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11111nata11111 [884]
2 years ago
10

A piece of nickel at 25 °C is dropped into a glass of water at 25 °C. Which statement is correct?

Physics
1 answer:
Makovka662 [10]2 years ago
5 0
The answer is C the temp of water in both beakers will decrease since the metal is flowing heat into the water. Therefore we can say the metal losses heat and the water gains the heat lost by the metal minus any heat loss to surroundings! Hope this explanation helps you understand the concept! Please rate if I helped you! Thank you so much!
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There is 4000 balls in the earth of the world why is that so hard for you
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Which of the following would not be useful in trying to correct spherical aberration a. using a combination lens made up of lens
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A. Using a combination lens made up of lenses, each of which has a different index of refraction. Is the correct answer.
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How can light energy solve our real life problem?
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It gives our light which we need for probably everything.

Explanation:

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A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0
2 years ago
An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and dire
dem82 [27]

Answer:

<em>1,378.9ms²</em>

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

Initial velocity = 0m/s

Required

Acceleration

Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

<em>Hence the acceleration is 1,378.9ms²</em>

7 0
2 years ago
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