Question

1. a nerinx student in a 1907 kg suv is traveling at 11 m/s (which is about 25 mph). what is the kinetic energy of the car?$k=\frac{1}{2}mv^2LXB0k=115373.5j$13. in gym class, nellie nerinx throws a baseball straight up in the air. when she releases the ball, it has velocity of 2.5 m/s upward. the ball has a mass of 0.065 kg. what is the kinetic energy of the ball when it is released?$k=\frac{1}{2}mv^2LXB1k=0.203125j$14. sasha the cat is running through the garden chasing a squirrel. if sashas mass is 6.5 kg and she has 64.3 joules of kinetic energy while she is running, calculate sashas velocity as she is running.$k=\frac{1}{2}mv^2LXB2\frac{64.3}{3.25}=\frac{3.25v^2}{3.25}$ $(19.78461538)=v^2$$v=4.45m/s(\hat{x})$15. mrs.basir is on a hot air balloon ride. while trying to take a selfie, she accidentally drops her iphone-16 over the side of the basket. the mass of the phone is 0.3 kg.a) if the phone is dropped from a height of 125 m, what is the potential energy of the phone the instant it was released?$u=mgh$$u=0.3(9.81)(125)$$u=367.88j$b) when the phone is about to hit the ground, all that energy has been converted into kinetic energy. what is the velocity of the phone at this point (when it is about to hit the ground)?$k=\frac{1}{2}mv^2$$7.88=\frac{1}{2}(125)v^2$$\frac{7.88}{62.5}=\frac{62.5v^2}{62.5}$$5.88608=v^2 \to \sqrt{5.88608}$$v=2.43m/s(\hat{x})$

Explanation:

Problem 12:

Step1: Recall kinetic energy formula

$K = \frac{1}{2}mv^2$

Step2: Substitute given values

$m=1907\ \text{kg}, v=11\ \text{m/s}$
$K = \frac{1}{2}(1907)(11)^2 = \frac{1}{2}(1907)(121)$

Step3: Calculate final value

$K = 1907 \times 60.5 = 115373.5\ \text{J}$

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Problem 13:

Step1: Recall kinetic energy formula

$K = \frac{1}{2}mv^2$

Step2: Substitute given values

$m=0.065\ \text{kg}, v=2.5\ \text{m/s}$
$K = \frac{1}{2}(0.065)(2.5)^2 = \frac{1}{2}(0.065)(6.25)$

Step3: Calculate final value

$K = 0.0325 \times 6.25 = 0.203125\ \text{J}$

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Problem 14:

Step1: Rearrange kinetic energy formula for $v$

$v = \sqrt{\frac{2K}{m}}$

Step2: Substitute given values

$K=64.3\ \text{J}, m=6.5\ \text{kg}$
$v = \sqrt{\frac{2(64.3)}{6.5}} = \sqrt{\frac{128.6}{6.5}}$

Step3: Calculate final value

$v = \sqrt{19.7846} \approx 4.45\ \text{m/s}$

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Problem 15a:

Step1: Recall potential energy formula

$U = mgh$

Step2: Substitute given values

$m=0.3\ \text{kg}, g=9.81\ \text{m/s}^2, h=125\ \text{m}$
$U = (0.3)(9.81)(125)$

Step3: Calculate final value

$U = 0.3 \times 1226.25 = 367.875 \approx 367.88\ \text{J}$

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Problem 15b:

Step1: Set kinetic energy = potential energy

$\frac{1}{2}mv^2 = U$

Step2: Rearrange for $v$

$v = \sqrt{\frac{2U}{m}}$

Step3: Substitute given values

$U=367.875\ \text{J}, m=0.3\ \text{kg}$
$v = \sqrt{\frac{2(367.875)}{0.3}} = \sqrt{\frac{735.75}{0.3}}$

Step4: Calculate final value

$v = \sqrt{2452.5} \approx 49.52\ \text{m/s}$

Answer:

1. $\boldsymbol{115373.5\ \text{J}}$
2. $\boldsymbol{0.203125\ \text{J}}$
3. $\boldsymbol{4.45\ \text{m/s}}$

15a. $\boldsymbol{367.88\ \text{J}}$
15b. $\boldsymbol{49.52\ \text{m/s}}$ (corrected from the handwritten error)