What is the mass of 1 l of water? what is its weight in newtons?

A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum

Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , $m_1= 0.473 kg$

mass of puck 2 , $m_2= 0.819 kg$

initial speed of puck 1 , $u_1=2.76\frac{m}{s}$

initial speed of puck 2 , $u_2=0.00\frac{m}{s}$

Final speed of puck 1 and puck 2 be $v_1\, and\, v_2$ respectively

Apply conservation of linear momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

=> $0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2$

=> $1.594=0.5775\times v_1+ v_2$ -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

$u_2-u_1=v_1-v_2$

=> $0-2.76=v_1-v_2$ ------(B)

From equation (A) and (B)

$v_1=-0.739\frac{m}{s}$

and $v_2=2.02\frac{m}{s}$

Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

3 0

3 years ago

There. That is better.

mihalych1998 [28]

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This will really help you learn a lot.

6 0

3 years ago

Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter

valentinak56 [21]

Wee can use here kinematics

as we know that

$y = v*t + \frac{1}{2} at^2$

for shorter tree we know that

$y = 0 + \frac{1}{2}*9.8 * 2^2$

$y = 19.6 meter$

now since we know that other tree is twice high

So height of other tree is y = 39.2 m

now again by above equation

$y = v*t + \frac{1}{2} at^2$

$39.2 = 0 + \frac{1}{2}*9.8 * t^2$

$t = 2.83 s$

so the time taken is 2.83 s

4 0

2 years ago

Using the formula F = M* A. What is the acceleration of a .5 kg

Katyanochek1 [597]

Answer:32 m/s/s

Explanation: since F=M*A, F=16N, M=0.5kg, A= F/M

A=16/0.5

A=32 m/s/s

7 0

2 years ago

En un rio una Onda viaja con una velocidad de propagación de 50 m/s con una longitud de Onda de 40 metros. Hallar la frecuencia

sattari [20]

Answer:

Frequencia = 1.25 Hz

Explanation:

<u>Dados los siguientes datos;</u>

Velocidad = 50 m/s
Longitud de onda = 40 metros

Para encontrar la frecuencia de la onda;

Matemáticamente, la velocidad de una onda viene dada por la fórmula;

$Velocidad = Longitud \; de \; onda * Frequencia$

Haciendo de la frecuencia el tema de la fórmula, tenemos;

$Frequencia = \frac {Velocidad}{Longitud \; de \; onda}$

Sustituyendo en la fórmula, tenemos;

$Frequencia = \frac {50}{40}$

<em>Frequencia = 1.25 Hz</em>

7 0

3 years ago