Answer:
51.2 mi/h
Explanation:
Total distance, d = 100 miles
First 60 miles with speed 55 mi/h
Next 40 miles with speed 75 mi/h
Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h
Time taken for 40 miles, t2 = 40 / 75 = 0.533 h
Time spent to get stuck, t3 = 20 min = 0.33 h
Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h
The average speed is defined as the ratio of total distance traveled to the total time taken.
Average speed =
Thus, the average speed of the journey is 51.2 mi/h.
the height of the building is H=36 m.
<h3>What is The Law of Gravity?</h3>
According to Newton's law of gravity, every particle of matter in the universe is attracted to every other particle with a force that varies directly as the product of their masses and inversely as their distance from one another.
Properties of Gravity -
- It is a universal attractive force. It is directly proportional to the product of the masses of the two bodies.
- It obey inverse square law.
- It is the weakest force known in nature.
Examples of Gravity -
- The force that holds the gases in the sun.
- The force that causes a ball you throw in the air to come down again.
- The force that causes a car to coast downhill even when you aren't stepping on the gas.
v₀=0 m/s
H₀=0 m
g=10 m/s²
t=7,2 s
H - ?
H = 0 +0 × 7.2 + 10(7.2)²/2
H = 36m
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The kinetic energy of the ejected electrons will be 2782.5 ×10⁻²² eV.
<h3>What is threshold frequency?</h3>
The threshold frequency of incoming radiation is the lowest frequency at which photoelectric emission or electron emission is impossible.
The threshold frequency is the light frequency that causes an electron to dislodge and emit from the metal's surface.
From the photoelectric effect, the equation obtained as;
Hence,the kinetic energy of the ejected electrons will be 2782.5 ×10⁻²² eV.
To learn more about the threshold frequency, refer to the link;
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Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M = 1.99 × 10³⁰ kg
Mass of the neutron star
M = 2( M )
M = 2( 1.99 × 10³⁰ kg )
M = ( 3.98 × 10³⁰ kg )
Radius of neutron star R = 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω.
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM = / R² = mRω²
ω² = GM = / R³
ω = √(GM = / R³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω = √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω = √ 120831133.3636777
ω = 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s