To find the speed of the sweep second hand tip at the 6 o'clock position, we need to determine its angular speed.
The angular speed is given by the formula:
angular speed (ω) = linear speed (v) / radius (r)
The length of the sweep second hand (r) is given as 2.0 cm.
At the 6 o'clock position, the tip of the sweep second hand experiences a linear speed equal to the circumference of a circle with a radius of 2.0 cm, or 2πr.
Substituting these values into the formula, we get:
angular speed (ω) = 2πr / r = 2π
Therefore, the angular speed of the sweep second hand at the 6 o'clock position is 2π radians per unit of time.
Since no unit of time is specified in the question, we can assume it is per second. Therefore, the speed of the sweep second hand at the 6 o'clock position is 2π radians per second.
The sweep second hand of a clock has a length of 2.0 cm. a) What is the speed of the sweep second hand tip at the 6 o’clock position?
5 answers
What is the velocity of the sweep second hand tip at the 6 o’clock position?
To find the velocity of the sweep second hand tip at the 6 o'clock position, we need to consider both the angular speed of the hand and its radius.
The angular speed of the hand at the 6 o'clock position is 2π radians per unit of time, as we calculated in the previous response.
The velocity of the tip of the sweep second hand is given by the formula:
velocity (v) = angular speed (ω) × radius (r)
Substituting the values we have:
velocity (v) = 2π × 2.0 cm
= 4π cm (or approximately 12.57 cm)
Therefore, the velocity of the sweep second hand tip at the 6 o'clock position is 4π cm (or approximately 12.57 cm) per unit of time.
The angular speed of the hand at the 6 o'clock position is 2π radians per unit of time, as we calculated in the previous response.
The velocity of the tip of the sweep second hand is given by the formula:
velocity (v) = angular speed (ω) × radius (r)
Substituting the values we have:
velocity (v) = 2π × 2.0 cm
= 4π cm (or approximately 12.57 cm)
Therefore, the velocity of the sweep second hand tip at the 6 o'clock position is 4π cm (or approximately 12.57 cm) per unit of time.
Two friends see each other in a grocery store. Initially, they are 50 m apart. The first friend starts walking toward the second friend at a constant speed of 0.50 m/ s. At the same instant, the second friend accelerates uni- formly from rest at a rate of 1.0 m/ s 2 toward the first friend. How long before the two friends can shake hands?
To find out how long it takes for the two friends to shake hands, we need to determine the time it takes for the second friend to catch up to the first friend.
Let's assume t is the time it takes for the two friends to shake hands.
For the first friend:
Distance traveled = speed × time
Distance traveled = 0.50 m/s × t
For the second friend:
Distance traveled = initial velocity × time + (1/2) × acceleration × time^2
Distance traveled = 0.5 × t × t + (1/2) × 1.0 m/s^2 × t^2
Distance traveled = (0.5 + 0.5) × t^2
Distance traveled = t^2
Since they meet when they have traveled the same distance, we can set up the equation:
0.50t = t^2
Rearranging the equation:
t^2 - 0.50t = 0
This is a quadratic equation. We can solve it by factoring or using the quadratic formula.
If we factor the equation:
t(t - 0.50) = 0
From this, we can see that t = 0 or t = 0.50
Since time cannot be negative, we discard the t = 0 solution.
Therefore, it takes 0.50 seconds for the two friends to shake hands.
Let's assume t is the time it takes for the two friends to shake hands.
For the first friend:
Distance traveled = speed × time
Distance traveled = 0.50 m/s × t
For the second friend:
Distance traveled = initial velocity × time + (1/2) × acceleration × time^2
Distance traveled = 0.5 × t × t + (1/2) × 1.0 m/s^2 × t^2
Distance traveled = (0.5 + 0.5) × t^2
Distance traveled = t^2
Since they meet when they have traveled the same distance, we can set up the equation:
0.50t = t^2
Rearranging the equation:
t^2 - 0.50t = 0
This is a quadratic equation. We can solve it by factoring or using the quadratic formula.
If we factor the equation:
t(t - 0.50) = 0
From this, we can see that t = 0 or t = 0.50
Since time cannot be negative, we discard the t = 0 solution.
Therefore, it takes 0.50 seconds for the two friends to shake hands.
1 answer