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2 years ago

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in the history of astronomy, which of the following is true? a. galileo galilei was a student of tycho brahe. b. johannes kepler

built one of the first observatories. c. ptolemy argued that earth was the center of the universe. d. isaac newton was criticized for his work on the geocentric model.

1 answer:

bija089 [108]2 years ago

7 0

C. Ptolemy argued that earth was the center of the universe

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Relationship between voltage, resistance and current in a circuit

Lena [83]

Answer:

V = I×R

where -

V = potential difference across

I = current flowing in the circuit

R = Equivalent Resistance in the circuit

8 0

2 years ago

Test Due! Which statement is true?

Lubov Fominskaja [6]

The correct answer is C. Mercury and Mars have the same gravitational force

Explanation:

This chart compares the different features of two planets in our solar system (Mercury and Mars). In this chart, the only numerical value or feature that is the same for both planets is gravity because for both planets gravity is 1.7 m/s2. This implies the gravitational force or the force that attracts objects towards the center of the planet is the same or that objects are pulled with the same force in both planets. Moreover, this factor depends on others such as mass, density, among others.

8 0

2 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

2 years ago

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

4 0

2 years ago

As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis

klio [65]

Hi there!

We can begin by finding the acceleration of the block.

Use the kinematic equation:

$d = v_0t + \frac{1}{2}at^2$

The block starts from rest, so:

$d = \frac{1}{2}at^2\\\\12 = \frac{1}{2}a(4^2)\\\\\frac{24}{16} = a = 1.5 m/s^2$

Now, we can do a summation of forces of the block using Newton's Second Law:

$F = ma = m_bg - T$

mb = mass of the block

T = tension of string

Solve for tension:

$T = m_bg - ma = 8.2(9.8) - 8.2(1.5) = 68.06 N$

Now, we can do a summation of torques for the wheel:

$\Sigma \tau = rF\\\\\Sigma\tau = rT$

Rewrite:

$I\alpha = rT$

We solved that the linear acceleration is 1.5 m/s², so we can solve for the angular acceleration using the following:

$\alpha = a/r\\\\\alpha = 1.5/.42= 3.57 rad/sec^2$

Now, plug in the values into the equation:

$I(3.57) = (0.42)(68.06)\\\\I = (0.42)(68.06)/(3.57) = \boxed{8.00 kgm^2}$

8 0

2 years ago