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balandron [24]
3 years ago
9

What is the potential energy of a 2,000-kg car parked at the top of a 30-m hill?

Physics
2 answers:
babymother [125]3 years ago
5 0
Potential energy is a relative measure, so the answer is dependent on the assumptions we make. The potential energy in the car is going to be gravitational potential energy(PE). PE = mgh, where m is the mass, g is 9.8 m/s^2, and h is the height. So PE = 2000*9.8*h = 19600h. The final answer obviously depends on h. Most likely the problem is assuming that 30 meters under the top of the hill is considered 0 meters. Then h would be 30m and PE would equal 588 kJ.
vampirchik [111]3 years ago
5 0

Answer:

the answer is 588,000 J

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When can a theory be modified if a new type of technology allows for new observations that raise new questions?
olga nikolaevna [1]

Answer:

id say B, but not sure

Explanation:

4 0
3 years ago
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Which of the following is characteristic of proficient catching?
uysha [10]

Answer:

The correct answer is option D i.e. A and C

Explanation:

The correct answer is option D i.e. A and C

for proficient catching player must

- learn to absorbed the ball force

- moves the hang according to ball direction to hold the ball

- to catch ball at high height move the finger at higher position

- to catch ball at low height move the finger at lower position

5 0
2 years ago
A soccer player kicks a soccer ball with a force of 1.8 N. If the mass of the ball is .43 kg. How fast will the ball accelerate?
Vesnalui [34]

Answer:

The acceleration of the ball is 4.18 [m/s^2]

Explanation:

By Newton's second law we can find the acceleration of the ball

F = m*a\\where:\\F = force applied [N] or [kg*m/s^2]\\m = mass of the ball [kg]\\a = acceleration [m/s^s]

Now we have:

a = F/m\\a = \frac{1.8 [kg*m/s^s]}{0.43[kg]} \\a = 4.18 [kg]

4 0
3 years ago
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A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The ang
olga55 [171]

Answer:

9.22 s

Explanation:

One-quarter of a turn away is 1/4 of 2π, or π/2 which is approximately 1.57 rad

Let t (seconds) be the time it takes for the child to catch up with the horse. We would have the following equation of motion for the child and the horse:

For the child: s_c = \omega_ct = 0.233t

For the horse: s_h = s_0 + a_ht^2/2 = 1.57 + 0.0136t^2/2 = 1.57 + 0.0068t^2

For the child to catch up with the horse, they must cover the same angular distance within the same time t:

s_c = s_h

0.233t = 1.57 + 0.0068t^2

0.0068t^2 - 0.233t + 1.57 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{0.233\pm \sqrt{(-0.233)^2 - 4*(0.0068)*(1.57)}}{2*(0.0068)}

t= \frac{0.233\pm0.11}{0.0136}

t = 25.05 or t = 9.22

Since we are looking for the shortest time we will pick t = 9.22 s

6 0
3 years ago
Fiziksel büyüklüklerin skaler ve vektörel olarak ayrı ayrı sınıflandırılmasının nedeni nedir? ​
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Uhm sir what language is this
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