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kkurt [141]
3 years ago
6

A baseball is thrown a velocity of 20 m/s. It takes the 20 kg ball 10 seconds 5 points

Physics
1 answer:
liberstina [14]3 years ago
7 0

Answer: 40 Newton

Explanation:

Mass of baseball = 20 kg

Velocity of baseball = 20 m/s.

Time taken for throwing = 10 seconds

Since acceleration is the rate of change of velocity per unit time

i.e Acceleration = velocity / time taken

Acceleration = (20 m/s / 10seconds)

Acceleration = 2m/s^2

Now, force required to throw the baseball is a product of mass and acceleration

i.e Force = Mass x acceleration

Force = 20 kg x 2m/s^2

Force = 40Newton

Thus, 40 newton of force is required to throw baseball to reach the plate

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is described as the sum of all the individual resistances in a series.1) Equivalent resistance2) Parallel circuit3) Series circu
irga5000 [103]

The sum of all resistances in the series circuit is known as Equivalent resistance.

R=r_1+r_2+.\ldots.r_n

where R is the equivalent resistance.

Hence, option 1 is the correct answer.

6 0
1 year ago
(1) A positive charge +3 C is separated from another positive charge of +5 C by a distance of 7m. What is the magnitude of the e
Aneli [31]

1. The magnitude of the electric force between the two charges is 2.8×10⁹ N (Option B)

2. The net charge on the molecule is -8×10⁻¹⁹ C (Option D)

3. The magnitude of the force between the charges is 16000 N (Option C)

4. The correct statement is: A neutral object has equal numbers of protons and electrons. (Option C)

<h3>1. How to determine the force</h3>
  • Charge 1 (q₁) = +3 C
  • Charge 2 (q₂) = +5 C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 7 m
  • Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 3 × 5) / (7)²

F = 2.8×10⁹ N

<h3>2. How to determine the net charge on the molecule</h3>
  • Electron = 223 electrons
  • Proton = 218 protons
  • Net Charge =?

Charge = Proton - Electron

Charge = 218 - 223

Charge = -5 electrons

But

1 electron = 1.6×10⁻¹⁹ C

Thus,

Net Charge = -5 × 1.6×10⁻¹⁹ C

Net Charge = -8×10⁻¹⁹ C

<h3>3. How to determine the force</h3>
  • Charge 1 (q₁) = 2×10⁻⁴ C
  • Charge 2 (q₂) = 8×10⁻⁴ C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 0.3 m
  • Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 2×10⁻⁴ × 8×10⁻⁴) / (0.3)²

F = 16000 N

<h3>4. What is a neutral object?</h3>

A neutral object is an object having equal numbers of protons and electrons. For example, an object with 4 protons and 4 electrons is said to be neutral as illustrated below

  • Electron = 4 electrons
  • Proton = 4 protons
  • Net Charge =?

Charge = Proton - Electron

Charge = 4 - 4

Charge = 0 (neutral)

Thus, the correct statement about neutral object, given in the question is: A neutral object has equal numbers of protons and electrons (Option C)

Learn more about Coulomb's law:

brainly.com/question/506926

#SPJ1

6 0
2 years ago
A jewellery shop owner has two identical clear gemstones but cannot remember which one is diamond and which one is rutile. Rutil
pantera1 [17]

We know

\boxed{\sf n_{21}=\dfrac{C}{V}}

How he find:-

The owner will measure speed of light through the index(Diamond or rutile).Then using calculations he fill find which is the velocity of diamond or rutile

<h3> For diamond</h3>

\\ \sf\longmapsto n_D=\dfrac{C}{V_D}

\\ \sf\longmapsto 2.4=\dfrac{3\times 10^8ms^{-1}}{V_D}

\\ \sf\longmapsto V_D=\dfrac{3\times 10^8ms^{-1}}{2.4}

\\ \sf\longmapsto V_D=1.25\times 10^8ms^{-1}

\\ \sf\longmapsto V_D=125\times 10^6ms^{-1}

<h3>For rutile</h3>

\\ \sf\longmapsto n_R=\dfrac{C}{V_R}

\\ \sf\longmapsto V_R=\dfrac{C}{n_R}

\\ \sf\longmapsto V_R=\dfrac{3\times 10^8ms^{-1}}{2.9}

\\ \sf\longmapsto V_R=1.03\times 10^8

\\ \sf\longmapsto V_R=103\times 10^6ms^{-1}

5 0
3 years ago
A 1420-kg car is traveling with a speed of 12.4 m/s. What is the magnitude of the horizontal net force that is required to bring
zzz [600]

Answer:

1419.01436 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-12.4^2}{2\times 78}\\\Rightarrow a=-0.98564\ m/s^2

The force on the car

F=ma\\\Rightarrow F=1420\times -0.98564\\\Rightarrow F=-1419.01436\ N

Magnitude of the horizontal net force that is required to bring the car to a halt is 1419.01436 N

3 0
3 years ago
Read 2 more answers
Gravity ____.
vitfil [10]

I am like 98% sure it is C

4 0
4 years ago
Read 2 more answers
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