Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course :
= 0.75 m/s²,
= 20 m,
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s,
= -1.15 m/s²,
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) ×
)
0 = 129.96 - 2.3
2.3
= 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √(
² +
² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
Answer:
Glow
Explanation:
Actually, it is the air in front of the meteoroid that heats up. The particle is traveling at speeds between 20 and 30 kilometers per second. It compresses the air in front, causing the air to get hot. The air is so hot it begins to glow — creating a meteor - the streak of light observed from Earth.
Hope this helped!
Expression to calculate energy from voltage: E= V*Q where E= energy, V= voltage, and Q= charge
Additional help:
-To find the Voltage ( V )
[ V = I x R ] V (volts) = I (amps) x R (Ω)
-To find the Current ( I )
[ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
-To find the Resistance ( R )
[ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
I hope that helps to some extent-
Answer:
c) 2.02 x 10^16 nuclei
Explanation:
The isotope decay of an atom follows the equation:
ln[A] = -kt + ln[A]₀
<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>
[A] = Our incognite
k is constant decay:
k = ln 2 / Half-life
k = ln 2 / 4.96 x 10^3 s
k = 1.40x10⁻⁴s⁻¹
t is time = 1.98 x 10^4 s
[A]₀ = 3.21 x 10^17 nuclei
ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]
ln[A] = 37.538
[A] = 2.01x10¹⁶ nuclei remain ≈
<h3>c) 2.02 x 10^16 nuclei</h3>
Answer:
1.3823 rad/s
20.7345 m/s
28.66129935 m/s²

2006.29095 N radially outward
Explanation:
r = Radius = 15 m
m = Mass of person = 70 kg
g = Acceleration due to gravity = 9.81 m/s²
Angular velocity is given by

Angular velocity is 1.3823 rad/s
Linear velocity is given by

The linear velocity is 20.7345 m/s
Centripetal acceleration is given by

The centripetal acceleration is 28.66129935 m/s²
Acceleration in terms of g


Centripetal force is given by

The centripetal force is 2006.29095 N radially outward
The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.