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3 years ago

Give an example of a theory scientists believed for a long time that was eventually proved false

1 answer:

6 0

There are so many to choose from, I'll give you two. One is a big one. The other so/so. People used to believe the earth was flat and you would fall off, until Christopher Columbus sailed across. And another interesting one is the vulcan planet theory, which is that there was an extra planet between Mercury and T sun. however it turned out to be false. So there, need anymore just tell me.

You might be interested in

The proper IUPAC naming convention for the compound symbol 'KI' is

Darya [45]

Proablyly a or c idk

6 0

2 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

3 years ago

A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T

Serga [27]

Answer:

a) $k_{e}$ = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m, e) v = 15.23 m / s

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

$k_{e}$ = ½ k x²

$k_{e}$ = ½ 2900 0.80²

$k_{e}$ = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

U = m h and

U = 8 9.8 (-0.80)

U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

K = ½ m v²

K = 0

d) write the energy at two points, maximum compression and maximum height

Em₀ = ke = ½ m x²

$E_{mf}$ = mg y

Emo = $E_{mf}$

½ k x² = m g y

y = ½ k x² / m g

y = ½ 2900 0.8² / (8 9.8)

y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

Y = y- 0.80

Y = 11.8367 -0.80

Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

Emo = ½ k x² = 928 J

$E_{mf}$ = ½ m v²

Emo = $E_{mf}$

Emo = ½ m v²

v =√ 2Emo / m

v = √ (2 928/8)

v = 15.23 m / s

8 0

3 years ago

A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth

svetoff [14.1K]

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

$\omega = \sqrt{\dfrac{k}{m}}$

now from equation (3)

$a_x = \dfrac{k}{m}x$

$k = \dfrac{m a_x}{x}$

$k = \dfrac{0.45 \times (-14)}{0.24}$

k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

8 0

3 years ago

Read 2 more answers

A stationary boat in the ocean is experiencing waves from a storm. the waves move at 56 km/h and have a wavelength of 160 m, bot

Lera25 [3.4K]

The wavelength $\lambda$ of the wave is 160 m, and this is the distance between two consecutive crests. The boat is located at a crest of the wave, this means that the first trough is located 80 meters from the boat (because the distance between a crest and a trough is half the wavelength).

The speed of the wave is
$v=56 km/h = 15.6 m/s$
so the time the boat takes to reach the first trough is
$t= \frac{S}{v} = \frac{80 m}{15.6 m/s}=5.1 s$

5 0

3 years ago