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bekas [8.4K]
3 years ago
10

Gold has a density of 19.3 g/cm3. What is the mass of a 5 cm3 block of gold?

Physics
1 answer:
Tems11 [23]3 years ago
5 0
Do not forget that mass = <span>volume x density
</span>Mass of 1 cm^3 = Density[/tex]
mass of 2 cm^3 = 19.3 g + 19.3 g = 2*19.3 g
Then eventually we can find <span>mass of 5 cm^3 : = 
</span>19.3 g + 19.3 g+19.3g+19.3g+19.3g= 5*19.3 g
So the answer is D
<span>And that's it. I'm sure it will help.</span>
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3 years ago
Two identical traveling waves, moving in the same direction, are out of phase by π/5.0 rad. What is the amplitude of the resulta
andreev551 [17]

Answer:

Therefore the amplitude of the resultant wave is =0.95 y_m

Explanation:

The equation of wave:

y=A sin (kx-ωt)

For wave 1:

y₁=A sin (kx-ωt) = y_{m}sin (kx-ωt)

For wave 2:

y₂=A sin (kx-ωt+Φ) = y_{m}sin (kx-ωt+Φ)

Where A= amplitude=y_m

The angular frequency \omega=\frac{2\pi}{T}

k=\frac{2\pi}{\lambda} , \lambda= wave length.

t= time

T= Time period

\phi = phase difference =  \frac{\pi}{5}

The resultant wave will be

y = y₁ + y₂

 =y_m sin (kx-ωt) + y_m sin (kx-ωt+Φ)

 =y_m {sin (kx-ωt) + sin (kx-ωt+Φ)}

 =y_m\  sin(\frac{kx-\omega t +\phi + kx-\omega t }2)\ cos(\frac{kx-\omega t  +\phi -kx+\omega t}2)

 =y_m\  sin({kx-\omega t +\frac\phi 2)\ cos(\frac{\phi }2)

=y_m\ cos(\frac{\phi }2) sin({kx-\omega t +\frac\phi 2)

Therefore the amplitude of the resultant wave is

=y_m\ cos(\frac{\phi }2)

=y_m\ cos(\frac{\pi }{10})

=0.95 y_m

6 0
4 years ago
Whats the answer? Will give brainliest
kupik [55]

Answer:

The answer to your question is:  letter D

Explanation:

Fussion is combine light nuclei to form heavy nuclei

a)  This is wrong the definition says, combine light nuclei to form hevier nulcei

b)  Both fussion and fission are about radioactive compounds

c) Both fussion and fission use radioactive reactants

d) Yes, this is a characteristic of fussion

6 0
4 years ago
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 20 mph and half the d
Gnoma [55]

Answer:

On the way to grandmother´s, the average speed was 30 mph. On the way back, the average speed was 40 mph.

Explanation:

The average speed is given by the variation of the position over time.

Mathematically:

ΔX / Δt = v

where:

ΔX = distance (final position - initial position)

Δt = time (final time - initial time)

v = speed

On the way to Grandmother´s, we can calculate how much time Julie drove at each speed:

ΔX / Δt = v

ΔX / v = Δt

50 mi / 20 mph = 2.5 h

In the same way, we can calculate how much time she drove at 60 mph:

50 mi / 60 mph = 0.83 h

In total, she drove a distance of 100 mi in (2.5 h + 0.83 h) 3.33 h. Then, the average speed on the way to Grandmother´s was:

<u>ΔX / Δt = v = 100 mi / 3.33 h = 30 mph</u>

In the return trip, we do not know the distance nor the time that she drove at each speed, but we know that for each part of the trip, the time is the same (Δt)  and we also know that the total distance is 100 mi and the total time is 2Δt:

v1 = ΔX1 / Δt

v2 = ΔX2 / Δt

ΔX2 + ΔX1  = 100

where

v1 = speed during the first part of the trip (20 mph)

v2 = speed during the second part of the trip (60 mph)

ΔX1 = distance driven at 20 mph

ΔX2 = distance driven at 60 mph

Δt = time

If we divide v2/v1, we will get:

v2/v1 = (ΔX2 / Δt) / (ΔX1 / Δt)

60 mph / 20 mph = ΔX2 / ΔX1

3 = ΔX2 / ΔX1

3ΔX1 = ΔX2

Then we can replace ΔX2 for 3ΔX1 in this equation:

ΔX2 + ΔX1  = 100 mi

3ΔX1 + ΔX1 = 100 mi

4ΔX1 = 100 mi

ΔX1 = 25 mi

And now, we can solve Δt from the equation of v1:

v1 = ΔX1 / Δt

Δt = ΔX1 / v1 = 25 mi / 20 mph = 1.25 h

The average speed on the return trip is then:

<u>v = 100 mi / 2Δt = 100 mi / 2.5 h = 40mph</u>

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Answer:

c

Explanation:

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5 0
3 years ago
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