The velocity at the maximum height will always be 0. Therefore, you will count your final velocity as 0, and your initial velocity as 35 m/s. Next, we know that the acceleration will be 9.8 m/s^2. How? Because the ball is thrown directly upward, and the only force acting on it will be the force of gravity pushing it back down.
The formula we use is h = (Vf^2 - Vi^2) / (2*-9.8m/s^2)
Plugging everything in, we have h = (0-1225)/(19.6) = 62.5 meters is the maximum height.
Answer:
Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.
Explanation:
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The specific gravity of the object’s material is 5.09.
<h3>To calculate the specific gravity of the object:</h3>
Weight difference = 9 - 7.2 = 1.8 N = Buoyant force of water
Buoyant Force in water(Fb) = density of water x g x volume of the body(Vb)
1.8 = 1000 x 9.81 x Vb
Vb = 1.8/9810 cubic meter
Now, in the air;
Weight of body = mg = 9 N
Mass of body,m = 9/9.81 Kg
So,
Density of body = m/ Vb
= 9/9.81 ÷ 1.8/9810
= 5094.44 kg per cubic meter
The specific gravity of body = density of body ÷ density of water
= 5094.44 ÷ 1000
= 5.09
Therefore, Specific gravity of body = 5.09
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<span>net work = change in kinetic energy
for Block B, we just have the force from block A acting on it
F(ab)d= .5(1)vf² - .5(1)(2²)
F(ab)d= .5vf² - 2
Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction
10-F(ba)d = .5(4)vf² - .5(4)(2²)
10-F(ba)d = 2vf² - 8
F(ba)d = 18 - 2vf²
now we have two equations:
F(ba)d = 18 - 2vf²
F(ab)d= .5vf² - 2
since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab)
10-.5vf² + 2 = 2vf² - 8
12 - .5vf² = 2vf² - 8
20 = 2.5vf²
vf² = 8
they both will have the same velocity
KE of block A= .5(4)(2.828²) = 16 J
KE of block B=.5(1)(2.828²) = 4 J</span>
