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Norma-Jean [14]
2 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]2 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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Velocity vs Time
Grace [21]

Answer:

A and C

C

40 m/s

zero

Explanation:

8 0
3 years ago
A ski starts from rest and slides down a 22 incline 75 m long. if the coefficient of friction is 0.090, what is the ski's speed
miskamm [114]

L = length of the incline = 75 m

θ = angle of incline = 22 deg

h = height of skier at the top of incline = L Sinθ = (75) Sin22 = 28.1 m

μ = Coefficient of friction = 0.090

N = normal force by the surface of incline

mg Cosθ = Component of weight of skier normal to the surface of incline opposite to normal force N

normal force "N" balances the component of weight opposite to it hence we get

N = mg Cosθ

frictional force acting on the skier is given as

f = μN

f = μmg Cosθ

v = speed of skier at the bottom of incline

Using conservation of energy

potential energy at the top of incline = kinetic energy at the bottom + work done by frictional force

mgh = f L + (0.5) m v²

mgh = μmg Cosθ L + (0.5) m v²

gh = μg Cosθ L + (0.5) v²

(9.8 x 28.1) = (0.09 x 9.8 x 75) Cos22 + (0.5) v²

v = 20.7 m/s

8 0
3 years ago
Answer this please.____
ad-work [718]

Answer:

Explanation:

When a liquid is heated, the particles are given more energy. They start to move faster and further apart. At a certain temperature, the particles break free of one another and the liquid turns to gas. This is the boiling point.

5 0
3 years ago
A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. what is the angular acceleration?
SCORPION-xisa [38]

Answer:

α = -π/3 rad/s²

θ = 1.5π rad ≈ 4.71 rad

θ = 0.75 rev

Explanation:

30 rev/min (2π rad/rev) / (60 s/min) = π rad/s

α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²

θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad

θ = 1.5π rad / 2π rad/rev = 0.75 rev

5 0
3 years ago
13.4. Young's modulus for iron is 1.9 x 10' Pa. When an iron wire 1.0 m long with a cross-sectional area of 4.0 mm
Viefleur [7K]

Answer:

0.0129 m

Explanation:

ΔL = FL / (EA)

where ΔL is the deflection,

F is the force,

L is the initial length,

E is Young's modulus,

and A is the cross sectional area.

F = mg = 100 kg × 9.8 m/s² = 9800 N

A = 4.0 mm² × (1 m / 1000 mm)² = 4×10⁻⁶ m²

ΔL = (9800 N) (1.0 m) / ((1.9×10¹¹ Pa) (4×10⁻⁶ m²))

ΔL = 0.0129 m

6 0
3 years ago
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