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AleksandrR [38]
2 years ago
15

URGENTTT PLEASE HELPPPP. You put m1 = 1 kg of ice cooled to -20°C into mass m2 = 1 kg of water at 2°C. Both are in a thermally i

nsulated chamber. For water L = 3.33 x 105 J/kg. The specific heat of ice is 2090 J/(kg°C) and of water 4186 J/(kg°C). How much does the ice heat up in order to bring the water down to 0°C?
A. 0.04°C


B. 0.4°C


C. 4°C


D. 10°C


E. 20°C
Physics
1 answer:
STatiana [176]2 years ago
5 0

Answer:

Explanation:

heat lost by water will be used to increase the temperature of  ice

heat gained by ice

= mass x specific heat  x rise in temperature

1 x 2090 x t

heat lost by water in cooling to 0° C

= mcΔt  where m is mass of water , s is specific heat of water and Δt is fall in temperature .

= 1 x 2 x 4186  

8372

heat lost = heat gained

1 x 2090 x t  = 8372

t = 4°C

There will be a rise of  4 degree in the temperature of ice.  

 

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A spherical conductor is inside a uniform electric field with a magnitude of 1000 N/C to the right. What is the magnitude and di
Lyrx [107]

Answer:

1.-E=1000N/C to the LEFT

2.-The electric field inside a conductor in electrostatic state is always zero (conductor proprieties).

3.-The voltmeter read 0V as differential voltage between two points from the conductor

Explanation:

1.The electric field inside the conductor must be zero (conductor proprieties). Then the charges create a electric field equal an opposite to the external electric field. In other words E=1000N/C to the LEFT

2. The electric field inside a conductor in electrostatic state is always zero. As shown in the figure the electric field induced by the charges in the sphere surface cancelled the EXTERN electric field.

3.If the Electric field inside the conductor is zero, that means that the Voltage in the hole conductor is constant (conductor proprieties). In other words the the voltmeter read 0v as differential voltage between two points from the conductor.

3 0
2 years ago
Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher
FrozenT [24]

Answer: Option (C) is the correct answer.

Explanation:

As we know that metals are able to conduct electricity so, when a negatively charges rod is kept closer to the left sphere then electrons will enter the sphere.

Since, like charges repel each other. Hence, some of the negative changes from the rod will repel the negative charges of left sphere.

As both left and right spheres are touching each other so, the electrons will move towards the right sphere. As a result, there will be too many electrons (negative charge) present on the right sphere and very less electrons present in the left sphere.

Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.

7 0
2 years ago
Read 2 more answers
[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
As SCUBA divers go deeper underwater, the pressure from the weight of all the water above them increases tremendously which comp
slavikrds [6]

Answer: A.

As a diver rises, the pressure on their body decreases which allows the volume of the gas to decrease.

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The problem is that a diver, experiences an increased pressure of water compresses nitrogen and more of it dissolves into the body. Just as there is a natural nitrogen saturation point at the surface, there are saturation points under water. Those depend on the depth, the type of body tissue involved, and also how long a diver is exposed to the extra pressure. The deeper a diver go, the more nitrogen the body absorbs.

The problem is getting rid of the nitrogen once you ascend again. As the pressure diminishes, nitrogen starts dissolving out of the tissues of the diver's body, a process called "off-gassing." That results in tiny nitrogen bubbles that then get carried to the lungs and breathed out. However, if there is too much nitrogen and/or it is released too quickly, small bubbles can combine to form larger bubbles, and those can do damage to the body, anything from minor discomforts all the way to major problems and even death.

4 0
3 years ago
T or F :)
balandron [24]

That's true.

And I'll go ya a better one:

If the object is moving or not moving, at a constant or changing speed, in a straight or curvy line, and the forces on it do not cancel out and add up to zero, the object will accelerate.

4 0
3 years ago
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