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3 years ago

17. Suppose you were standing on a scale in an elevator in free fall. What would the scale read?

Physics

2 answers:

lianna [129]3 years ago

5 0

What would the scale read? zero

18 In free fall you are being pulled by a gravity. "Truly" weightless presumably happens in deep space where there is nothing to pull you.

19 coasters accelerate down to simulate weight loss/zeroised. As do NASA planes,

Roller coasters are for fun seekers. NASA is for science

geniusboy [140]3 years ago

5 0

What would the scale read? zero

18 In free fall you are being pulled by a gravity. "Truly" weightless presumably happens in deep space where there is nothing to pull you.

19 coasters accelerate down to simulate weight loss/zeroised. As do NASA planes,

Roller coasters are for fun seekers. NASA is for science

You might be interested in

A block of a plastic material floats in water with 35.6% of its volume under water. What is the density of the block in kg/m3

sasho [114]

Answer:

The density is $\rho_{block} = 356 \ kg/m^3$

Explanation:

From the question we are told that

The proportion of the plastic under water is p = 0.356

Generally at equilibrium the buoyant force is equal to the weight of the block i.e

$B = mg$

Generally the Buoyant force is mathematically represented as

$B = \rho * V * g$

Here $\rho$ density of the water with value

$\rho = 1000 kg/m^3$ ,

V is the volume of water displaced by the block which is $p * V_{block$ ,

$1000 * 0.356 V_{block} * g = m * g$

Here m is the mass of the block which is mathematically represented as

$m = \rho_{block} * V_{block}$

$1000 * 0.356 V_{block} * g = \rho_{block} * V_{block}* g$

$1000 * 0.356 V_{block} = \rho_{block} * V_{block}* g$

=> $\rho_{block} = 1000 * 0.356$

=> $\rho_{block} = 356 \ kg/m^3$

7 0

3 years ago

A sled slides down a hill 42 meters high with a slope of 27 o. The total mass of the sled plus contents is 256 kg. It starts fro

kkurt [141]

Answer:

v = 28.7 m/s

Explanation:

According to the Work Energy Theorem, the work done by external forces on the system, is equal to the change in the kinetic energy of the system.
In absence of friction, the only force that does work , producing a displacement in the direction of the force, is the component of gravity parallel to the slide:

$Fg_{p } = m*g* sin \theta (1)$

The displacement d,along the slide, can be found from the definition of the sine of an angle, as follows:

$sin \theta =\frac{h}{d}$

Solving for d, we get:

$d = \frac{h}{sin \theta} (2)$

Now, the work done by Fgp, is just the product of the force times the displacement, as follows:

$W = Fgp * d$

From (1) and (2) we can find W as follows:

$W =Fg_{p }* d = m*g* sin \theta * \frac{h}{sin \theta} = m*g*h$

This expression must be equal to ΔK, as follows:

$\Delta K = \frac{1}{2} * m *v^{2} = m*g*h$

Simplifying common terms, we can solve for v (the velocity of the sled at the bottom of the slide), as follows:

$v =\sqrt{2*g*h} = \sqrt{2*9.8 m/s2*42 m} = 28.7 m/s$

The velocity of the sled at the bottom of the slide is 28.7 m/s, taking as positive the direction down the slide.

7 0

3 years ago

How many coulombs of positive charge are there in 47.0 gm of plutonium, given its atomic mass is 244 and that each plutonium ato

enot [183]

Answer:

1.78×10⁶ C

Explanation:

Using the atomic mass of pluonium atoms (244 g/mol), you can calculate the number of atoms in 47.0 g. Then, knowing that each plutonium atom has 96 protons, you calculate the number of protons in the 47.0 g sample. Finally, using the positive charge of one proton, you calculate the total positive charge in the 47.0 g of plutonium.

1. Number of atoms of plutonium in 47.0 g

Number of moles = mass / atomic mass = 47.0 g / 244 = 0.1926 moles

Number of atoms = number of moles × 6.022 × 10²³ atoms/mol

Number of atoms = 0.1926 mol × 6.022 × 10²³ atoms/mol = 1.15998×10²³ atoms

2. Number of protons

Number of protons = 1.15998×10²³ atoms × 96 protons/atom = 1.11385×10²⁵ protons

3. Charge

Charge = charge of one proton × number of protons

Charge = 1.602×10⁻¹⁹ C/proton × 1.11385×10²⁵ protons = 1.78×10⁶C

7 0

3 years ago

A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of

Paul [167]

Answer:

The height of water in the column = 348.14 cm

Explanation:

Pressure:This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of pressure is N/m²

p = Dgh............... Equation 1

Where p = pressure, D = density, g = acceleration due to gravity, h = height.

From the question, the same pressure will support the column of mercury and water.

p₁ = p₂

Where p₁ = pressure of mercury, p₂ = pressure of water

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

h₂ = 348.14 cm

The height of water in the column = 348.14 cm

6 0

3 years ago

true or false? destructive interference occurs when a trough meets up with another trough given location along the medium

disa [49]

Answer:Poopy-di scoop

Scoop-diddy-whoop

Whoop-di-scoop-di-poop

Poop-di-scoopty

Scoopty-whoop

Whoopity-scoop, whoop-poop

Poop-diddy, whoop-scoop

Poop, poop

Scoop-diddy-whoop

Whoop-diddy-scoop

Whoop-diddy-scoop, poop

Explanation:

4 0

3 years ago