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Kryger [21]
3 years ago
12

Question 2 - If Juan starts out at 10m/s, and in 15 s speeds up to 60 m/s, what is his acceleration?

Physics
1 answer:
34kurt3 years ago
5 0

Answer:

B) 3.33 m/s²

Explanation:

Given that,

The initial velocity of Juan, u = 10 m/s

Final velocity of Juan,  v = 60 m/s

Time taken, t = 15 s

The acceleration is given by the relation

                          a = v-u/t  m/s²

Substituting in the above equation

                           a = 60 - 10 / 15

                           a = 3.33 m/s²

Hence, the acceleration of the Juan is, a = 3.33 m/s²

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3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
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Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
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