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3 years ago

Under what condition will the results of an experiment based on a hypothesis most likely lead to new experimentation?

Physics

1 answer:

MakcuM [25]3 years ago

7 0

Failed experiments, uncontrolled variables, invalid data, and generalized human error

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3 years ago

4) T F If two objects have the same moment of inertia, they must have the same mass.

lisov135 [29]

Answer:

False

Explanation:

The moment of inertia for a rigid body is given by

$I=\int \rho(r) r^2 dV$

where

$\rho(V)$ is the density distribution of the object

r is the distance from the axis of rotation of the object

Essentially, the moment of inertia does not depend only on the mass of the object, but also on its shape. For example: for a solid cylinder, the moment of inertia derived from the formula above is

$I=\frac{1}{2}MR^2$

where M is the mass of the cylinder and R is its radius. As we see, I (moment of inertia) does not depend on the mass only: therefore, if two objects have same moment of inertia, this does not imply that they also have the same mass.

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3 years ago

Electric charges come in two forms: positive and negative. What combinations of charges will attract and repel each other?

sammy [17]

Opposites attract while like charges repel. :)

5 0

3 years ago

Water drips from the nozzle of a shower onto the floor 200 cm below. The drops fall at regular (equal) intervals of time, the fi

Delicious77 [7]

Answer:

(A) 88.92 cm

(B) 22.22 cm

Explanation:

distance (s) = 200 cm = 0.2 m

initial velocity (v) = 0 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the time (T) it takes for the first drop to strike the floor

from s = ut + $0.5at^{2}$

200 = 0 + $0.5 x 9.8 x T^{2}$

200 = $4.9 x T^{2}$

200 / 4.9 = $T^{2}$

T = 6.4

(A) When the first drop strikes the floor, how far below the nozzle is the second drop.

we can find how far the second drop was when the first drop hits the ground from the formula s = ut + $0.5at^{2}$

where

s = distance
u = initial velocity = 0
t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the second drop = 2/3 of the time it takes the first drop to strike the ground ( $\frac{2}{3}.T$ )
a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + ( $0.5 x 9.8 x (\frac{2}{3}T)^{2}$ )

s = 0 + ( $0.5 x 9.8 x (\frac{2}{3} x 6.4)^{2}$ )

s = 88.92 cm

(B) When the first drop strikes the floor, how far below the nozzle is the third drop.

we can find how far the third drop was when the first drop hits the ground from the formula s = ut + $0.5at^{2}$

where

s = distance
u = initial velocity = 0
t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the third drop = 1/3 of the time it takes the first drop to strike the ground ( $\frac{2}{3}.T$ )
a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + ( $0.5 x 9.8 x (\frac{1}{3}T)^{2}$ )