 m/s. Thus, the initial velocity of the second ball is,

$v_{2f}=5.82+3.44+v_{1f}\\v_{2f}=2.38+v_{1f}$

Let v1f is the final velocity of first ball. Thus, the initial velocity of the first ball is,

$v_{1f}=\dfrac{(0.060)(5.82) + (0.090)(3.44-2.38)}{(0.060)+(0.090)}\\v_{1f}=2.964\rm\; m/s$

Thus, final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

Learn more about the conservation of momentum here;

brainly.com/question/7538238

#SPJ4

4 0

2 years ago