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DENIUS [597]
3 years ago
5

You release a pendulum of mass 1 kg from a height of 0.75 m. If there is no air resistance, how fast is the pendulum going when

it reaches the bottom
Physics
1 answer:
puteri [66]3 years ago
8 0

Explanation:

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × 9.8 m/s² × 0.75 m)

v = 3.83 m/s

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A dog starts at the origin and runs forward at 6m/s for 1.5s and then turns around to fetch the ball by running backward at 7m/s
Vera_Pavlovna [14]

Answer:

Total time elapsed between the start and when he returns with the ball is 7.5s

Explanation:

From the question,

- The dog starts at the origin and runs forward at 6m/s for 1.5s. First, we will determine the distance covered while running forward.

From

Speed = Distance / Time

Distance = Speed × Time

Speed = 6m/s

Time = 1.5s

∴ Distance = 6m/s × 1.5s

Distance = 9m

That is, the dog covered a distance of 9m while running forward.

- The dog turns around and runs backward at 7m/s for 3s. Now, we will also determine the distance the dog covered backwards.

Distance = Speed × Time

Speed = 7m/s

Time = 3s

Distance = 7m/s × 3s

Distance = 21m

The dog's displacement from the origin is 21m - 9m = 12m

Now, to calculate how much time has elapsed between the start if the dog runs back to the origin at 4m/s, we will first determine the time the dog spent back to the origin and then add to the time spent for the first two distances.

To get back to the origin, the dog needs to cover 12m

From

Speed = Distance / Time

Time = Distance / Speed

Distance = 12m

Speed = 4m/s

∴ Time = (12m) / (4m/s)

Time = 3s

Therefore, the dog spent 3s to run back to the origin.

Hence, total time elapsed = 1.5s + 3s + 3s

Total time elapsed = 7.5s

8 0
3 years ago
A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two
stepladder [879]

Answer:

ΔT = 0.02412 s

Explanation:

We will simply calculate the time for both the waves to travel through rail distance.

FOR THE TRAVELING THROUGH RAIL:

T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s

FOR THE WAVE TRAVELING THROUGH AIR:

T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s

The separation in time between two pulses can now be given as follows:

\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\

<u>ΔT = 0.02412 s</u>

3 0
3 years ago
The relationship among speed, distance, and time is
alexdok [17]
Distance= speed (multiplied by) time
3 0
3 years ago
Read 2 more answers
If an ideal fluid element moves from a region of lower pressure to a region of higher pressure, is the work done on it by the pr
nikklg [1K]

Answer:

Positive

Explanation:

An ideal fluid is on which is incompressible and non-viscous.

In an ideal fluid element when a fluid element moves from a region of high pressure to region of low pressure then the work done on the fluid element is positive because the force acts in the direction from high pressure towards low pressure and displaces the fluid element, so the work-done is positive.

6 0
3 years ago
When you jump upward, your hang time is the time your feet are off the ground. Does hang time depend on the vertical component o
hjlf

Answer:

It only depends on the vertical component

Explanation:

Hello!

The horizontal component will tell you how much you travel in that direction.

You could have a large horizontal velocity, but if the vertical velocity is zero, you will never be out of the ground. Similarly, you could have a zero horizontal velocity, but if you have a non-zero vertical velocity you will be some time off the ground. This time can be calculated by two means, one is using the equation of motion (position as a function of time) and the other using the velocity as a fucntion of time.

For the former you must find the time when the position is zero.

Lets consider the origin of teh coordinate system at your feet

y(t) = vt - (1/2)gt^2

We are looking for a time t' for which y(t')=0

0 = vt' - (1/2)gt'^2

vt' = (1/2)gt'^2

The trivial solution is when t'=0 which is the initial position, however we are looking for t'≠0, therefore we can divide teh last equation by t'

v = (1/2)gt'

Solving for t'

t' = (2v/g)

7 0
3 years ago
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