PHYSICS HIGH SCHOOL

After landing from your skydiving experience, you are so excited that you throw your helmet upward. The helmet rises 5.8 m above your hands. What was the initial speed of the helmet when it left your hands? Express your answer to three significant figures and include the appropriate units How long was it moving from the time it left your hands until it returned? Express your answer to three significant figures and include the appropriate units.

Answers

Answer 1

Answer:

Answer:

Initial velocity: approximately $10.7\; {\rm m\cdot s^(-1)}$ .

Time taken before return to initial height: approximately $2.17\; {\rm s}$ .

(Assumptions: $g = 9.81\; {\rm m\cdot s^(-2)}$ ; air resistance is negligible.)

Explanation:

Under the assumption, acceleration of the helmet would be constantly $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ .

During the interval between being thrown upward and reaching maximum height:

Initial velocity immediately after the person threw the helmet upward, $u$ , needs to be found.
Final velocity when the helmet reached maximum height is $v = 0\; {\rm m\cdot s^(-1)}$ .
Change in position over this interval is $x = 5.8\; {\rm m}$ .
Acceleration is $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ .

Apply the following SUVAT equation to find initial velocity $u$ :

$\begin{aligned} u &= \sqrt{v^(2) - 2\, a\, x} \n &= \sqrt{(0)^(2) - 2\, (-9.81)\, (5.8)} \; {\rm m\cdot s^(-1)}\n &\approx 10.668\; {\rm m\cdot s^(-1)} \n &\approx 10.7\; {\rm m\cdot s^(-1)}\end{aligned}$ .

(Round to three significant figures for the final result, but keep more significant figures for intermediary values.)

In other words, the velocity of the helmet was approximately $10.7\; {\rm m\cdot s^(-1)}$ immediately after the person threw the helmet upward.

Right before returning to the initial height, the velocity of the helmet would be the opposite of its initial velocity: $(-u) \approx (-10.7)\; {\rm m\cdot s^(-1)}$ .

The change in velocity would be:

$\begin{aligned}((-u) - u)/(a) &\approx ((-10.668) - 10.668)/((-9.81))\; {\rm s} \n &\approx 2.17\; {\rm m\cdot s^(-1)}\end{aligned}$ .

(Rounded to three significant figures.)

Answer 2

Answer:

Final answer:

The initial speed of the helmet was 10.7 m/s and it was in the air for a total of 2.18 s.

Explanation:

This problem involves concept from physics specifically kinematics. Kinematics helps us study the motion of objects. To solve this problem, we need to use the second equation of motion: v²=u²+2as. In this case, the final speed (v) is 0 (when the helmet reaches the highest point, its velocity becomes 0), acceleration (a) is -9.8 m/s² (gravity acts downwards), and the distance (s) is 5.8 m.

Plugging in these values we get: 0 = u² - (2 * 9.8 * 5.8). Solving for u (initial velocity), we get u = √(2 * 9.8 * 5.8) = 10.7 m/s. This is the initial speed of the helmet when it left your hands.

To find out how long the helmet was in the air, we can use the first equation of motion: v = u + at. Solving for t (time), we get: t = (v - u) / a = (0 - 10.7) / -9.8 = 1.09 s going up. Because the time going up and coming down is the same, the total time the helmet was moving is 2 * 1.09 = 2.18 s.

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Answers

Answer:

0.432kg or 432g

Explanation:

Copper block is removed from 310C oven.

Mass of water = 1.10kg

Initial temperature of water (θ1) = 23.0C

Final temperature of water (θ2 ) = 33.0C

Specific heat of copper= 385

Specific heat of water = 4190

Let M (copper) be the mass of copper

The amount of heat lost will be absorbed by water

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For water,

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= 1.1 * 4190 * (33.0 -23.0)

= 1.1 *4190 * 10

= 46090J

For copper,

Q(copper) = M(copper) * C(copper) * (33.0 - 310)...........(3)

from equation 1,

Q(copper) = -Q(water)

M(copper) * C(copper) * (33.0 - 310) = - 46090

M(copper) * 385 * (-277) = -46090

M(copper) * 106645 = -46090

M(copper) = -46090 / -106645

M(copper) = 0.432kg

M(copper) = 432g

The mass of copper =432g

Final answer:

The mass of the copper block is 3.69kg.

Explanation:

To find the mass of the copper block, we can use the equation q1 = q2. The heat gained by the water is given by q1 = mcΔT, where m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature. The heat lost by the copper block is given by q2 = mcΔT, where m is the mass of the copper block, c is the specific heat of copper, and ΔT is the change in temperature. Since the water reaches a final temperature of 33.0°C, the change in temperature is ΔT = 33.0°C - 23.0°C = 10.0°C.

Using the equation q1 = q2, we have mcΔT = mcΔT. Rearranging the equation to solve for the mass of the copper block, we get m = (m1c1ΔT2) / (c2ΔT1), where m1 is the mass of the water, c1 is the specific heat of water, ΔT2 is the change in temperature of the copper block, c2 is the specific heat of copper, and ΔT1 is the change in temperature of the water. Substituting the given values, we have m = (1.10kg * 4190 J/(kg?K) * 10.0°C) / (385 J/(kg?K) * 10.0°C), which simplifies to m = 3.69kg.

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Answers

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Have a nice day!

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