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3 years ago

Who did not have experimental evidence to support his theory of the atom? Dalton Thomson Rutherford Democritus

Physics

1 answer:

Aloiza [94]3 years ago

3 0

Democritus was the one who did not have experimental evidence to support his theory of the atom.

Answer: Option 4

Explanation:

The discovery of atoms were first stated by Democritus but due to the absence of any experimental proof, his statement was not noted as significant at that time.

After this, Dalton made the specific assumptions formulating some postulates for the atomic theory with proof. Then the cathode rays tube experiments performed by Thomson lead to the formation of plum pudding models of atom.

This is followed by Rutherford’s gold foil experiment discovering the presence of nucleus inside the atoms. So, Democritus first stated but due to absence of experimental evidences, his theory of atoms were not supported at that time.

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If a 20 N object has been lifted 5 meters above the ground, how much gravitational potential energy does it have?

Andreyy89

The gravitational potential energy of the object is 100 J.

Gravitational potential energy stored in an object is the work done in raising the object to a height h against the gravitational force acting on it.

The gravitational force acting on a body is its weight mg, where m is its mass and g, the acceleration due to gravity.

Work done by a force is equal to the product of the force and the displacement made by the point of application of the force.

$GPE = mgh$

The weight of the object is given as 20 J and it is raised to a height of 5 m.

$GPE =(mg)*h = (20 N)*(5m)=100 J$

The gravitational potential energy of the object is 100 J.

5 0

3 years ago

Read 2 more answers

A block weighs 15 n and is suspended from a spring that is attached to the ceiling. the spring stretches by 0.075 m from its uns

Illusion [34]

We can salve the problem by using the formula:

$F=kx$

where F is the force applied, k is the spring constant and x is the stretching of the spring.

From the first situation we can calculate the spring constant, which is given by the ratio between the force applied and the stretching of the spring:

$k=\frac{F}{x}=\frac{15 N}{0.075 m}=200 N/m$

By using the value of the spring constant we calculated in the first step, we can calculate the new stretching of the spring when a force of 33 N is applied:

$x=\frac{F}{k}=\frac{33 N}{200 N/m}=0.165 m$

4 0

3 years ago

A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati

Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

$\tau = I * \alpha (1)$

Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
τ = F*r (2)
For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
Replacing (2) and (3) in (1), we can solve for α, as follows:

$\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)$

Since the angular acceleration is constant, we can use the following kinematic equation:

$\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)$

Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

$0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)$

Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

$\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)$

Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

$v = \omega * r (8)$

where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
Replacing this value and (7) in (8), we get:

$v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)$

There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

$a_{t} = \alpha * r (9)$

where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
Replacing this value and (4), in (9), we get:

$a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)$

Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

$a_{c} = \omega^{2} * r (11)$

Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
Replacing this value and (7) in (11) we get:

$a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)$

The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

$a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)$