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netineya [11]
3 years ago
7

Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr

ound. When the balls collide, they are moving n opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur? (Use any variable or symbol stated above as necessary.) Both balls experience constant acceleration once they are �n flight. Choose a coordinate system with the origin at the ground and the upward direction positive. When the balls collide they are at the same height above the ground.
Physics
1 answer:
kramer3 years ago
3 0

Answer:

2h / 3

Explanation:

Let after time t , collision occurs

distance covered by A

= 1/2gt²

Let u be the initial velocity with which B was thrown upwards

distance by B

= ut - 1/2 gt²

total distance by A and B

h = ut - 1/2 gt² +  1/2gt²

h = ut

t = h / u

velocity of A after t

= gt

= g x h/u

velocity of B after t

= u - gt

= u - g x h/u

given

g x h/u = 2 (u - g x h/u)

2u² - 2gh = gh

u = \sqrt{\frac{3gh}{2} }

t = h / u

= \sqrt{\frac{2h}{3g} }

Distance covered by A

= 1/2 gt²

= 1/2 x 2h/3

= h/3

Height at which collision occured

= h - h/3

= 2h /3

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Answer:

True The grid with more slits gives more angle separation increases

True. The grating with 10 slits produces better-defined (narrower) peaks

Explanation:

Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is

      d sin θ = m λ

where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.

For network with 5 slits

      d = 1/5 = 0.2

For the network with 10 slits

      d = 1/10 = 0.1

let's calculate the separation (teat) for each one

      θ = sin⁻¹ (m λ / d)

for 5 slits

     θ₅ = sin⁻¹ (m λ 5)

for 10 slits

     θ₁₀ = sin⁻¹ (m λ 10)

we can appreciate that for more slits the angle increases

the intensity of a series of slits is

       I = I₀ sin²2 (N d/2) / sin² d/2)

when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)

let's analyze the claims

False

True The grid with more slits gives more angle separation increases

False

True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases

False

5 0
3 years ago
How far will a man travel in 15 min driving his car down the highway at an average speed of 24 m/?s
emmasim [6.3K]

Answer:

21.6 km

Explanation:

15×60×24 = 21600 m

\frac{21600}{1000}= 21.6 km

7 0
3 years ago
A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially
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Answer:

Explanation:

I will try to use newton’s second law and its’ concept of spring. The detailed solution is shown it the documents and I also used some mathematical concept which is highlighted.

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3 years ago
Se deja caer una moneda desde cierta altura. Si se desprecian los efectos del aire, ¿cómo varía la fuerza neta sobre la moneda a
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Answer:

Ok, primero pensemos en una situación normal.

La moneda comienza a caer, pero la moneda esta inmersa en una sustancia, el aire. El aire comienza a aplicar una resistencia al movimiento de la moneda, y esta resistencia incremente a medida que la velocidad de la moneda incremente. Llega un punto en el que esta nueva fuerza es igual a la fuerza gravitatoria, y en sentido opuesto, lo que causa que la fuerza neta sea 0, y que la moneda caiga a velocidad constante hasta que esta impacta con el suelo.

Ahora, en este caso tenemos que ignorar los efectos del aire, entonces no hay ninguna fuerza que se oponga a la fuerza gravitatoria, entonces la fuerza neta no cambia a medida que cae (La fuerza neta cambia cuando la moneda impacta el suelo).

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3 0
3 years ago
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Answer:

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Time \Delta t = 0.2\ s

The velocity is no more than a 14 % error in the speed of light.

So,

Velocity v= c\times 86\%

We need to calculate the distance

Using formula of speed

v = \dfrac{d}{t}

d = v\times t

Where, v = speed

d = distance

t = time

Put the value into the formula

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d=516\times10^{5}\ m

We know that,

The one side distance d' is

d'=\dfrac{d}{2}

d'=\dfrac{516\times10^{5}}{2}

d'=2.58\times10^{7}\ m

Hence, The distance is 2.58\times10^{7}\ m.

5 0
3 years ago
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