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Dima020 [189]
2 years ago
6

Describe a well-known hypothesis that was discarded because it was found to be untrue.earth-centered model of the universe. the

movement of sun, moon, and earth
Physics
1 answer:
polet [3.4K]2 years ago
6 0
The widely accepted hypothesis before that turned out wrong was the Earth-Centered theory or the Geocentric Theory. This was proposed by the philosopher Ptolemy. He came about to this hypothesis from hi observation that from the Earth's perspective, the celestial bodies like the Sun, stars and the moon, look like they rotate around the Earth each day and night. However, this was disproved by Galileo Galelei by his Heliocentric Theory. He observed through the telescope that the Venus also changes phases like the moon. However, he deduced that this is not possible from the positions of the Venus, Earth, Moon and Sun. 
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Use the terms "force", "weight", "mass", and "inertia" to explain why it is easier to tackle a 220 lb football player than a 288
Tomtit [17]
<span><u>Answer </u>
The mass of 220 lb football has less than 288 lb football. So, it will be easier to move it since it will require less force. The heavy football will have a bigger momentum. Since 288 lb has more weight than 220 lb, it will have bigger inertia making it difficult for the players to stop it.
This makes it easier to tackle 220 lb football than 288 lb football. 
</span>
7 0
2 years ago
Read 2 more answers
Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli
Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

a. Equation for minima is defined as: sin \theta=\frac{m\lambda}{\alpha}

Given m=3,\lambda=6.33\times 10^-^7 and \alpha=0.00015:

#Substitute our variable values in the minima equation to obtain \theta:

\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad

#draw a triangle to find the relationship between \theta, y \ and L.

tan(\theta)=y/L               #where y=4.05cm

L=y/tan(\theta)=3.20

Hence the screen is 3.20m from the split.

b.  To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

sin \theta=\frac{m\lambda}{\alpha}, where m=4 given that it's the minima with the smallest distance.

sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad

#we then use tan(\theta)=y/L to calculate L=4.5cm

Then from the equation subtract y_3 from y:

4.50cm-4.05cm=0.45cm

Hence, the distance \bigtriangleup y is 0.45cm

8 0
2 years ago
In an experiment you are performing, your lab partner has measured the distance a cart has traveled: 28.4inch. You need the dist
alexgriva [62]

As per the question the distance travelled by  a car is 28.4 inch.

we are asked to determine the  conversion factor in centimeter  which when multiplied with 28.4 inch will give a unit.

we know that one inch =2.54 centimeter.

Hence 28.4 inch = 2.54  ×28.4 cm

                            =72.136 cm.

Now we have to determine the conversion factor .The multiplication factor is calculated as    P =\frac{28.4 inch*2.54cm}{1 inch}

                         p= 72.136 cm        [p is the multiplication factor.]

Hence the multiplication factor is 72.137 cm which will give unit conversion when multiplied with 28.4 inch.

                 

                             


4 0
2 years ago
Read 2 more answers
A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters​ [cm]. The rover picks up the rock and lifts it
Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid SG (also called relative density) is the ratio of the density of that solid \rho_{rock} to the density of water \rho_{water}=1 kg/m^{3} (normally at 4\°C):

SG=\frac{\rho_{rock}}{\rho_{water}} (1)

On the other hand, the density of the rock is calculated by:

\rho_{rock}=\frac{m_{rock}}{V_{rock}} (2)

Where:

m_{rock} is the mass of the rock

V_{rock}=\frac{4}{3} \pi r^{3} is the volume of the rock, since is spherical

Well, we already know the value of \rho_{water}, but we need to find \rho_{rock} in order to find the rock's specific gravity; and in order to do this, we firsly have to find m_{rock} and then calculate V_{rock}:

In the case of the mass of the rock, we can calclate it by the following equation:

W_{rock}=m_{rock}g_{mars} (3)

Where:

W_{rock} is the weight if the rock in mars

g_{mars}=3.7 m/s^{2} is the acceleration due gravity in Mars

Isolating m_{rock}:

m_{rock}=\frac{W_{rock}}{g_{mars}} (4)

m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}} (5)

To find W_{rock} we can use the following equation of the potential gravitational energy U:

U=W_{rock}H (6)

Where:

U=2 J=2 Nm is the potential energy

H=20 cm \frac{1m}{100 cm}=0.2 m is the height at which the rock has the mentioned potential energy

Isolating W_{rock}:

W_{rock}=\frac{U}{H} (7)

W_{rock}=\frac{2 Nm}{0.2 m} (8)

W_{rock}=10 N (9)

Substituting (9) in (5):

m_{rock}=\frac{10 N}{3.7 m/s^{2}} (10)

m_{rock}=2.702 kg (11)

Substituting (11) in (2):

\rho_{rock}=\frac{2.702 kg}{V_{rock}} (12) At this point we only need to find the volume of the rock, knowing its diameter is d=10 cm, hence its radius is r=\frac{d}{2}=5 cm

V_{rock}=\frac{4}{3} \pi (5 cm)^{3} (13)

V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3} (14)

Substituting (14) in (12):

\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}} (15)

\rho_{rock}=5166.34 kg/m^{3} (16)

Substituting (16) in (1):

SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}} (17)

Finally we obtain the specific gravity of the​ rock:

SG=5166.347

7 0
2 years ago
When kicking a football, the kicker rotates his leg about the hip joint. (a) If the velocity of the tip of the kicker’s shoe is
Maurinko [17]

Answer:

Part a)

\omega = 33.33 rad/s

Part b)

F = 500 N

Part c)

R_{max} = 40.8 m

Explanation:

Part a)

As we know that the velocity of the tip of the kicker's shoe is given as

v = 35 m/s

also the length of the tip of the shoe from his hip joint is given as

L = 1.05 m

now the angular speed is given as

\omega = \frac{v}{L}

\omega = \frac{35}{1.05}

\omega = 33.33 rad/s

Part b)

As we know that force on the ball is given as rate of change in momentum of the ball

so it is given as

F = \frac{\Delta P}{\Delta t}

so we have

F = \frac{m(v_f - v_i)}{\Delta t}

F = \frac{0.500(20 - 0)}{20 \times 10^{-3}}

F = 500 N

Part c)

As we know that the formula of range is given as

R = \frac{v^2 sin(2\theta)}{g}

now for maximum range we know

\theta = 45 degree

R_{max} = \frac{v^2}{g}

R_{max} = \frac{20^2}{9.8}

R_{max} = 40.8 m

6 0
2 years ago
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