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torisob [31]
3 years ago
11

What is density of gold

Physics
2 answers:
Mademuasel [1]3 years ago
4 0

Answer:

the density of gold is 19.3 g/cc

ivann1987 [24]3 years ago
4 0

Answer:

Gold has density of 19.3 g/cc.

Explanation:

The density of an element is the amount of mass it has per unit volume.

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Which of the following stages follows a protostar?
goldfiish [28.3K]
The answer is a White Dwarf.
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3 years ago
An object in equilibrium has a net force of
Butoxors [25]

Answer:

An object in equilibrium has a net force of zero  

Static equilibrium describes an object at  rest having equal and balanced forces acting upon it.

Dynamic equilibrium describes an object in  motion having equal and balanced forces acting upon it.

Explanation:

An object is said to be in equilibrium when a net force of zero is acting on it. When this condition occurs, the object will have zero acceleration, according to Newton's second law:

F=ma

where F is the net force, m the mass of the object, a the acceleration. Since F=0, then a=0. As a result, we have two possible situations:

- If the object was at rest, then it will keep its state of rest. In this case, we talk about static equilibrium.

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8 0
3 years ago
Please help!!! Thank you!!!!!
MrRa [10]
The amoeba is having a asexual reproduction and the two birds are not. The two birds are having a sexual reproduction.
7 0
3 years ago
A convex lens with a focal length of 10 cm is located 30 cm away from a
zloy xaker [14]

Answer: 15 cm

Explanation:

According to the Lens Equation we have the following:

\frac{1}{f}=\frac{1}{v}+\frac{1}{u} (1)

Where:

f=10 cm is the focal length

u=30 cm is the distance between the candle (the object) and the lens

v is the distance between the image and the lens

Isolating v:

v=\frac{uf}{u-f}} (2)

Solving:

v=\frac{(30 cm)(10 cm)}{30 cm-10 cm}} (3)

Finally:

v=15 cm This is where the image is located

8 0
3 years ago
A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T
Mamont248 [21]

Let y_0 be the height of the building and thus the initial height of the ball. The ball's altitude at time t is given by

y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity.

The ball reaches the ground when y=0 after t=2.9\,\mathrm s. Solve for y_0:

0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2

\implies y_0\approx16.258\,\mathrm m

so the building is about 16 m tall (keeping track of significant digits).

3 0
3 years ago
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