A baseball travels 50 meters in 4 seconds what is the average velocity of the baseball?

A 27.0-g object moving to the right at 21.5 cm/s overtakes and collides elastically with a 11.0-g object moving in the same dire

7nadin3 [17]

Answer:

final speed of the first object: 17.73 cm/s

final speed of the second object: 24.23 cm/s

Explanation:

Data:

mass of the first object, m1 = 27.0 g

mass of the second object, m2 = 11.0 g

initial speed of the first object, v1i = 21.5 cm/s

initial speed of the second object, v2i = 15 cm/s

final speed of the first object, v1f = ? cm/s

final speed of the second object, v2f = ? cm/s

Elastic collisions formula:

v1f = (m1 - m2)/(m1 + m2) * v1i + 2*m2/(m1 + m2) * v2i

v1f = (27 - 11)/(27 + 11) * 21.5 + 2*11/(27 + 11) * 15

v1f = 17.73 cm/s

v2f = 2*m1/(m1 + m2) * v1i + (m2 - m1)/(m1 + m2)*v2i

v2f = 2*27/(27 + 11) * 21.5 + (11 - 27)/(27 + 11)*15

v2f = 24.23 cm/s

8 0

3 years ago

Consider the model above. It represents the electrical force. As r increases, the attractive force decreases. How would this mod

aivan3 [116]

Answer:

As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.

Explanation:

Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.

Gravitation Force= $\frac{Gm_{1}m_{2} }{r^{2}}$

Coulombs's Law= $\frac{Kq_{1}q_{2} }{r^{2}}$

6 0

4 years ago

Read 2 more answers

Given the initial wavefunction Ψ (x, 0) = Axexp (-k x) withx> 0 andk> 0, and Ψ (x, 0) = 0 forx <0, what value must A ta

madam [21]

Answer with explanation:

The Normalization Principle states that

$\int_{-\infty }^{+\infty }f(x)dx=1$

Given

$f(x)=xe^{-kx}(x>0\\\\0(x$

Thus solving the integral we get

$\int_{0 }^{+\infty }A\cdot xe^{-kx}dx=1\\\\A\int_{0 }^{+\infty }\cdot xe^{-kx}dx=1$

The integral shall be solved using chain rule initially and finally we shall apply the limits as shown below

$I=\int xe^{-kx}dx\\\\x\int e^{-kx}dx-\int \frac{d(x)}{dx}\int e^{-kx}dx\\\\-\frac{xe^{-kx}}{k}-\int 1\cdot \frac{-e^{-kx}}{k}\\\\\therefore I=\frac{e^{-kx}}{k}-\frac{xe^{-kx}}{k}$

Applying the limits and solving for A we get

$I=\frac{1}{k}[\frac{1}{e^{kx}}-\frac{x}{e^{kx}}]_{0}^{+\infty }\\\\I=-\frac{1}{k}\\\\\therefore A=-k$

3 0

3 years ago

Write the mathemetical relation between work force and displacement

il63 [147K]

Answer:

Work is measured as the product of force and the displacement in the direction of the force. Work = force × displacement in the direction of the force.

8 0

2 years ago

A person walks in the following pattern: 3.1 km north, then 2.4 km west, and finally 5.2 km south. a) Sketch the vector diagram

zysi [14]

Answer:

$d = 3.19 km$