With the use of below formula, at 879 °C, velocity will be double the velocity at 15 °C.
<h3>
What is the relationship between Velocity and sound ?</h3>
The velocity of sound waves in air is proportional to the square root of Thermodynamic temperature. That is, V = K
Given that the temperature at which the velocity of sound in air is twice its velocity at 15°C, Let us make use of the formula;
(v2/v1) = √(T2 / T1)
Where
- T2 = final absolute temperature
- T1 = initial temperature.
Recall that absolute temperature = °C + 273.
If v2 = 2 × v1 and temperature in degree Celsius = 15°C, then,
Temperature in Kelvin K = 15 + 273 = 288
Substitute all the parameters into the formula
(2 × v1)/v1 = √(T2/288)
2 = √ (T2 /288)
Square both sides
4 = (T2/288)
T2 = 4 × 288
T2 = 1152K
Temperature in degrees Celsius = 1152 - 273 = 879 °C.
Therefore, at 879 °C, velocity will be double the velocity at 15 °C.
Learn more about sound waves here: brainly.com/question/13105733
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<span>Molecules at higher temperatures have more energy,
thus they can vibrate faster. Since the molecules vibrate faster, sound
waves can travel more quickly. The speed of sound in room temperature
air is 346 meters per second. so that would be answer A</span>
Answer:
Explanation:
Given that,
The magnitude of vector A, 
The magnitude of vector B, 
Scalar product of A and B, 
The formula for the scalar product is given by :
Where, 
is the angle between A and B.
The formula for the vector product is given by :
So, the vector product between these two vectors is 
.
Answer:The velocity of the object will be 5
m/s or 13.23m/s
Explanation:
force exerted by the object= 30N
distance displayed by the object by the action of force=6.0m
mass of object=10kg
velocity gained by the object=?
Answer:
(a) a= 0.139 m/s²
(b) d= 4.45 m
(c) vf= 1.1 m/s
Explanation:
a) We apply Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass (kg)
a : acceleration (m/s²)
Data
F₁= +2.05 * 10³ N : forward push by a motor
F₂= -1.87* 10³ N : resistive force due to the water.
m= 1300 kg
Calculation of the acceleration of the boat
We replace data in the formula (1):
∑F = m*a
F₁+F₂= m*a
a= 0.139 m/s²
b) Kinematics of the boat
Because the boat moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+at Formula (3)
Where:
d:displacement in meters (m)
t : time interval (s)
v₀: initial speed (m/s)
vf: final speed (m/s)
a: acceleration (m/s²
)
Data
v₀ = 0
a= 0.139 m/s²
t = 8 s
Calculation of the distance traveled by the boat in 8 s
We replace data in the formula (2)
d= v₀t+ (1/2)*a*t²
d= 0+ (1/2)*(0.139)*(8)²
d= 4.45 m
c) Calculation of the speed of the boat in 8 s
We replace data in the formula (3):
vf= v₀+at
vf= 0+( 0.139)*(8)
vf= 1.1 m/s
