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Molodets [167]
3 years ago
15

You have an aquarium which holds 2.65 gallons of water and you would like to place it on an antique stand. Before placing the aq

uarium on the stand, as a precaution, you decide to determine the weight of the water in the aquarium. Determine the weight of water you will be placing in the aquarium.
Physics
2 answers:
snow_tiger [21]3 years ago
8 0

Answer:

10.0 kg

Explanation:

1 gallon of water is equal to 3.79 liters of water. Then, 2.65 gallons of water are:

2.65 gal water × (3.79 L water / 1 gal water) = 10.0 L water

The density of water is about 1.00 kg/L. The mass corresponding to 10.0 L of water is:

10.0 L × (1.00 kg/L) = 10.0 kg

The mass of 2.65 gallons of water is 10.0 kg.

qwelly [4]3 years ago
6 0

Answer:

The weight of the water is 10.0 kg

Explanation:

Hi there!!

Since the density of water is 1 g/ml, knowing the volume of water, we can determine its weight. First, let´s convert the unit gallon into liters:

1 gallon = 3.785 l

Then:

2.65 gallons · (3.785 l / 1 gallon) = 10.0 l

Since 1 l = 1000 ml, 10.0 l will be 10000 ml

The density of water is 1 g/ml, then the weight of 10.0 l is 10000 g or 10.0 kg.

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What is the acceleration of the object in the graph?
viktelen [127]

Answer:

-0.8 m/s²

Explanation:

Acceleration is the slope of a velocity vs. time graph.

a = Δv / Δt

a = (0 m/s − 12 m/s) / (15 s − 0 s)

a = -0.8 m/s²

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3 years ago
The air that flows over the top part of an airplane's wing moves faster than the air that flows across the bottom. This faster m
kaheart [24]
As an airplane moves through the air, its wings cause changes in the speed and pressure of the air moving past them. These changes result in the upward force called lift.

The Bernoulli principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in the pressure exerted by the fluid.

A wing is shaped and tilted so the air moving over it moves faster than the air moving under it. As air speeds up, its pressure goes down. So the faster-moving air above exerts less pressure on the wing than the slower-moving air below. The result is an upward push on the wing—lift!
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3 years ago
A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb
NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

\sum \tau = F*d

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

F*(27-6)= 6*600

F = \frac{6*600}{21}

F= 171.42 lb

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

4 0
3 years ago
When you stretch a spring 20 cm past its natural length, it exerts a force of 8
Nastasia [14]

Answer:

40 N/m

Explanation:

F = -kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = -k (0.2m)

-k = 8N / 0.2 m

k = -40 N/m

6 0
3 years ago
Read 2 more answers
A simple harmonic oscillator has amplitude 0.43 m and period 3.9 sec. What is the maximum acceleration?
rjkz [21]

Answer:

Maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2              

Explanation:

We have given amplitude of simple harmonic motion is A = 0.43 m

Time period of the oscillation is T = 3.9 sec

We have to find the maximum acceleration

For this we have to find the angular frequency

Angular frequency will be equal to \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{3.9}=1.61rad/sec

Maximum acceleration is given by a_{max}=\omega ^2A=1.61^2\times 0.43=0.854rad/sec^2

So maximum acceleration in the simple harmonic motion will be 0.854rad/sec^2

8 0
3 years ago
Read 2 more answers
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