Which projectile will be visibly affected

A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric

Liula [17]

The electric field produced by a single-point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.

1) The first charge is $q=-3.00 nC=-3.00 \cdot 10^{-9} C$ , and it is located at x=0, so its distance from the point x=0.200 m is

$r=0.200 m-0=0.2 m$

Therefore, the electric field is

$E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C$

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.

2) The second charge is $q=-5.50 nC=-5.5 \cdot 10^{-9}C$ and it is located at x=0.800 m, so its distance from the point is

$r=0.800 m-0.200 m=0.6 m$

Therefore, the electric field is

$E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N$

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.

3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

$E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C$

and the sign tells us that the field is directed toward negative x-direction.

7 0

3 years ago

In which of these situations is convection most likely the main form of heat transfer? A.Warm air from a heater on the first flo

Sergeeva-Olga [200]

All points ate true

But option A is perfect

6 0

3 years ago

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Can someone please help me with these. I don't understand any of it.

ValentinkaMS [17]

1. electrical energy: electrical energy that is caused by moving electrons

2. coolant: a mixture of antifreeze and water that removes excess heat from an internal engine

3. electric compressor: a device that acts as a pump, circulating refrigerant throughout the refrigerator

4. the inside of the fridge and the food becomes colder

5. the coolants becomes a hot, high-pressure gas

6. as coolant transfers thermal energy to the air outside, it turns back into a liquid

6 0

3 years ago

The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the

Rainbow [258]

Answer:

3 seconds

Explanation:

Since h(t) represents the height and t represents the time, we can set the equation equal to 150 to find t.

-16t^2+96t+6=150

Subtract 150 from both sides to set the equation equal to 0, to find the solutions.

-16t^2+96t-144=0

Factor out -16, because all of the terms are divisible by it.

-16(t^2+6t+9)=0

Now we can focus on the terms inside the parenthesis and factor it again.

t^2-6t+9=0

We need to find two value that can be multiplied to get 9 and added to get -6.

-3 and -3 works.

Thus, we get (x-3)(x-3).

Now solve for 0.

x-3=0

x=3

The object reaches its maximum height after 3 seconds.

6 0

3 years ago

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A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

BlackZzzverrR [31]

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by

∑F = ma (1)

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:

a_max = -w^2A (2)

3- The angular frequency w of a wave is related to the frequency f by:

w = 2π f (3)

Given Data

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)= 0.04 m.

- The frequency of oscillation of the piston is steadily increased.

Required Data

In part (a), we are asked to determine the point at which the penny first loses contact with the piston.

In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.

Solution

(a)

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:

∑F_y -mg= ma

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So

0 = m(g + a)

a = —g

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):

a = —w^2A

where a = —g at the highest point. So

g = w^2A

Solve for w:

w =√g/A

Substitute for w from Equation (3):

2πf = √g/A

Solve for f :

f = 1/2π√g/A

Substitute numerical values:

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

6 0

3 years ago