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Ostrovityanka [42]

4 years ago

5

Imagine that someone pushes one marble toward a motionless marble. Would there still be action-reaction forces involved in the c

ollision? How might the marbles’ motions be changed?? HELP

1 answer:

Finger [1]4 years ago

8 0

All the energy that was in the first marble was transfers into the second marble

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Is the range of the projectile dependent or independent of the projectile's mass? Explain.

Artist 52 [7]

Answer:

independent of the projectiles mass

Explanation:

the range only depends on velocity of projection,angle of projection,acceleration due to gravity.

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3 years ago

List two ways that economics relates to the environment

rjkz [21]

Answer:

it relates to the economic in these ways

1.unlimited resources

2 wants

Explanation:

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3 years ago

A uniform steel rod has mass 0.400 kg and length 50.0 cm and is horizontal. A uniform sphere with radius

Eddi Din [679]

Explanation:

Center of gravity , Let center of rod is origin

= $\frac{(0.400*0)+(0.800*\frac{50}{2}+10)+ (0.200 *(-(25+5)) }{0.400+0.800+0.200}$

=15.71cm from the center of rod

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3 years ago

What's the concept of mass ?

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6 0

3 years ago

A simple pendulum has a period of 3.45 second, when the length of the pendulum is shortened by 1.0m, the period is 2.81 second c

den301095 [7]

Answer:

Original length = 2.97 m

Explanation:

Let the original length of the pendulum be 'L' m

Given:

Acceleration due to gravity (g) = 9.8 m/s²

Original time period of the pendulum (T) = 3.45 s

Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, $L_1=L-1$

New time period of the pendulum is, $T_1=2.81\ s$

We know that, the time period of a simple pendulum of length 'L' is given as:

$T=2\pi\sqrt{\frac{L}{g}}$ -------------- (1)

So, for the new length, the time period is given as:

$T_1=2\pi\sqrt{\frac{L_1}{g}}$ ------------ (2)

Squaring both the equations and then dividing them, we get:

$\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1$

Now, plug in the given values and calculate 'L'. This gives,

$L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m$

Therefore, the original length of the simple pendulum is 2.97 m

4 0

3 years ago