<span>What's the speed of a sound wave through water at 25° Celsius?
<span> <span><span> <span>A. </span>1,250 m/s</span><span> <span>B. </span>1,500 m/s</span><span> <span>C. </span>750 m/s</span><span> <span>D. </span>1,000 m/s
</span></span></span></span>
The volume of the tire at the given diameter and thickness of tube is determined as 1,128.2 cubic inch.
<h3>What is volume?</h3>
Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.
<h3>
Volume of the tire</h3>
The volume of the tire is the measure of the product of area and thickness of the tire.
The volume of the tire is calculated as follows;
Radius of the tire = 0.5 x 26" = 13"
Volume of the tire = Area x thickness
Volume of the tire = πr² x h
where;
- r is the radius of the tire
- h is the thickness of the tube
Volume of the tire = π(13)² x (2.125)
Volume of the tire = 1,128.2 cubic inch
Thus, the volume of the tire at the given diameter and thickness of tube is determined as 1,128.2 cubic inch.
Learn more about Volume of tire here: brainly.com/question/1972490
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Answer:
24,000 m
Explanation:
First find the rocket's final position and velocity during the first phase in the y direction.
Given:
v₀ = 75 sin 53° m/s
t = 25 s
a = 25 sin 53° m/s²
Find: Δy and v
Δy = v₀ t + ½ at²
Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²
Δy = 7736.8 m
v = at + v₀
v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)
v = 559.0 m/s
Next, find the final position of the rocket during the second phase (as a projectile).
Given:
v₀ = 559.0 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 15945.5 m
The total displacement is:
7736.8 m + 15945.5 m
23682.2 m
Rounded to two significant figures, the maximum altitude reached is 24,000 m.
Explanation:
Given that,
Mass of the rocket, m = 2150 kg
At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :
Here F = 888.93 N when t = 1.25 s
(c) We can find the value of A first as :
The value of A is 
.
(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :
Limits will be from 2 s to 2+ 4 = 6 s
It implies :
(b) Impulse is also equal to the change in momentum as :
Hence, this is the required solution.
