Ask question

Ask question

All categories

alexandr402 [8]

3 years ago

6

How are gravity and air resistance alike? How are they different?

1 answer:

Iteru [2.4K]3 years ago

8 0

Air resistance is a force that slows down objects moving through the air. ... Objects are affected by gravity the same, but air resistance can affect the speed of an object's descent. An object with more surface area will have more air resistance.

You might be interested in

Identify the type of chemical reaction that is described.

Mashcka [7]

If its out and has been rained on it will rust if its like metal or iron stuff like that it will rust
the answer is rust
but no so quickly

5 0

3 years ago

Jacki evaluated the expression below. 2 cubed (3 minus 1) + 4 (8 minus 12) = 2 cubed (2) + 4 (4) = 8 (2) + 16 = 16 + 16 = 32. Wh

Naddika [18.5K]

Answer:

Her error was that she did not subtract 12 from 8 correctly

Explanation:

Jackie did 8-12 instead of 12-8

8 0

3 years ago

A system consists of a disk of mass 2.33 kg and radius 50 cm upon which is mounted an annular cylinder of mass 2.20 kg with inne

Gekata [30.6K]

Answer:

Kinetic energy of the system = 2547.41 Joules.

Explanation:

Given:

Disk:

Mass of the disk (m) = $2.33$ kg

Radius of the disk (r) = $50$ cm = $\frac{50}{100} =0.5$ m

Cylinder:

Mass of the annular cylinder (M) = $2.20$ kg

Inner radius of the cylinder $(R_i)$ = $0.2$ m

Outer radius of the cylinder $(R_o)$ = $0.3$ m

The angular speed of the system $(\omega)$ = $15.1$ rev/s

Angular speed in in terms of Rad/sec = $15.1\times 2\pi =94.876$ rad/sec

Formula to be used:

Rotational Kinetic energy, $(KE)_r$ = $\frac{I\times \omega^2}{2}$

So, before that we have to work with the moment of inertia (MOI) of the system.

⇒ MOI of the system = MOI of the disk + MOI of the cylinder

⇒ MOI (system) = $\frac{mr^2}{2} +\frac{M(R_i+R_o)^2}{2}$

⇒ MOI (system) = $\frac{2.33\times (0.5)^2}{2} + \frac{2.20\times (0.2+0.3)^2}{2}$

⇒ MOI (system) = $0.566$ kg.m^2

Now

The rotational Kinetic energy.

⇒ $(KE)_r =\frac{I\omega^2}{2}$

Plugging the values.

⇒ $(KE)_r=\frac{0.566\times (94.876)^2}{2}$

⇒ $(KE)_r=2547.41$ Joules

Then

The kinetic energy of the rotational system is 2547.41 J.

5 0

3 years ago

These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the neg

Kay [80]

Answer:

Following are the solution to the given question:

Explanation:

For charging plates that are connected in a similar manner:

Calculating the total charge:

$\to q =q_1 + q_2 = C_1V_1 +C_2V_2 =1320 + 2714 = 4034 \mu C$

Calculating the common potential:

$\to V = \frac{q}{C}= \frac{q}{(C_1 + C_2)} =\frac{4034}{6.8} = 593 \ V\\\\$

Calculating the charge after redistribution:

$When: \\\\q = q_{1}' + q_{2}' = q_1 + q_2$

$\to q_{1}' = C_1V = 2.2 \times 593 = 1305\ \mu C\\ \\ \to q_{2}' = C_2V = 4.6 \times 593 = 2729 \ \mu C$

6 0

3 years ago

An ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

Bond [772]

The y-component of the acceleration is $0.33 m/s^2$

Explanation:

The y-component of the acceleration is given by:

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

$u_y = u sin \theta_1 = (2.25 m/s)(sin 50.0^{\circ})=1.72 m/s$

where

u = 2.25 m/s is the initial velocity

$\theta_1 = 50.0^{\circ}$ is the initial direction

$v_y = v sin \theta_2 = (4.65)(sin 120^{\circ})=4.03 m/s$ , where

v = 4.65 m/s is the final velocity

$\theta_2 = 120.0^{\circ}$ is the final direction

The time elapsed is

t = 8.33 s

Therefore, we can find the y-component of the acceleration:

$a_y=\frac{4.03-1.72}{8.33}=0.33 m/s^2$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0

3 years ago