Question

1)Earth rotates around Sun at distance of 150 000 000 km. Calculate tangential speed of Earth.Period of rotation of Earth is 1 year.Why do not we feel this speed? 2)Discus throwers throw discus with a rate of 20 m/s. Length of arm is about 1 m.Calculate period and frequency of thrower.How many turns does he make in 3 seconds?помогите пж срочно​

Answer 1

Правильный ответик это 1) я это делала

Answer 2

1) Calculation of Tangential Speed of Earth: The tangential speed (\(v\)) of an object moving in a circular path can be calculated using the formula: \[ v = \frac{2 \pi r}{T} \] where: \( r \) is the radius of the circular path (distance from the Sun to Earth), \( T \) is the period of rotation (time it takes for one complete rotation).

Given that the distance from the Sun to Earth (\( r \)) is 150,000,000 km and the period of rotation (\( T \)) is 1 year, we need to convert the units to meters and seconds to be consistent in the formula.

\[ r = 150,000,000 \, \text{km} \times 1000 \, \text{m/km} = 150,000,000,000 \, \text{m} \] \[ T = 1 \, \text{year} \times 365 \, \text{days/year} \times 24 \, \text{hours/day} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 31,536,000 \, \text{seconds} \]

Now, plug these values into the formula: \[ v = \frac{2 \pi \times 150,000,000,000}{31,536,000} \]

Calculate this to find the tangential speed (\( v \)).

Although the tangential speed is quite high, we don't feel it because we are in constant motion with the Earth. Everything on Earth is moving together, so there's no relative motion to give us a sensation of speed.

2) Calculation of Period, Frequency, and Turns: The period (\( T \)) is the time it takes for one complete revolution, and it is the reciprocal of the frequency (\( f \)), which is the number of revolutions per unit time. \[ T = \frac{1}{f} \]

Given the rate of throwing (\( v \)) is 20 m/s and the length of the arm (\( r \)) is 1 m, the formula for the period is: \[ T = \frac{2 \pi r}{v} \]

Once you find \( T \), you can find the frequency \( f \) using the relationship \( f = \frac{1}{T} \).

Finally, to find the number of turns in 3 seconds (\( N \)), you can use the formula: \[ N = f \times t \] where \( t \) is the time in seconds.

Plug in the values and calculate to find \( T \), \( f \), and \( N \).