Answer:8.3m/sec 30 sec,
Explanation:
A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
Answer:
We know that Force = mass × acceleration
By substituting the values we get,
30 N = 15 kg × a (where a is acceleration)
Or we can write it as
15 kg × a = 30 N
Transposing 15 to RHS,
a = 30 ÷ 15 m/s²
Therefore, acceleration = 2 m/s²
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Answer:
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Explanation:
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Answer:
<u>We are given:</u>
u = 2.5 m/s
a = 0.2 m/s/s
t = 25 seconds
v = v m/s
<u>Solving for 'v':</u>
From the first equation of motion:
v = u + at
Replacing the values
v = 2.5 + (0.2)(25)
v = 2.5 + 5
v = 7.5 m/s