Those who support the alien theory believe that the ancient Egyptians where very primitive. What evidence about the ancient Egyptians suggests that claim is inaccurate?

Answers

Answer 1

Answer:

Answer:

They had a well-developed agricultural system , they raised domesticated animals , they developed a writing system, and they used early forms of tools.

Answer 2

Answer:

Final answer:

The claim that the ancient Egyptians were primitive and relied on aliens to build their monuments is inaccurate. Evidence from archaeology and history shows that the ancient Egyptians had advanced knowledge and skills in various fields. Their construction techniques and use of mathematics in building the pyramids are well-documented.

Explanation:

The claim that the ancient Egyptians were primitive and that their accomplishments, such as building the pyramids, were assisted by aliens is inaccurate. There is evidence from archaeology and history that ancient monuments were built by ancient people using their own ingenuity and capabilities. The ancient Egyptians had advanced knowledge in various fields including architecture, engineering, astronomy, and mathematics. For example, their construction techniques and use of mathematics in building the pyramids and other structures are well-documented.

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7. An engineer is using a wire that has a resistance of 1.5 . This resistance is too high for the application he is designing. The wire must be exactly 2.5 cm long. What two things could he do to reduce the wire's resistance
A rock is thrown vertically upward with a speed of 14.0 m/sm/s from the roof of a building that is 70.0 mm above the ground. Assume free fall.A) In how many seconds after being thrown does the rock strike the ground? B) What is the speed of the rock just before it strikes the ground?
1 microgram equals how many milligrams?
. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.You should obtain e/m = 2V/(B^2)(r^2)3. The magnetic field on the axis of a circular current loop a distance z away is given byB = mu I R^2 / 2(R^2 + z^2)^ (3/2)where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/Rwhere B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters

What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Answers

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

A. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 45m?b. How long will it be in the air?

Answers

Answer:

A. 29.7 m/s

B. 6.06 s

Explanation:

From the question given above, the following data were obtained:

Maximum height (h) = 45 m

A. Determination of the initial velocity (u)

Maximum height (h) = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 m/s (at maximum height)

Initial velocity (u) =.?

v² = u² – 2gh (since the ball is going against gravity)

0² = u² – (2 × 9.8 × 45)

0 = u² – 882

Collect like terms

0 + 882 = u²

882 = u²

Take the square root of both side

u = √882

u = 29.7 m/s

Therefore, the ball must be thrown with a speed of 29.7 m/s.

B. Determination of the time spent by the ball in the air.

We'll begin by calculating the time take to reach the maximum height. This can be obtained as follow:

Maximum height (h) = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) to reach the maximum height =?

h = ½gt²

45 = ½ × 9. 8 × t²

45 = 4.9 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the time spent by the ball in the air. This can be obtained as follow:

Time (t) to reach the maximum height = 3.03 s

Time (T) spent by the ball in the air =?

T = 2t

T = 2 × 3.03

T = 6.06 s

Therefore, the ball spent 6.06 s in the air.

Nichrome wire, often used for heating elements, has resistivity of 1.0 × 10-6 Ω ∙ m at room temperature. What length of No. 30 wire (of diameter 0.250 mm) is needed to wind a resistor that has 50 ohms at room temperature?

Answers

Answer:

Length = 2.453 m

Explanation:

Given:

Resistivity of the wire (ρ) = 1 × 10⁻⁶ Ω-m

Diameter of the wire (d) = 0.250 mm = 0.250 × 10⁻³ m

Resistance of the wire (R) = 50 Ω

Length of the wire (L) = ?

The area of cross section is given as:

$A=(1)/(4)\pi d^2\n\nA=(1)/(4)*\ 3.14* (0.250* 10^(-3))^2\n\nA=0.785* 6.25* 10^(-8)\n\nA=4.906* 10^(-8)\ m^2$

We know that, for a constant temperature, the resistance of a wire is directly proportional to its length and inversely proportional to its area of cross section. The constant of proportionality is called the resistivity of the wire. Therefore,

$R=\rho (L)/(A)$

Expressing the above in terms of length 'L', we get:

$L=(RA)/(\rho)$

Plug in the given values and solve for 'L'. This gives,

$L=(50* 4.906* 10^(-8))/(1* 10^(-6))\ m\n\nL=(2.453)/(1)=2.453\ m$

Therefore, length of No. 30 wire (of diameter 0.250 mm) is 2.453 m.

2. Annealed low-carbon steel has a flow curve with strength coefficient of 75000 psi and strain-hardening exponent of 0.25. A tensile test specimen with a gauge length of 2 in. is stretched to a length of 3.3 in. Determine the flow stress and the average flow stress that the metal experienced during this deformation.

Answers

Answer:

Flow stress= 9390Psi

Average flow stress= 4173.33Psi

Explanation:

Given:

Strength Coefficient = 75000psi

Strain hardening Exponent = 0.25

Gauge length = 2inches

Stretch length = 3.3 inches

Flow stress,Yf = 75000 × ln(3.3/2) × 0.25

Yf = 75000× ln(1.65) × 0.25

Yf = 75000× 0.5008 × 0.25

Flow stress = 9390Psi

Average flow stress = 75000× 0.5008 × (0.25/2.25)

Average flow stress= 4173.33psi

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of a star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius R that is initially rotating at 2.4 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is 4.3R. Assume that all of the star's original mass is contained in the shell.

Answers

Answer:

0.0768 revolutions per day

Explanation:

R = Radius

$\omega$ = Angular velocity

As the mass is conserved the angular momentum is conserved

$I_1\omega_1=I_2\omega_2\n\Rightarrow (I_1)/(I_2)=(\omega_2)/(\omega_1)$

Moment of intertia for solid sphere

$I_1=(2)/(5)MR^2\n\Rightarrow I_1=0.4MR^2$

Moment of intertia for hollow sphere

$I_2=(2)/(3)M(4.3R)^2\n\Rightarrow I_2=12.327MR^2$

Dividing the moment of inertia

$(I_1)/(I_1)=(0.4MR^2)/(12.327MR^2)\n\Rightarrow (I_1)/(I_2)=0.032$

From the first equation

$\omega_2=\omega_1(I_1)/(I_2)\n\Rightarrow \omega_2=2.4* 0.032\n\Rightarrow \omega_2=0.0768\ rev\day$

The angular velocity, in revolutions per day, of the expanding supernova shell is 0.0768 revolutions per day

Final answer:

To find the angular velocity of the expanding supernova shell, we can use the principle of conservation of angular momentum. The initial angular momentum of the star can be equated to the final angular momentum of the shell. By substituting the given information and solving the equation, we can find the angular velocity of the shell.

Explanation:

When a star undergoes a supernova explosion, a large amount of its mass is blown outward in the form of a rapidly expanding shell. To find the angular velocity of the expanding shell, we can use the principle of conservation of angular momentum. Assuming that all of the star's original mass is contained in the shell, we can equate the initial angular momentum of the star to the final angular momentum of the shell.

The angular velocity of the star before the explosion can be calculated using the equation:

angular velocity before = 2 * pi * initial frequency

where the initial frequency is given as 2.4 revolutions per day.

After the explosion, the radius of the expanding shell is given as 4.3 times the radius of the star. Using the principle of conservation of angular momentum, we can set the initial angular momentum of the star equal to the final angular momentum of the shell:

initial angular momentum of the star = final angular momentum of the shell

Since the final angular momentum of the shell is given by:

final angular momentum of the shell = moment of inertia of the shell * angular velocity of the shell

where the moment of inertia of the shell is given by:

moment of inertia of the shell = 2/5 * mass of the shell * (radius of the shell)^2

and the angular velocity of the shell is what we are trying to find, we can rewrite the equation as:

initial angular momentum of the star = 2/5 * mass of the shell * (radius of the shell)^2 * angular velocity of the shell

By substituting the expression for the initial angular momentum of the star and solving for the angular velocity of the shell, we can find the answer.

Learn more about Angular velocity of a supernova shell here:

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Please help really easy

Answers

Answer:

I am sure it is A because no chemical change occurs and it is a physical change. If you can Brainllest than that would be great but if you wanna you don't have to. Hope this helps!! If wrong sorry.

Explanation:

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