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tankabanditka [31]
3 years ago
4

Three different experiments are conducted that pertain to the oscillatory motion of a pendulum. For each experiment, the length

of the pendulum and the mass of the pendulum are indicated. In all experiments, the pendulum is released from the same angle with respect to the vertical. If the students collect data about the kinetic energy of the pendulum as a function of time for each experiment, which of the following claims is true?
a. The data collected from Experiment 1 will be the same as the data collected from Experiment 2.
b. The data collected from Experiment 1 will be the same as the data collected from Experiment 3.
c. The data collected from Experiment 2 will be the same as the data collected from Experiment 3.
d. The data collected from each experiment will be different.
Physics
1 answer:
ryzh [129]3 years ago
0 0

Answer:

d. The data collected from each experiment will be different.

Explanation:

The time period of simple pendulum depends on length of pendulum and independent of mass of pendulum.

For three experiments data is collected. Kinetic energy depends on mass and square of velocity.

SO FOR THREE EXPERIMENTS THE DATA COLLECTED WILL BE DIFFERENT.

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Explain what happens to the particles in a substance during a physical change.
omeli [17]

Answer:

In physical changes no new materials are formed and the particles do not change apart from gaining or losing energy. ... Particles stay the same unless there is a chemical change whether the matter is solid, liquid or gas. Only their arrangement, energy and movement changes.

Explanation:

Hope this helps

7 0
3 years ago
Read 2 more answers
A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on
lorasvet [3.4K]

Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

Final velocity= Vf= 20 m/s

time = t= 5 s

Using kinematic equation,

Vf= Vi + at

20= 0 + a (5)

5 a=20

a= 20/5

a= 4 m/s²

Now force is given by F = ma

F= 1.2 x 10³ (4)

F=4.8 x 10³ N

7 0
3 years ago
What is the resistance of the coil A at 600 kelvin if its resistance at 300 kelvin is 50 ohms? (Assume the temperature coefficie
leonid [27]

155Ω

Explanation:

R = R ref ( 1 + ∝ ( T - Tref)  

where R = conduction resistance at temperature T

R ref = conductor resistance at reference temperature

∝ = temperature coefficient of resistance for conductor

T = conduction temperature in degrees Celsius

T ref = reference temperature that ∝ is specified at for the conductor material

T = 600 k - 273 k = 327 °C

Tref = 300 - 273 K = 27 °C

R = 50 Ω ( 1 + 0.007 ( 327 - 27) )

R = 155Ω

8 0
3 years ago
Read 2 more answers
You are on a sled at the top of a hemispherical, snowy hill of radius 13 m. You begin to slide down the hill. How fast are you m
8_murik_8 [283]

Answer:

Explanation:

There will be loss of potential energy due to loss of height and gain of kinetic energy .

loss of height = R - R cos 14 ,    R is radius of hemisphere .

R ( 1 - cos 12 )

= 13 ( 1 - .978 )

h = .286 m

loss of potential energy

= mgh

= m x 9.8 x .286

= 2.8 m

gain of kinetic energy

1/2 m v ² = mgh

v² = 2 g h

v²  = 2 x 9.8 x 2.8

v = 7.40 m /s

4 0
3 years ago
Read 2 more answers
) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffi
natali 33 [55]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

4 0
3 years ago
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