Question

Физика: 1) За какое время автомобиль двигаясь с ускорением 0,4 м/с увеличит свою скорость с 12 м/с до 20 м\с 2) Мальчик бросил вертикально вверх мяч и поймал его через 2 сек, на какую высоту поднялся мяч, какова его начальная скорость? 3) Найти центростремительное ускорение точек колеса автомобиля соприкасающееся с дорогой, если автомобиль движется 72 км/с, при этом частота вращения колеса 8с 4) Найти частоту обращения луны вокруг земли.

Answer 1

1) dV=V1-V1 =>dV=20-12=8м/c    t=V/a=> t=8/0,4=20c

Answer 2

Physics Questions

1) Time to Increase Speed: To calculate the time it takes for a car to increase its speed from 12 m/s to 20 m/s with an acceleration of 0.4 m/s^2, we can use the following kinematic equation:

v = u + at

Where: - v = final velocity (20 m/s) - u = initial velocity (12 m/s) - a = acceleration (0.4 m/s^2) - t = time

Solving for t: 20 m/s = 12 m/s + (0.4 m/s^2) * t t = (20 m/s - 12 m/s) / 0.4 m/s^2 t = 8 s

So, it will take the car 8 seconds to increase its speed from 12 m/s to 20 m/s.

2) Vertical Motion of the Ball: When a ball is thrown vertically upward and caught after 2 seconds, we can calculate the height it reaches and its initial velocity using the following kinematic equations:

v = u + gt s = ut + (1/2)gt^2

Where: - v = final velocity (0 m/s at the peak) - u = initial velocity (upward, to be calculated) - g = acceleration due to gravity (-9.81 m/s^2) - t = time (2 s) - s = height reached

Using the first equation, we can solve for the initial velocity: 0 m/s = u - 9.81 m/s^2 * 2 s u = 19.62 m/s

Using the second equation, we can solve for the height reached: s = (19.62 m/s) * 2 s + (1/2)(-9.81 m/s^2)(2 s)^2 s = 19.62 m * 2 s - 4.905 m/s^2 * 4 s^2 s = 39.24 m - 19.62 m s = 19.62 m

So, the ball reaches a height of 19.62 meters, and its initial velocity is 19.62 m/s upward.

3) Centripetal Acceleration of Car Wheel: The centripetal acceleration of a point on the car wheel in contact with the road can be calculated using the formula:

a = rω^2

Where: - a = centripetal acceleration - r = radius of the wheel - ω = angular velocity

First, we need to convert the speed of the car from km/s to m/s: 72 km/s = 72000 m/s

Then, we can calculate the angular velocity using the formula: ω = 2πf Where f is the frequency of rotation.

Given that the frequency of rotation is 8 s, we can calculate ω: ω = 2π / 8 ω = π / 4 rad/s

Now, we can calculate the centripetal acceleration using the formula: a = (72000 m) * (π / 4 rad/s)^2 a ≈ 56548.19 m/s^2

So, the centripetal acceleration of the point on the car wheel in contact with the road is approximately 56548.19 m/s^2.

4) Lunar Orbital Frequency: The frequency of revolution of the Moon around the Earth can be calculated using the formula:

f = 1 / T

Where: - f = frequency - T = time period

The time period for one complete revolution of the Moon around the Earth is approximately 27.3 days.

Converting the time period to seconds: 27.3 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute ≈ 2.36 × 10^6 seconds

Now, we can calculate the frequency: f = 1 / (2.36 × 10^6 seconds) f ≈ 4.24 × 10^-7 Hz

So, the frequency of revolution of the Moon around the Earth is approximately 4.24 × 10^-7 Hz.

I hope this helps! If you have further questions or need additional assistance, feel free to ask.