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krek1111 [17]
3 years ago
10

A 26.5 g object is thrown straight up into the air. If the object's initial speed is 1.60 m/s, determine how high the object wil

l rise.
Physics
1 answer:
fenix001 [56]3 years ago
8 0

Answer:

0.131 m

Explanation:

Applying,

v² = u²+2gh................... Equation 1

Where, v = final velocity, u = initial velocity, g = acceleration due to gravity, h =  maximum height of the object

make h the subject of the equation

h = (v²-u²)/2g........... Equation 2

From the question,

Given: v = 0 m/s (at the maximum height), u = 1.60 m/s

Constant: g = -9.8 m/s² (thrown up)

Substitute these values into equation 2

h = (0²-1.6²)/[2×(-9.8)]

h = -2.56/-19.6

h = 0.131 m

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The molecular weight of oxygen gas (o2) is 32 while that of hydrogen gas (h2) is 2.
anzhelika [568]

use the formula: v^2=(3kT)/m

Where:

<em>v is the velocity of a molecule</em>

<em>k is the Boltzmann constant (1.38064852e-23 J/K)</em>

<em>T is the temperature of the molecule in the air</em>

<em>m is the mass of the molecule</em>

For an H2 molecule at 20.0°C (293 K):

v^2 = 3 × 1.38e-23 J/K × 293 K / (2.00 u × 1.66e-27 kg/u)

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v = 1.91e+3 m/s

For an O2 molecule at same temp.:

v^2 = 3 × 1.38e-23 J/K × 293 K / (32.00 u × 1.66e-27 kg/u)

v^2 = 2.28e+5 m^2/s^2

v = 478 m/s

Therefore, the ratio of H2:O2 velocities is:

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7 0
3 years ago
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Explanation:

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3 years ago
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6. 25 grams of sodium chloride is placed into a beaker. The reading on the digital balance says that
eduard

Answer:

<u>149 grams to 3 sig figs</u>

Explanation:
Beaker + NaCl = 155 grams

                NaCl = 6.25 grams

Subtract the NaCl

 155 g - 6.25 g = 148.75 grams is the mass of the beaker.  But there are only 3 sig figs, so the mass should be reported as 149 grams.

6 0
1 year ago
Determine the value of V3 in volts where Vs = 15V, V1 = 5V, and V2 = 3V. Put the value of V3 in the box below without the units.
marshall27 [118]

Answer:

The value of V₃ is 7 V

enter value =7

Explanation:

Given that,

Source voltage V_{s}=15 V

First voltage V_{1}=5V

Second voltage V_{2}=3V

According to figure,

We need to calculate the value of third voltage

Using Kirchhoff voltage law

V_{s}+V_{1}+V_{2}+V_{3}=0

-15+5+3+V_{3}=0

v_{3}=7\ V

Hence, The value of V₃ is 7.

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3 years ago
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