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3 years ago

5

Work out this problem if an object weighs 34.8 grams and has a volume of 22.8ml , what is this object's density?

2 answers:

Stels [109]3 years ago

5 0

We know, density = Mass / volume
Here, mass = 34.8 g
v = 22.8 ml = 22.8 cm³

Substitute their values,
d = 34.8/22.8
d = 1.52 g/cm³

Hope this helps!

pishuonlain [190]3 years ago

3 0

Density = Mass / Volume

= 34.8 g / 22.8 ml ≈ 1.5263 g/mL

Density is ≈ 1.5263 g/mL

Hope this explains it.

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3 years ago

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Someone help please by providing work and answers please :)

Nastasia [14]

First we gotta use an equation of motion:

$d = ut + \frac{1}{2} a {t}^{2}$

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

$100 = 0 + \frac{1}{2} (9.8) {t}^{2}$

Now we just need to solve for t:

${t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}$

Hit the calculators, and you'll get 4.5 seconds!

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3 years ago

Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1

harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum $p_{i}$ (before the elastic collision) must be equal to the final momentum $p_{f}$ (after the elastic collision):

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=m_{1}V_{i} + m_{2}U_{i}$

$p_{f}=m_{1}V_{f} + m_{2}U_{f}$

Where:

$m_{1}=200 kg +100 kg=300 kg$ is the combined mass of Tubby and Libby with the car

$V_{i}=10 m/s$ is the velocity of Tubby and Libby with the car before the collision

$m_{2}=25 kg + 100 kg=125 kg$ is the combined mass of Flubby with its car

$U_{i}=0 m/s$ is the velocity of Flubby with the car before the collision

$V_{f}=4.12 m/s$ is the velocity of Tubby and Libby with the car after the collision

$U_{f}$ is the velocity of Flubby with the car after the collision

So, we have the following:

$m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f}$ (2)

Finding $U_{f}$ :

$U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}}$ (3)

$U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg}$ (4)

Finally:

$U_{f}=14.1 m/s$

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3 years ago

A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After th

Zolol [24]

Answer:

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we note that the distance between B and C is 0.5 m

Then we have

Sum of initial momentum = Sum of final momentum

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Therefore from the conservation of linear momentum we also have

(m₁ + m₂)×v₂ = (m₁ + m₂ + m₃)×v₃

Where:

m₃ = Mass of the carrier = 30 kg

Therefore

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The velocity of the carrier = 0.40872 m/s.

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3 years ago

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Answer:

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Explanation:

omnicalculator.com/physics/wavelength

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