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Vedmedyk [2.9K]

3 years ago

9

The y-intercept of a position-time graph of an object gives the average velocity of the object.

1 answer:

finlep [7]3 years ago

4 0

<em>The y-intercept of a position-time graph of an object gives the average velocity of the object is False.</em>

<u>Answer:</u> <em>False. </em>

<u>Explanation:</u>

The initial position of the object can be obtained by calculating the Y intercept of a position-time graph. In a position-time graph, the independent variable is time and dependent variable is position. Y axis is the position axis and x axis is the time axis.

The equation of the graph is given by $y =mx + c$ . m is the slope of the graph and c is the y intercept. When a particle starts from the origin its y intercept is zero.

The nature of the graph gives us an idea about velocity. When the velocity is positive, the position- time graph has positive slope and when the velocity is positive the graph has negative slope.

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When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot

gayaneshka [121]

Answer:

The value is $E = 1.35 *10^{14} \ J$

Explanation:

From the question we are told that

The mass of matter converted to energy on first test is $m = 1 \ g = 0.001 \ kg$

The mass of matter converted to energy on second test $m_1 = 1.5 \ g = 1.5 *10^{-3} \ kg$

Generally the amount of energy that was released by the explosion is mathematically represented as

$E = m * c^2$

=> $E = 1.5 *10^{-3} * [ 3.0 *10^{8}]^2$

=> $E = 1.35 *10^{14} \ J$

7 0

3 years ago

Which two fields of study do the principles of biomechanics come from? A. science and technology B. physics and engineering C. b

kati45 [8]

Answer:

B. physics and engineering

Explanation:

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8 0

3 years ago

Is the time in Tennessee the same as florida

geniusboy [140]

I would say the same thing as the first answer

5 0

3 years ago

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25 mL of water is poured into a graduated cylinder. Several coins are dropped into the water, and the new volume is measured to

ra1l [238]

5 cubic cm is the answer

6 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

3 years ago