Answer:
The acceleration at the moment the train speed reaches 52 kilometers per hour is approximately 1.826 meters per square second.
Explanation:
According to Rotational Physics, the total acceleration of the train rounding the horizontal turn is a combination of tangential (
) and radial accelerations (
), measured in meters per square second. The former one represents the change in the magnitude of the velocity, whereas the latter one represents the change in its direction. By definition of magnitude and Pythagorean Theorem we get that magnitude of total acceleration (
), measured in meters per square second, is:
(Eq. 1)
Magnitudes of tangential and radial accelerations are determined by using the following formulas:
(Eq. 1)
(Eq. 2)
Where:
, 
- Initial and final speeds, measured in meters per second.
- Time, measured in seconds.
- Radius, measured in meters.
If we know that 
, 
, 
and 
, then the magnitude of the total acceleration when the train speed reaches 52 kilometers per hour is:
The acceleration at the moment the train speed reaches 52 kilometers per hour is approximately 1.826 meters per square second.
1 would use less energy. Please vote my answer brainliest! Thanks.
Answer:
Bi. Current in 15.4 Ω (R₁) is 7.14 A.
Bii. Current in 21.9 Ω (R₂) is 5.02 A.
Biii. Current in 11.7 Ω (R₃) is 9.40 A.
C. Total current in the circuit is 21.56 A.
Explanation:
Bi. Determination of the current in 15.4 Ω (R₁)
Voltage (V) = 110 V
Resistance (R₁) = 15.4 Ω
Current (I₁) =?
V = I₁R₁
110 = I₁ × 15.4
Divide both side by 15.4
I₁ = 110 / 15.4
I₁ = 7.14 A
Therefore, the current in 15.4 Ω (R₁) is 7.14 A.
Bii. Determination of the current in 21.9 Ω (R₂)
Voltage (V) = 110 V
Resistance (R₂) = 21.9 Ω
Current (I₂) =?
V = I₂R₂
110 = I₂ × 21.9
Divide both side by 21.9
I₂ = 110 / 21.9
I₂ = 5.02 A
Therefore, the current in 21.9 Ω (R₂) is 5.02 A
Biii. Determination of the current in 11.7 Ω (R₃)
Voltage (V) = 110 V
Resistance (R₃) = 11.7 Ω
Current (I₃) =?
V = I₃R₃
110 = I₃ × 11.7
Divide both side by 11.7
I₃ = 110 / 11.7
I₃ = 9.40 A
Therefore, the current in 11.7 Ω (R₃) is 9.40 A.
C. Determination of the total current.
Current 1 (I₁) = 7.14 A
Current 2 (I₂) = 5.02 A
Current 3 (I₃) = 9.40 A
Total current (Iₜ) =?
Iₜ = I₁ + I₂ + I₃
Iₜ = 7.14 + 5.02 + 9.40
Iₜ = 21.56 A
Therefore, the total current in the circuit is 21.56 A
Answer:
bromine
Explanation:
Any element in the halogen group will have seven valence electrons. These elements include fluorine, chlorine, bromine, iodine, and astatine
