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Sunny_sXe [5.5K]
3 years ago
14

If a 4.5 kg object is dropped from a height of 6.0 m, what will be its velocity when it is halfway toward the ground? (Use g = 9

.80 m/s2, and ignore air resistance.) 7.7 m/s 11 m/s 16 m/s 29 m/s
Physics
2 answers:
dangina [55]3 years ago
7 0
When an object falls or is dropped from rest it's initial velocity is zero.
Using the equations for a motion in straight line. I can find the time it takes to reach 3.0 m down (half way).
x = vt - 4.9t²
-3 = 0 - 4.9t²
-3/-4.9 = t²
0.6122 = t²
0.7825 sec = t

v = v - gt
v = 0 - 9.8(0.7825)
v = -7.67 m/s
the negative denotes downward direction.

You  could also solve the problem using potential and kinetic energy.

Since it starts with maximum PE and gets converted to KE when it hits the ground. mgh = mv²/2
mass cancels, use 3 meters for the halfway distance
-9.8(-3) = v²/2
29.4 * 2 = v²
√(58.8) = 7.67 m/s downwards
notsponge [240]3 years ago
6 0

Answer:

7.7 rounded

Explanation:

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4 years ago
A 1500 kg car travels 50 m north in 20 seconds. What is the magnitude of the average velocity of the car during the 20 second in
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a. 2.5 m/s

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50 / 20 = 2.5 m/s

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a student conducts an experiment to determine how the additional of salt to water affects the density of the water. the student
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4 years ago
The enthalpy of fusion of mercury is 2.292 kJ mol-1 at its normal freezing point of 234.3 K; the change in molar volume on melti
aev [14]

Answer:

T_2= 234.37 K

Explanation:

According to Claperyon, we know that

P_2-P_1= \frac{\Delta H_{fus}}{\Delta V}\times\frac{T_1}{T_2}

P_1= Atmospheric pressure 760 mm Hg

P_2 = pressure at the bottom of the column

= 10×10^3 mm of Hg+ 760 mm of Hg

= 10760 mm of Hg

now,

P_2-P_1= 10760-760= 10^4 mm

P_2-P_1 ( in pascals) = 10^4× 133.322= 1333220 mm

the enthalpy of fusion (ΔH-fus) of mercury is 2.292 KJ/mol

use the above equation to calculate ΔT as follows

1333220= \frac{2292}{0.517\times10^{-6}}\times ln\frac{T_2}{234.3K}

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3 years ago
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