I need some help with my science homework about energy, work, and power. It would be greatly appreciated :)

In a science lab, Thomas used "universal indicator" to investigate the pH of some household products. He created a reference tab

dolphi86 [110]

Usually

pH scale measures from 1 to 14

Purple colour indicates pH as 14

Hence it's an base

But it doesn't react with metals to form salts and hydrogen gas.

It's sure it's non metal but not base

One possibility is that it's Carbon compound or Graphite compound which reacts like base but a metal in real sometimes

6 0

3 years ago

PLEASE HELP!! DUE IN 40 MINUTES!!

Irina18 [472]

A frame of reference can be thought of as the state of motion of the observer of some event. For example, if you're sitting on a lawn chair watching a train travel past you ... you could even watch a glass of water sitting on a table inside the train move ... This seems like a simple and obvious example, yet when you take a step

6 0

3 years ago

A 23.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 45.0 deg

Romashka [77]

Answer:

a. 139.748J

b.3.486m/s

c.zero

Explanation:

a. Given the mass of the child as 23.0kg, rope length is 2.1mand incline is 45°

Potential energy during release is calculated as: $PE=mgh$

#Find vertical difference of when the swing is at rest (2.1m) and when the child is pulled back.

Find the height when the child is pulled back:

$cos 45\textdegree=y/2.10\\\\y=1.48m$

#therefore,vertical difference is 2.1m-1.48m=0.62m

$\therefore PE=mgh\\\ \ =23.0kg\times 9.8m/s^2\times 0.62m\\\ \ =139.748J$

#Hence the potential energy during release is 139.748J

b. From a, above, we have PE=139.748J, M=23.0kg.

At the bottom, all the PE will be transferred into KE. Potential energy is calculated as:

$KE=0.5mv^2\\\\mv^2=2KE\\\\v=\sqrt{2KE/m}\\\\v=\sqrt{2\times 139.748J/23.0}\\\\v=3.486m/s$

#Hence the velocity at the bottom of the swing is 3.486m/s

c. Work is calculated as the product of force by distance.

From a, b above we have mass as 23.0kg .

-since the distance of the ropes remained constant the change in distance is zero:

$W=mgd\\=23.0\times 9.8m/s^2\times 0\\=0$

Therefore the work in the ropes is 0,zero.

8 0

3 years ago

As a descending elevator approaches the correct floor, it slows to a stop with a constant acceleration of magnitude 1 ms21\,\dfr

vovangra [49]

Answer:

Explanation:

Given that,

The elevator slow to stop, this shows that it is decelerating and the final velocity is 0

v = 0m/s

Constant deceleration

a = -1m/s²

It is negative because it is deceleration

The distance the elevator descend is 4.5m

S=4.5m

Then, we want to find the time the elevator spent before stopping

Using the equation of motion

v = u + at

v² = u² + 2as

Where

v is final velocity

u Is initial velocity

a is the deceleration

s- is distance traveled

From here we can find the initial velocity of the elevator

v² = u² + 2as

0² = u² - 2 × 1 × 4.5

0 = u² - 9

u² = 9

u = √9

u = 3m/s

The initial speed is 3m/s

Then, to find the time taken, we can use the first equation

v = u + at

0 = 3 - 1 × t

0 = 3 - t

t = 3 seconds

The time taken before the elevator stop is 3 secs

6 0

3 years ago

Read 2 more answers

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land

andrey2020 [161]

Answer:

a).

The magnitude force impulse is

$F=656.6 kg*m/s$

the average force is

$F=656.6 kg*m/s$

Explanation:

Using the conservation energy the potential energy in the high is equal to the kinetic energy just before the collision so:

$E_K=E_p$

$\frac{1}{2}*m*v^2=m*g*h$

Notice the mass can be cancel as a factor so:

$v^2=2*g*h$

$v=\sqrt{2*g*h}=\sqrt{2*9.8m/s^2*0.50m}$

$v=9.8 m/s$

Now the impulse is determinate by:

$F=m*(v_f-(-v_i))$

$kg*F=m*v_i=67*9.8m/s$

$F=656.6 kg*m/s$

The average force is also

$F'=\frac{F}{t}=\frac{656.6}{0.095s}$

$F'=6911.57 N$

8 0

4 years ago