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ehidna [41]
3 years ago
13

Imagine an isolated positive point charge Q (many times larger than the charge on a single proton). There is a charged particle

A (whose charge is much smaller than charge Q) at a distance from the point charge Q. On which of the following quantities does the magnitude of the electric force on this charged particle A depend:_____.
1. the size and shape of the point charge Q
2. the specific location of the point charge Q (while the distance between Q and A is fixed)
3. the mass of the charged particle A
4. the size and shape of the charged particle A
5. the distance between the point charge Q and the charged particle A
6. the type of charge on the charged particle A
7. the mass of the point charge Q
8. the amount of the charge on the point charge Q
9. the magnitude of charge on the charged particle A
10. the specific location of the charged particle A (while the distance between Q and A is fixed)
11. the relative orientation between Q and A (while the distance between Q and A is fixed)
Physics
1 answer:
egoroff_w [7]3 years ago
7 0

Answer:

Explanation:

The magnitude of the electric force on this charged particle A depends upon the following

5. the distance between the point charge Q and the charged particle A

8. the amount of the charge on the point charge Q

9. the magnitude of charge on the charged particle A

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The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

5 0
3 years ago
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

8 0
3 years ago
a paper airplane gliding down towards the ground will experience the force of air resistance pushing up. the weight of the paper
Oduvanchick [21]

The net force acting on the airplane is 25N.

Forces acting on the paper airplane when it is in the air:

  • The forward force generated by the engine, propeller, or rotor is called thrust. It resists or defeats the drag force. It operates generally perpendicular to the longitudinal axis. However, as will be discussed later, this is not always the case.
  • Drag is an airflow disruption generated by the wing, rotor, fuselage, and other projecting surfaces that causes a backward, decelerating force. Drag acts backward and perpendicular to the relative wind, opposing thrust.
  • Weight is the total load carried by airplane, including the weight of the crew, fuel, and any cargo or baggage. Due to the influence of gravity, weight pulls the airplane downward.
  • Lift—acts perpendicular to the flight path through the center of lift and opposes the weight's downward force. It is produced by the air's dynamic influence on the airfoil.

Given.

Weight of the paper airplane, F1 = 16N

The force of air resistance, F2 = 9N

Net force = F1 + F2

Net force = 25N

Thus, the net force acting on the airplane is 25N.

Learn more about the net force here:

brainly.com/question/18109210

#SPJ1

3 0
1 year ago
There is less oxygen in air at higher altitudes than at sea level. Scientists have shown that people living at sea level have 5
Ivanshal [37]

Answer with Explanation:

"Red Blood Cells" <em>(RBCs)</em> contain <em>Hemoglobin</em> that is responsible for carrying oxygen into the body. When people are exposed to higher altitudes, <u>the number RBCs in the body increases</u>. This is because the body has a hard time taking in oxygen due to <u>low atmospheric pressure</u>. It makes it hard for oxygen to pass through the lung membranes. This is called "hypoxia." Such condition deprives the body from oxygen, thus, it creates more red blood cells in order to compensate the condition.

When it comes to people living at sea level,<em> the oxygen can easily pass through the lung membranes</em> due to <u>higher atmospheric pressure.</u> This doesn't require the body to build new RBCs. Therefore, the numbers of RBCs needed by people to thrive is lower than living at higher altitudes.

5 0
3 years ago
The drawing shows four sheets of polarizing material, each with its transmission axis oriented differently. Light that is polari
Brrunno [24]

Answer:

Explanation:

When sheet A is removed

I_B=32\cos^230=24W/m^2\\\\I_C=24 \cos^260=6W/m^2\\\\I_D=6\cos^230=4.5W/m^2

When sheet B is removed

I_A=32\cos^20=32W/m^2\\\\I_C=24 \cos^290=0W/m^2\\\\I_D=0\cos^230=0W/m^2

When sheet C is removed

I_A=32\cos^20=32W/m^2\\\\I_D=32 \cos^230=24W/m^2\\\\I_B=24\cos^290=0W/m^2

When sheet D is removed

I_A=32\cos^20=32W/m^2\\\\I_B=32\cos^230=24W/m^2\\\\I_C=24\cos^260=6W/m^2

8 0
3 years ago
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