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Norma-Jean [14]
3 years ago
15

You are investigating how objects move when they are dropped from different heights. To collect your data, you drop a 1 kg weigh

t from different heights and record the time it takes for the object to hit the ground when dropped from different heights. The data you collect is shown below:
Height dropped (m)

Time to fall (s)

1.0

0.45

2.0

0.63

3.0

0.78

4.0

0.89



Next, you plan to drop a 5 kg weight from the same heights. How will your time values in your new data table for the 5 kg weight compare to the time values in the old data table for the 1 kg weight?
Physics
1 answer:
ASHA 777 [7]3 years ago
3 0

The time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

The given parameters;

<em>Mass of the first object, m1 = 1 kg</em>

<em>Mass of the second object, m2 = 5 kg</em>

The final velocity of the objects during the downward motion is calculated as follows;

v_f = v_0 + gt\\\\v_f = 0 + gt\\\\\v_f = gt

The time of motion of the object from the given height is calculated as;

h = v_0 t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} }

The time of motion of each object is independent of mass of the object.

Thus, the time of motion of the 5 kg object will be the same as 1 kg since both objects are dropped from the same height.

Learn more about time of motion here: brainly.com/question/2364404

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Two ropes are attached to a heavy box to pull it along the floor. one rope applies a force of 475 newtons in a direction due wes
vekshin1

Answer:570 N

Explanation:

Given

one rope applies a force of 475 N in west direction

other applies a force of 315 N in south direction

So both forces are at an angle of 90^{\circ}

Net resultant of two forces is

R=\sqrt{315^2+475^2}

R=\sqrt{99225+225625}

R=569.95 N \approx 570

and the resultant will be an angle of \theta with 475 Newton force

Rcos\theta =475

cos\theta =\frac{475}{569.95}=0.833

\theta =cos^{-1}(0.833)=33.591

So replacing two force we can apply a force of magnitude of 570 N at angle of 33.6 ^{\circ} with west

3 0
3 years ago
Generally blank friction is greater than sliding friction
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4 0
3 years ago
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tankabanditka [31]
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6 0
4 years ago
-In the Bohr model, as it is known today, the electron is imagined to move in a circular orbit about a stationary proton. The fo
vampirchik [111]

Answer:

4.3859007196\times 10^{-9}\ m

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

v = Velocity of electron = 2.4\times 10^5\ m/s

q = Charge of electron = 1.6\times 10^{-19}\ C

m = Mass of electron = 9.11\times 10^{-31}\ kg

r = Radius

The electrical and centripetal force will balance each other

\dfrac{kq^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{kq^2}{mv^2}\\\Rightarrow r=\dfrac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times (2.4\times 10^5)^2}\\\Rightarrow r=4.3859007196\times 10^{-9}\ m

The radius of the orbital is 4.3859007196\times 10^{-9}\ m

4 0
3 years ago
A highway patrolman traveling at the speed limit is passed by a car going 20 mph faster than the speed limit. After one minute,
Oksi-84 [34.3K]

Answer:

It will take 40 seconds to catch the speeding car

Explanation:

Initial speed of the patrolman = 55 mph

after one minute speed of patrol man = 115 mph

now acceleration of patrolman is given by

a = \frac{v_f - v_i}{t}

a = \frac{115 - 55}{1/60} = 3600 m/h^2

now at this acceleration the distance covered by patrolman in "t" time is given as

d = v_i t + \frac{1}{2}at^2

d = 55t + 1800t^2

now we know the speed of the speeding car is given as

v' = (55+20) mph

now in the same time distance covered by it

d = 75 t

now since the distance covered is same

75 t = 55t + 1800 t^2

t = \frac{20}{1800} = \frac{1}{90} h

t = 40 seconds

7 0
4 years ago
Read 2 more answers
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