What is the net displacement of the particle between 0 seconds and 80seconds?
1 answer:
Answer:
Option (A)
Explanation:
Displacement of a particle on a velocity time graph is represented by the area between the line representing velocity and x-axis (time).
Displacement of a particle from t = 0 o t = 40 seconds = Area of ΔAOB
Area of triangle AOB =
=
= 80 m
Similarly, displacement of the particle from t = 40 to t = 80 seconds = Area of ΔBCD
Area of ΔBCD =
= 80 m
Total displacement of the particle from t = 0 to t = 80 seconds,
= 80 + 80
= 160 m
Option (A) will be the answer.
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Answer:
The final temperature of the element = 262.67°C
The power dissipated in the heating element initially = 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W
Explanation:
Our given parameters include;
A Nichrome heating element operates on 120 V.
Voltage (V) = 120V
Initial Current (I₁) = 1.36 A
Initial Temperature (T₁) = 28°C
Final Current (I₂) = 1.23 A
Final Temperature (T₂) = unknown ????
Temperature dependencies of resistance is given by:
---------------------- (1)
in which R₁ is the resistance at temperature T₁
is the resistance at temperature T₂
Given that V= IR
R =
Therefore, the resistance at temperature 28°C is;
= 88.24Ω
=
= 97.56Ω
From (1) above;
97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]
1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)
0.1056 = 4.5×10⁻⁴(T₂-28°C)
T - 28° C = 234.67
T = 234.67 + 28° C
T = 262.67 ° C
(b)
What is the power dissipated in the heating element initially and when the current reaches 1.23 A
The power dissipated in the heating element initially can be calculated as:
P = I²₁R₂₈
P = (1.36A)²(88.24Ω)
P = 163.209 W
P ≅ 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:
P = (1.23)²(97.56Ω)
P = 147.598524
P ≅ 147.60 W
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 = / R
² = mR
ω
²
ω² = GM
= / R
³
ω = √(GM
= / R
³)
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