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lbvjy [14]
3 years ago
15

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 17.3 m/s directly upward.

Choose upward as the positive direction in your analysis. (a) What is the maximum height above the ground that the ball reaches?
Physics
1 answer:
elena-s [515]3 years ago
7 0

Answer:

h=15.27m

Explanation:

Since at maximum height the vertical velocity must be null it's better to use the formula:

v_f^2=v_i^2+2ad

We will use this formula for the vertical direction, choosing the upward direction as the positive one, so we have:

0=v_i^2+2ah

or

h=-\frac{v_i^2}{2a}

which for our values is:

h=-\frac{(17.3m/s)^2}{2(-9.8m/s^2)}=15.27m

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Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

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Answer:

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<h3>Heat released by the water when it freezes</h3>

The heat released by the water when it freezes is calculated as follows;

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Explanation:

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