All categories

3 years ago

21. The force is the product of

Physics

1 answer:

KiRa [710]3 years ago

4 0

Answer:

D mass and acceleration

You might be interested in

The temperature of an object can be raised by adding _________ or doing _________ on a system.

Nookie1986 [14]

D is the answer because you cause heat by exerting energy

7 0

4 years ago

A rebuttal that Rousseau would use to counter the King’s argument?

ankoles [38]

Is this your Question:

Read the hypothetical situation below and answer the question that follows.

In response to criticisms that the King of Mordor has failed to gain the consent of his people, the King claims that he does in fact represent the consent of the people because without their King, society would crumble into chaos.

Which statement BEST represents a rebuttal that Rousseau would use to counter the King's argument?

A. Absolute chaos is preferable to the absolute power of the King.

B. The King of Mordor should establish a system of checks and balances.

C. If the King of Mordor were to represent the will of his people, then he would at least provide the public services demanded by his people.

D. It is impossible for the King of Mordor to have the consent of the people because the monarchy is not a freely formed government in which the people have a say in how it operates.

If this is your question.. Here is your Answer:

Answer: D. It is impossible for the King of Mordor to have the consent of the people because the monarchy is not a freely formed government in which the people have a say in how it operates.

Explanation:

Rousseau believed that sovereignty (the power to make laws) belonged to the people, not one person, and that it should be carried out by a representative body of the people who are elected by them on their behalf. While he agreed society requires a ruler(s), he thinks they should be chosen by the people.

5 0

3 years ago

Please help me for brainliest

mihalych1998 [28]

Answer:

was but its six days ago

Explanation:

6 0

3 years ago

The object has a redshift of 7.6 and the JWST observes the object at a wavelength of 2 microme-

olasank [31]

b. The wavelength of light emitted by the object is 233 nm

c. The type of radiation originally emitted by the object is ultraviolet radiation.

To find the wavelength, we need to know what redshift is.

<h3>What is redshift?</h3>

Redshift is the increase in wavelength and the corresponding decrease of frequency and photon energy of electromagnetic radiation.

Redshift is given by z = λ'/λ - 1 where

λ' = observed wavelength and
λ = emitted wavelength.

Making λ subject of the formula, we have

λ = λ'/(1 + z)

Given that has a redshift of 7.6 and the JWST observes the object at a wavelength of 2 micrometres (mid-infrared light).

So,

z = 7.6 and
λ' = 2μm

<h3>(b) What is the wavelength of the light emitted by the object?</h3>

Substituting the values of the variables into the equation, we have

λ = λ'/(1 + z)

λ = 2μm/(1 + 7.6)

λ = 2μm/8.6

λ = 0.233 μm

λ = 233 nm

So, the wavelength of light emitted by the object is 233 nm

<h3>c. What type of radiation was originally emitted by the object?</h3>

Since the wavelength is 233 nm and the wavelength is in the range of ultraviolet radiation 200 nm - 315 nm.

So, the type of radiation originally emitted by the object is ultraviolet radiation.

Learn more about redshift here:

brainly.com/question/27915180

#SPJ1

8 0

3 years ago

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

4 0

3 years ago