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Pavlova-9 [17]
3 years ago
11

An astronaut holds a rock 100 m above surface of Planet X. The rock is then thrown upwards with a sleek of 15m/s. The rock reach

es the ground 10s after it is thrown the atmosphere of Planet X has a negligible effect on the rock. Determined the acceleration due to gravity of the rock when it is in planet x
Physics
1 answer:
Gelneren [198K]3 years ago
5 0

Answer:5 m/s^{2}

Explanation:

This problem is related to vertical motion, and the equation that models it is:

y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2} (1)

Where:

y=0m is the rock's final height

y_{o}=100 m is the rock's initial height

V_{o}=15 m/s is the rock's initial velocity

\theta=90\° is the angle at which the rock was thrown (directly upwards)

t=10 s is the time

g is the acceleration due gravity in Planet X

Isolating g and taking into account sin(90\°)=1 :

g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t) (2)

g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s)) (3)

g=5 m/s^{2} (4) This is the acceleration due gravity in Planet X

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Q1. The answer is 54.629 metric tons

A mass (m) of a liquid is its density (D) multiplied by its volume (V): m = D · V
Step 1. Calculate the volume (V) of the swimming pool.
Step 2. Calculate the mass (m) of the water.
Step 3. Convert the mass into the metric tones.

Step 1. 
V = w * l * h             (w - weight, l - length, h - height)
w = 3.5 m
l = 9 m
h = 1.75 m
_______
V = 3.5 * 9 * 1.75
V = 55.125 m³

Step 2:
m = D * V
D = 0.991 g/cm³
1 g = 0.001 kg
1 cm³ = 0.000001 m³
D = 0.991g / 1cm³ = 0.991 * 0.001 kg / 0.000001 m³ = 991 kg/m³
V = 55.125 m³
_________
m = 991 * 55.125
m = 54628.875 kg ≈ 54629 kg

Step 3.
Since 1 metric ton is 1000 kg, how many metric tons are 54629 kg?
1 mt : 1000 kg = x : 54629 kg
x = 1 mt : 1000 kg * 54629 kg
x = 54.629 metric tons



Q2. The answer is 3.2 l.
<span>
Step 1. Convert g/cm</span>³ into kg/dm³ (because 1 l = 1 dm³)
Step 2. Calculate how many liters of the liquid have a mass of 3.75 kg using the proportion

Step 1.
D = <span>1.17 g/cm3
</span>1 g = 0.001 kg
1 cm³ = 0.001 dm³
D = 1.17 * 0.001 kg / 0.001 dm³ = 1.17 kg/dm³

Step 2.
If 1.17 kg is in 1 dm³, 3.75 kg are in how many dm³:
1.17 kg : 1 dm³ = 3.75 kg : x
x =  3.75 kg * 1 dm³ : 1.17 kg
x = 3.2 dm³
1 dm³ = 1 l
x = 3.2 l



Q3. The answer is 2757140 g/m³

Step 1. Calculate the increased volume and convert units into liters.
Step 2. Calculate the density of the alloy sample and convert units into g/m³

Step 1.
The difference in the volume of the water in cylinder is:
V = 19.5 ml - 16 ml = 3.5 ml
1 ml = 0.001 l
V = 3.5 * 0.001 l = 0.0035 l

Step 2. 
A density (D) of an object with mass m and volume V is:
D = m/V

We have:
D = ?
m = 9.65 g
V = 0.0035 l
______
D = 9.65/0.0035 g/l = 2757.14 g/l
1 l = 1 dm³ = 0.001 m³
D = 2757.14 g/0.001 m³ = 2757140 g/m³
5 0
3 years ago
Read 2 more answers
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