Every magnet has two poles which can be separated true or false

A sample of copper with a mass of 1.80 kg, initially at a temperature of 150.0°C, is in a well-insulated container. Water at a t

user100 [1]

Answer:

the mass of water is 0.3 Kg

Explanation:

since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:

Q water + Q copper = Q surroundings =0 (insulated)

Q water = - Q copper

since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature

and denoting w as water and co as copper :

m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)

m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]

We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C

if we assume that both specific heats do not change during the process (or the change is insignificant)

m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]

m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))

m w= 0.3 kg

7 0

4 years ago

Instead of rolling down the ramp over the step, imagine the ball falls off the back of the ramp directly to the ground. What is

lesya692 [45]

0J

According to the law of conservation of energy, the potential energy is converted to kinetic energy. Remember, potential energy is calculated using height and weight. If the ball is on the ground, height is 0.

8 0

3 years ago

Read 2 more answers

An elastic rope is tied securely to a wall. An upward-displaced pulse is introduced into the rope. The pulse travels through the

Natasha2012 [34]

Answer:

Inverted (displaced downwards)

Explanation:

The pulse becomes INVERTED upon reflecting off the boundary with the wall. That is, an upward-displaced pulse will reflect off the end and return with a downward displacement. This inversion behavior will always be observed when the end of the medium is fixed, like this wall in this instance. This INVERSION BEHAVIOR can also be observed when the medium is connected to another more heavy or more dense medium. And in this case, when the pulse reaches the end of the medium, a portion of the pulse will reflect off the end and return with an inverted displacement. The heavier medium acts like a fixed end to cause the pulse to be inverted.

Summary: a pulse reaching the end of a medium becomes inverted whenever it either:

i. reflects off a fixed end,

ii. is moving in a less dense medium and reflects off a more dense medium.

3 0

3 years ago

What is the final position of the object if its initial position is x = 0.40 m and the work done on it is equal to 0.21 J? What

r-ruslan [8.4K]

Answer:

a) Final position is x = 0.90 m

b) Final position is x = 0.133 m

Explanation:

The workdone between two points is usually approximated as the area under the force-distance curve between those two points.

From the graph,

As at the initial position, x = 0.40 m and the corresponding F = 0.8 N,

The area from that point onwards up to the end of that particular bar = 0.8 (0.5 - 0.4) = 0.08 J

The next bar has force = 0.4 N and the width of the bar = (0.75 - 0.50) = 0.25 m

Work done under this bar = 0.4 × 0.25 = 0.1 J

Total work done from the starting position up to this point now = 0.08 + 0.1 = 0.18 J, still less than 0.21 J

So, the final position has to be on the last bar. Let the position be x. The force on the last bar = 0.2 N

0.21 = 0.18 + 0.2 (x - 0.75)

0.03 = 0.2x - 0.15

0.2x = 0.18

x = 0.9 m

Therefore, the final position of the object, to do 0.21 J worth of work, starting from x = 0.4 m is 0.90 m.

b) For this part, negative work is done, this means, we will move in the negative direction to try and trace this total work done.

From the starting point where the initial position is 0.40 m, the force here is 0.80 N

The workdone under this bar to the left is

The workdone = 0.8 (0.25 - 0.4) = - 0.12 J

Since we're tracing -0.19 J, the final position has to be on the last bar (on the left), Let the position be x. The force on the last bar on the left (could also be referred to as the first bar) = 0.60 N

- 0.19 = -0.12 + 0.6 (x - 0.25)

-0.07 = 0.6x - 0.15

0.6x = 0.08

x = (0.08/0.6) = 0.133 m

Therefore, the final position of the object, after doing -0.19 J worth of work, starting from x = 0.4 m is 0.133 m.

Hope this Helps!!!