Joe runs 10 m north, 20 m south, 9m south, and then 15 m north. What is Joe's<br> displacement?

Ball A is allowed to fall and strike ball B. Assume that all of ball A's energy is transferred to ball B at point I, and that th

Alika [10]

Answer:

Explanation:

If no Energy is lost ( either as Sound energy or any other energy) the total energy of ball A equals the total energy of ball B

Therefore the Kinetic Energy of Ball B equals the Total Energy of Ball A minus the Potential Energy of Ball A

K. E = T. E - P.E

5 0

3 years ago

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a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleratio

Harman [31]

Answer:

The angular acceleration is $\alpha = 3.235 \ rad/s ^2$

Explanation:

From the question we are told that

The moment of inertia is $I = 0.034\ kg \cdot m^2$

The net torque is $\tau = 0.11\ N \cdot m$

Generally the net torque is mathematically represented as

$\tau = I * \alpha$

Where $\alpha$ is the angular acceleration so

$\alpha = \frac{\tau }{I}$

substituting values

$\alpha = \frac{0.1 1}{ 0.034}$

$\alpha = 3.235 \ rad/s ^2$

6 0

3 years ago

An experiment is designed to investigate the relation between the pressure and temperature in a tank that has a constant volume

viva [34]

Answer:

3rd order polynomial

Explanation:

Given that the increase in the order of the polynomial the error between the curve fit and measured data will decreases hence :

The polynomial order that is best to use is the 3rd order polynomial, this is because using a 3rd order polynomial will produce a less variance and a low Bias

4 0

3 years ago

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s$