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Blizzard [7]
2 years ago
14

? Why might the expejiment need to be repeated when the hypothesis

Physics
1 answer:
Marina86 [1]2 years ago
6 0
To make sure the answer is correct
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It is easy..plz help!
Aleks04 [339]
No because point A has to be higher for more KE to push the object and make the look
3 0
3 years ago
Feathers and a bowling ball are dropped in a vacuum, airless environment. Which one will hit the ground first?
Reil [10]

Answer: at the same time

Explanation: in a vacuum, there isnt air, right? so there isnt gravity pushing down on the heavier object, so they will both land at the same time.

good? :)

8 0
3 years ago
How can force-time and force-displacement graphs be used to find the impulse or work done?
balandron [24]

Answer:

A. Area under force-time graph & Area under force-displacement graph

Explanation:

To find the impulse or work done the area under force-time graph and area under force-displacement graph will give us these respective values.

 Impulse  = Force x time

 Work done  = Force x displacement

When we plot a graph of force and time, the area under it is the impulse.

When a graph of force and displacement is plotted, the area under is the work done.

7 0
2 years ago
lightning strikes in the distance and six seconds later Thunder is heard how far away was the lightning strike
marshall27 [118]
The speed of sound really depends on the temperature and moisture in the air,
and the sound of the thunder doesn't necessarily travel straight from the lightning
to where you are.  So we can't say exactly. 

But if we use the nominal speed of 340 m/s, then 6 seconds means 2,040 meters,
or about 6,700 feet, or about 1.27 miles.
8 0
3 years ago
A car runs at a constant speed of 15ms-1 for 30secs, and then accelerates uniformly to a speed of 25ms-1 over a time of 20secs.
Otrada [13]

Answer:

a. Please find the attached velocity time graph of the car's motion created with Microsoft Excel

b. The total distance traveled by the car is 8,725 meters

c. The average speed of the car is 22.9605263 m/s

Explanation:

The given parameters of the motion are;

The initial speed of the car, v₁ = 15 m/s

The time during which the car runs at the initial speed, t₁ = 30 seconds

The new speed the car then accelerates at 'a₁' to, v₂ = 25 m/s

The duration it takes for the car to accelerate to the new speed = 20 seconds

The time during which the car runs at the initial speed = 300 seconds

The time it takes the car to be brought to rest with a deceleration, 'a₂' from the new speed (20 m/s) = 30 seconds

The final speed of the car at rest, v₃ = 0 m/s

The acceleration, a₁ = (v₂ - v₁)/t₁ = (25 - 15)/20 = 1/2 m/s²

The deceleration , a₂ = (v₃ - v₂)/t₁ = (0 - 25)/30 = -5/6 m/s²

a. Please find attached the drawing of the velocity time graph of the motion created with Microsoft Excel

b. The total distance traveled by the car, 'Δx', is given b the area under the velocity time graph as follows;

Area of trapezoid, A₁ = (320 + 300)/2 × 10 = 3,100

Area of rectangle, A₂ = 15 × 350 = 5,250

Area of triangle, A₃ = 1/2×30×25 = 375

The total area under the velocity time graph = A₁ + A₂ + A₃ = 3,100 + 5,250 + 375 = 8,725

The total area under the velocity time graph = The total distance traveled by the car, Δx = 8,725 meters

c. The average speed of the car is given as follows;

The \ average \  speed  \ of  \ the \  car, \overline v =\dfrac{\Delta x}{\Delta t}  = \dfrac{The \  total \  distance  \ traveled by \  the \  car}{The  \ total  \  time \  the \  car \ travels}

Where;

Δt = The total time during which the car travels

∴ The average speed of the car = 8,725 m/(380 s) = 22.9605263 m/s

4 0
3 years ago
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