All of the following are categories of individual activities except?

The answer is not B.

mezya [45]

Answer:

D

Explanation:

15m/ s

i hope this helps

4 0

1 year ago

A student wants to determine the impulse delivered to the lab cart when it runs into the wall. The student measures the mass of

WINSTONCH [101]

The correct answer to the question is : A) The velocity of the cart after it hits the wall.

EXPLANATION:

Before answering this question, first we have to understand impulse.

Impulse of a body is defined as the change in momentum or the product of force with time.

Mathematically impulse = m ( v- u ).

Here, v is the final momentum and u is the initial momentum.

Hence, we need the velocity of the cart after it hits the wall in order to calculate the impulse of the lab cart.

4 0

3 years ago

Read 2 more answers

A baseball pitcher throws a ball horizontally at a speed of 34.0 m/s. A catcher is 18.6 m away from the pitcher. Find the magnit

Sidana [21]

To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

$y = \frac{1}{2} at^2+v_0t+y_0$

Where,

a = acceleration

t = time

$v_0$ = Initial velocity

$y_0$ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

$v = \frac{x}{t}$

x = Displacement

t = time

With the data we have we can find the time as well

$v = \frac{x}{t}$

$t = \frac{x}{v}$

$t = \frac{18.6}{34}$

$t = 0.547s$

With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,

$y = \frac{1}{2} at^2+v_0t+y_0$

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

$y = \frac{1}{2} gt^2$

$y = \frac{1}{2} 9.8*0.547^2$

$y = 1.46m$

Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.

6 0

3 years ago

You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica

Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have $m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}$ , where:

$l$ is the length

$m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3$

We divide the first equation by the second equation to get:

$\frac{m}{R} = \frac{d A^2}{\rho}$

$A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2$

Using this Area, we find the diameter of the wire:

$D = \sqrt{\frac{4A}{\pi}}$

$D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}$

$D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm$

To find the length, we multiply the two equations stated initially:

$mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}$

$l^2 = 8.534\\l = 2.92 \ m$

8 0

3 years ago

Read 2 more answers

Noise-canceling headphones are an application of destructive interference. Each side of the headphones uses a microphone to pick

exis [7]

The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.

4 0

3 years ago