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3 years ago

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A 0.250-kg ball is thrown upwards with an initial velocity of 12.0 m/s Determine the acceleration of the ball when it has reache

d the peak of it's trajectory. Assume negligible air resistance.

1 answer:

Nostrana [21]3 years ago

7 0

No math needed here. The only force acting on the ball is gravity. Gravity provides an acceleration of 9.8 m/s^2 downwards. This is the acceleration of the ball at every point along its trajectory, including the peak (when the VELOCITY is zero).

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When two atoms of the same nonmetal react, they often form a(an) A. ionic bond. B. polyatomic ion. C. diatomic molecule. D. pola

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3 years ago

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Consider the electric force between a pair of charged particles a certain

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Doubled

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F and q (Either q1 or q2) are directly proportional, so double the charge would also double the electruc force between the charges.

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What is the frequency of sound waves whose wavelength is 0.25 m on a day when the air temperature is 25 degree Celsius? Group of

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Answer:1384 Hz

Explanation:

Given

wavelength $(\lambda )$ =0.25 m

Temperature T $=25^{\circ}$

at $T=25^{\circ}$ velocity of sound is 346 m/s

and we know

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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

PtichkaEL [24]

Answer:

533.33 nm

Explanation:

Since dsinθ = mλ for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.

Since the fringes coincide,

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8 0

2 years ago

To understand how to find the velocities of objects after a collision.

trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = $\frac{v}{2}$

Part C: v_1 = $\frac{v}{3}$ ; v_2 = $\frac{4v}{3}$

Part D: v_1 = v_2 = $\frac{v}{4}$

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = $\frac{1}{2}$ m. $v^{2}$

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

$\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f} )$

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

$v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i} + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}$

$v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}$

For all the collisions, object 2 is static, i.e. $v_{2i}$ = 0

<u>Part A</u>: Both objects have the same mass (m), $v_{1i}$ = v and collision is elastic:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = 0

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.m}{m+m}.v$

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

$m_{1}$ . $v_{1i}$ + $m_{2}$ . $v_{2i}$ = $m_{1}.v_{1f} + m_{2}.v_{2f}$

$m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}$

$v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m.v}{m+m}$

v_1 = v_2 = $\frac{v}{2}$

<u>Part C:</u> Object 1 is 2m, object 2 is m and elastic collision:

v_1 = $\frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}$

v_1 = $\frac{2m - m}{2m + m } . v$

v_1 = $\frac{v}{3}$

v_2 = $\frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}$

v_2 = $\frac{2.2m}{2m+m}.v$

v_2 = $\frac{4v}{3}$

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = $v_{f} = \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }$

v_1 = v_2 = $\frac{m}{m+3m}.v$

v_1 = v_2 = $\frac{v}{4}$

5 0

3 years ago