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4 years ago

Please hurry! Only answer if you know the correct answer.

1 answer:

8 0

its c because The microscopically colliding particles, that include molecules, atoms and electrons, transfer disorganized microscopic kinetic and potential energy, jointly known as internal energy.

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Which best describes the purpose of a controlled experiment?

navik [9.2K]

A controlled experiment is best described as a safe, in depth, and insightful display that helps you understand the purpose of the experiment better

3 0

3 years ago

A 600 g model rocket is on a cart that is rolling to the right at a speed of 4.0 m/s. The rocket engine, when it is fired, exert

murzikaleks [220]

Answer:

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

Explanation:

Step 1: Data given

mass of the rocket = 600 grams

speed = 4.0 m/s

Step 2: Calculate weight

Fw = mg

with Fw = the weight (in Newton)

with m = the mass (in kg)

with g = the acceleration due to gravity (9.81 m/s²).

Fw = (0.600 kg)(9.81 m/s²) = 5.886 N

Step 3: Calculate force available to provide acceleration

The rocket engine, when it is fired, exerts a 8.0 AND vertical thrust on the rocket.

5.886 N of that force will be used to counter the rocket's weight, leaving 2.114 N of force available to provide acceleration.

Step 4: Calculate the rocket's upward acceleration:

Fnet = m*a

With Fnet = the net force (the force that remains after the rocket's weight is compensated)

with a = the rocket's acceleration (in m/s²)

2.114 N = (0.600 kg)*a

a = 3.52 m/s² = the rocket's upward acceleration

Step 5: Calculate how long it will take to rise 20 meters into the air.

Δy = v0*t + 1/2 at²

with v0 = 0m/s

Δy = 1/2 at²

20 m = 1/2(3.52)t²

20 m = (1.76 m/s²)t²

11.36 = t²

t = 3.37 s

This means the rocket will take 3.37 seconds to reach the hoop. It should be launched when the cart is 3.37 seconds away from being directly beneath the hoop.

Step 6: Calculate the distance

v = Δx / t

4.0 m/s = Δx / 3.37 s

Δx = 13.48 m

The rocket should be launched when the cart is 13.48m away from a point directly below the hoop.

8 0

4 years ago

Pure water has a pH of 7. Pure water _______. A. is a neutral substance B. could be either an acid or a base C. is a base D. is

malfutka [58]

A. is a neutral substance.

Only neutral substances (like pure water) can have a pH of 7.

5 0

3 years ago

Read 2 more answers

A worker at the top of a 588-m-tall television transmitting tower accidentally drops a heavy tool. If air resistance is negligib

liq [111]

The final velocity of the tool is 107.4 m/s

Explanation:

We can solve this problem by using the principle of conservation of energy.

In fact, if air resistance is negligible, the total mechanical energy of the tool is conserved during the fall, so we can write:

$K_i + U_i = K_f + U_f$

where

$K_i = 0$ is the kinetic energy of the tool at the top (zero since it is at rest)

$U_i = mgh$ is the gravitational potential energy of the tool at the top, where

m is the mass of the tool

g is the acceleration of gravity

h is the heigth of the tool

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the tool just before hitting the ground, where

v is the final speed of the tool

$U_f = 0$ is the gravitational potential energy of the tool at the bottom (zero since the height is zero)

Re-arranging the equation,

$mgh=\frac{1}{2}mv^2$

where we have

$g=9.8 m/s^2\\h=588 m$

And solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(588)}=107.4 m/s$

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0

3 years ago

A 10 kg ball strikes a wall with a velocity of 3 m/s to the left. The ball bounces off with a velocity of 3 m/s to the right. If

lord [1]

Answer:

The force is 272.73 newtons

Explanation:

We're going to use impulse-momentum theorem that states impulse is the change on the linear momentum this is:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (1)

Impulse is also defined as average force times the time the force is applied:

$\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t)$ (2)

By (2) on (1):

$\overrightarrow{F}_{avg}(\varDelta t)= \overrightarrow{p}_{f}-\overrightarrow{p}_{i}$

solving for $\overrightarrow{F}_{avg}$ :

$\overrightarrow{F}_{avg}=\frac{\overrightarrow{p}_{f}-\overrightarrow{p}_{i}}{\varDelta t}$ (3)

We already know Δt is equal to 0.22 s, all we should do now is to find $\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ and put on (3) ( $\overrightarrow{p_{i}}$ the initial momentum and $\overrightarrow{p_{f}}$ the final momentum). Linear momentum is defined as $\overrightarrow{p}=m\overrightarrow{v}$ , using that on (3):

$\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}}$ (4)

Velocity (v) are vectors so direction matters, if positive direction is the right direction and negative direction left $\overrightarrow{v_{i}}=+3\, \frac{m}{s}$ and $\overrightarrow{v_{f}}=-3\, \frac{m}{s}$ so (4) becomes:

$\varDelta\overrightarrow{p}=m(-3\frac{m}{s}- (+3\frac{m}{s}))=-(10kg)(6\frac{m}{s})$

$\varDelta\overrightarrow{p}=-60\, \frac{mkg}{s}$ (5)

Using (5) on (3):

$\overrightarrow{F}_{avg}=\frac{-60\, \frac{mkg}{s}}{0.22s}$

$F_{avg}=272.73N$

8 0

4 years ago