(a) An object with constant acceleration but zero velocity: Possible. Example - A ball thrown upwards momentarily stops at the highest point of its trajectory due to gravity, having zero velocity while experiencing constant acceleration. (b) An object moving with acceleration perpendicular to its diRead more
(a) An object with constant acceleration but zero velocity: Possible. Example – A ball thrown upwards momentarily stops at the highest point of its trajectory due to gravity, having zero velocity while experiencing constant acceleration.
(b) An object moving with acceleration perpendicular to its direction: Possible. Example – A car moving eastwards turns left, experiencing centripetal acceleration perpendicular to its velocity, enabling circular motion.
To find the speed of the artificial satellite moving in a circular orbit around the Earth, we can use the formula relating the circumference of the orbit to the time taken for one revolution. The formula for the circumference of a circle is 2 x π x radius Given: - Radius of the orbit r = 42250 km -Read more
To find the speed of the artificial satellite moving in a circular orbit around the Earth, we can use the formula relating the circumference of the orbit to the time taken for one revolution.
The formula for the circumference of a circle is 2 x π x radius
Given:
– Radius of the orbit r = 42250 km
– Time taken for one revolution T = 24 hours
Calculations:
The circumference of the circular orbit:
Circumference = 2 x π x radius = 2 x π x 42250 km
The speed of the satellite is given by the formula:
Speed = Circumference / Time taken for one revolution
First, let’s convert the time from hours to seconds because the speed is usually measured in distance per unit time in seconds.
Given: 1 hour = 3600 seconds
Time taken for one revolution in seconds = 24 hours 3600 seconds/hour = 86400 seconds
Now, calculate the speed:
Speed = (2 x π x 42250 km) / (86400 seconds)
Speed ≈ (2 x 3.1416 x 42250 km) / (86400 seconds)
Speed ≈ 265258 km) / (86400 seconds)
Speed ≈ 3.07 km/s
Therefore, the speed of the artificial satellite moving in a circular orbit of radius 42250 km, taking 24 hours to revolve around the Earth, is approximately 3.07 kilometers per second.
To plot the speed versus time graphs for the two cars, we'll first convert the speeds from km/h to m/s (since the time is given in seconds) and then illustrate the deceleration of the cars. Given: - Car 1: Initial speed v1 = 52 km/h, Time to stop t1 = 5 s - Car 2: Initial speed v2 = 3 km/h, Time toRead more
To plot the speed versus time graphs for the two cars, we’ll first convert the speeds from km/h to m/s (since the time is given in seconds) and then illustrate the deceleration of the cars.
Given:
– Car 1: Initial speed v1 = 52 km/h, Time to stop t1 = 5 s
– Car 2: Initial speed v2 = 3 km/h, Time to stop t2 = 10 s
Converting speeds to m/s:
– Car 1: v1 = 52km/h = ((52 x 1000) x (3600)) m/s ≈ 14.44 m/s
– Car 2: v2 = 3km/h = ((3 x 1000) x (3600))m/s}\) ≈ 0.83 m/s
Now, let’s plot the speed versus time graphs for both cars:
Graph:
– Car 1 (Deceleration):
– Starts at 14.44 m/s
– Decelerates uniformly until 0 m/s in 5 seconds.
– Car 2 (Deceleration):
– Starts at 0.83 m/s
– Decelerates uniformly until 0 m/s in 10 seconds.
The area under the speed-time graph represents the distance covered.
– Car 1’s Area: 1/2 x (initial speed + final speed) x time = 1/2 x (14.44m/s + 0 m/s) x 5 s = 36.1m
– Car 2’s Area: 1/2 x (initial speed + final speed) x time = 1/2 x (0.83 m/s + 0 m/s x 10 s = 4.15m
Conclusion:
Car 1, despite having a higher initial speed, covered a greater distance after the brakes were applied. Car 1 traveled approximately 36.1 meters, while Car 2 covered approximately 4.15 meters before coming to a stop.
State which of the following situations are possible and give an example for each of these:
(a) An object with constant acceleration but zero velocity: Possible. Example - A ball thrown upwards momentarily stops at the highest point of its trajectory due to gravity, having zero velocity while experiencing constant acceleration. (b) An object moving with acceleration perpendicular to its diRead more
(a) An object with constant acceleration but zero velocity: Possible. Example – A ball thrown upwards momentarily stops at the highest point of its trajectory due to gravity, having zero velocity while experiencing constant acceleration.
(b) An object moving with acceleration perpendicular to its direction: Possible. Example – A car moving eastwards turns left, experiencing centripetal acceleration perpendicular to its velocity, enabling circular motion.
See lessAn artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
To find the speed of the artificial satellite moving in a circular orbit around the Earth, we can use the formula relating the circumference of the orbit to the time taken for one revolution. The formula for the circumference of a circle is 2 x π x radius Given: - Radius of the orbit r = 42250 km -Read more
To find the speed of the artificial satellite moving in a circular orbit around the Earth, we can use the formula relating the circumference of the orbit to the time taken for one revolution.
The formula for the circumference of a circle is 2 x π x radius
Given:
– Radius of the orbit r = 42250 km
– Time taken for one revolution T = 24 hours
Calculations:
The circumference of the circular orbit:
Circumference = 2 x π x radius = 2 x π x 42250 km
The speed of the satellite is given by the formula:
Speed = Circumference / Time taken for one revolution
First, let’s convert the time from hours to seconds because the speed is usually measured in distance per unit time in seconds.
Given: 1 hour = 3600 seconds
Time taken for one revolution in seconds = 24 hours 3600 seconds/hour = 86400 seconds
Now, calculate the speed:
Speed = (2 x π x 42250 km) / (86400 seconds)
Speed ≈ (2 x 3.1416 x 42250 km) / (86400 seconds)
Speed ≈ 265258 km) / (86400 seconds)
Speed ≈ 3.07 km/s
Therefore, the speed of the artificial satellite moving in a circular orbit of radius 42250 km, taking 24 hours to revolve around the Earth, is approximately 3.07 kilometers per second.
See lessA driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
To plot the speed versus time graphs for the two cars, we'll first convert the speeds from km/h to m/s (since the time is given in seconds) and then illustrate the deceleration of the cars. Given: - Car 1: Initial speed v1 = 52 km/h, Time to stop t1 = 5 s - Car 2: Initial speed v2 = 3 km/h, Time toRead more
To plot the speed versus time graphs for the two cars, we’ll first convert the speeds from km/h to m/s (since the time is given in seconds) and then illustrate the deceleration of the cars.
Given:
– Car 1: Initial speed v1 = 52 km/h, Time to stop t1 = 5 s
– Car 2: Initial speed v2 = 3 km/h, Time to stop t2 = 10 s
Converting speeds to m/s:
– Car 1: v1 = 52km/h = ((52 x 1000) x (3600)) m/s ≈ 14.44 m/s
– Car 2: v2 = 3km/h = ((3 x 1000) x (3600))m/s}\) ≈ 0.83 m/s
Now, let’s plot the speed versus time graphs for both cars:
Graph:
– Car 1 (Deceleration):
– Starts at 14.44 m/s
– Decelerates uniformly until 0 m/s in 5 seconds.
– Car 2 (Deceleration):
– Starts at 0.83 m/s
– Decelerates uniformly until 0 m/s in 10 seconds.
The area under the speed-time graph represents the distance covered.
– Car 1’s Area: 1/2 x (initial speed + final speed) x time = 1/2 x (14.44m/s + 0 m/s) x 5 s = 36.1m
– Car 2’s Area: 1/2 x (initial speed + final speed) x time = 1/2 x (0.83 m/s + 0 m/s x 10 s = 4.15m
Conclusion:
See lessCar 1, despite having a higher initial speed, covered a greater distance after the brakes were applied. Car 1 traveled approximately 36.1 meters, while Car 2 covered approximately 4.15 meters before coming to a stop.