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2 years ago

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The mass of an object is determined by: weighing the object with a scale multiplying the height, width, and length measuring the

object with an equal-arm balance using the water displacement method

1 answer:

rosijanka [135]2 years ago

7 0

The true scientific way is the last: using the water displacement method

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For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee

Natasha2012 [34]

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity $e$ of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

$e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}$

Where:

$r_{a}$ is the apoapsis (the longest distance between the moon and its planet)

$r_{p}=0.27 r_{a}$ is the periapsis (the shortest distance between the moon and its planet)

Then:

$e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}$

$e=\frac{0.73 r_{a}}{1.27 r_{a}}$

$e=0.57$ This is the moon's orbital eccentricity

3 0

3 years ago

If ball 4 has a mass of 2 kg and it is 5m high, what will be its gravitational potential energy? (g=10 N/kg) *

Nastasia [14]

Explanation:

Gravitational potential energy

= mgh

= (2kg)(10N/kg)(5m)

= 100J.

3 0

3 years ago

Read 2 more answers

A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

The block's speed is 31.422 cm/s

4 0

3 years ago

A book has been lifted 1.5 meters into the air giving it 30J of potential energy. How much force was used to lift it

joja [24]

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

$U = F.s$

here we know that

U = 30 J

s = displacement = 1.5 m

so we have

$30 = F(1.5)$

$F = 20 N$

6 0

3 years ago

A box, compressing a spring by 4 m, is loaded and ready. Moments later the 8.0 kg box has been launched with

Mrac [35]

Answer:

50 N/m

Explanation:

Elastic energy = kinetic energy

EE = KE

½ kx² = ½ mv²

½ k (4 m)² = ½ (8.0 kg) (10.0 m/s)²

k = 50 N/m

3 0

3 years ago