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VMariaS [17]
3 years ago
10

A bowling ball collides with a tennis ball. Which object has the larger impact force on the other considering that they have the

same acceleration? Why?
This is for an exam could you please answer fast
Physics
1 answer:
Bezzdna [24]3 years ago
4 0

Answer:

The tennis ball experiences a larger impact force due to its smaller mass.

Explanation:

Considering that they have the same acceleration, and F = ma where F is force and m = mass, the object with the smaller mass has a greater impact force on it.

This is because since force F = ma and a = constant, the object with a larger mass (the bowling ball)has a larger impact force with which it acts on the object with the smaller mass (the tennis ball). So, the tennis ball experiences a larger impact force due to its smaller mass.

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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.
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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

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3 0
1 year ago
Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform el
kvv77 [185]

Answer:

The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

Explanation:

<u>Step 1:</u> calculate the charge on each electrode

Given;

Electric field strength = 2.0 X 10⁶ N/C

The distance between the electrode = 1mm = 1 X 10⁻³ m

Electric field strength (E) = Force (F)/Charge (q)

E =\frac{Kq}{r^2}

where;

E is the electric field strength = 2.0 X 10⁶ N/C

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

r is the distance between the electrodes = 1 X 10⁻³ m

q is the charge in each electrode = ?

q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9} = 0.2225 X 10⁻⁹ C

The charge on each electrode is 0.2225 X 10⁻⁹ C

<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.

1 electron contains 1.602 X 10⁻¹⁹ C

So, 0.2225 X 10⁻⁹ C will contain how many electrons ?

= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)

= 1.4 X 10⁹ electrons

Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

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