A wheel rotates at 24 revolutions every 3 minutes. This is equivalent to

How is velocity different from speed? Based on distance and/or direction

Mnenie [13.5K]

Answer:

Speed is solved with time and distance but has no direction

Average velocity is solved with Δx/Δt and has a direction

6 0

2 years ago

As shown in the diagram below, a rope attached to a 500.-kilogram crate is used to exert a force of 45 newtons at an angle of 65

kaheart [24]

Answer:

19.01 N

Explanation:

F = Force being applied to the crate = 45 N

$\theta$ = Angle at which the force is being applied = $65^{\circ}$

Horizontal component of force is given by

$F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}$

The horizontal component of the force acting on the crate is 19.01 N.

4 0

3 years ago

An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 a

sveticcg [70]

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222$

We have the final answer as

Hope this helps you

5 0

3 years ago

Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as far

marta [7]

a) Their angular speeds are the same

b) Andrea's tangential speed is twice the value of Chuck's tangential speed

Explanation:

a)

The angular speed of Andrea and Chuck is the same.

Let's call $\omega$ the angular speed at which the merry-go-round is rotating. We know that the angular speed is defined as:

$\omega= \frac{2\pi}{T}$

where

$2 \pi$ is the angular displacement covered in one revolution

T is the period of revolution

The merry go round is a rigid body, so all its point cover the same angular displacement in the same time: this means that it doesn't matter where Andrea and Chuck are located along the merry-go-round, their angular speed will still be the same.

b)

For an object in circular motion, the tangential speed is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance from the centre of rotation

Here let's call $r_c$ the distance at which Chuck is rotating, so his tangential speed is

$v_c = \omega r_c$

Now we know that Andrea is rotating twice as far from the centre, so at a distance of

$r_a = 2 r_c$

So his tangential speed is

$v_a = \omega r_a = \omega (2 r_c) = 2(\omega r_c) = 2 v_c$

So, Andrea's tangential speed is twice the value of Chuck's tangential speed.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

5 0

3 years ago

A rigid tank contains 7 kg of an ideal gas at 5 atm and 30c. a valve is opened, and half of mass of the gas can escape. the fin

Readme [11.4K]

The equation of state for an ideal gas is
$pV=nRT$
where p is the gas pressure, V the volume, n the number of moles, R the gas constant and T the temperature.

The equation of state for the initial condition of the gas is
$p_1 V_1 = n_1 R T_1$ (1)
While the same equation for the final condition is
$p_2 V_2 = n_2 R T_2$ (2)

We know that in the final condition, half of the mass of the gas is escaped. This means that the final volume of the gas is half of the initial volume, and also that the final number of moles is half the initial number of moles, so we can write:
$V_1 = 2 V_1$
$n_1 = 2 n_2$
If we substitute these relationship inside (1), and we divide (1) by (2), we get
$\frac{p_1}{p_2} = \frac{T_1}{T_2}$

And since the initial temperature of the gas is $T_1 = 30 C=303 K$ , we can find the final temperature of the gas:
$T_2 = T_1 \frac{p_2}{p_1}=(303 K) \frac{1.5 atm}{5.0 atm}=90.9 K$

6 0

3 years ago

Read 2 more answers