Can Someone Please Help Me For 10 Points.

As an tempature decreases what happens to the rate of radiation?

Firlakuza [10]

As the temperature decreases, the rate of radiation goes down, but the radiation exists as long as the temperature is above the absolute zero, which is actually 0 Kelvin. 0 Kelvin equals -273°C or -460°F. All objects in the world radiate if above that temperature.

5 0

3 years ago

The Outlaw Run roller coaster in Branson, Missouri, features a track that is inclined at 84 ∘ below the horizontal and that span

harina [27]

Answer:

Explanation:

a)

Ff = μmgcosθ

Ff = 0.28(1600)(9.8)cos(-84)

Ff = 458.9217...

Ff = 460 N

b) ignoring the curves required at top and bottom which change the friction force significantly, especially at the bottom where centripetal acceleration will greatly increase normal forces and thus friction force.

W = Ffd

W = 458.9217(-49.4/sin(-84)

W = 22,795.6119...

W = 23 kJ

c) same assumptions as part b

The change in potential energy minus the work of friction will be kinetic energy.

KE = PE - W

½mv² = mgh - (μmgcosθ)d

v² = 2(gh - (μgcosθ)(h/sinθ))

v = √(2gh(1 - μcotθ))

v = √(2(9.8)(49.4)(1 - 0.28cot84))

v = 30.6552...

v = 31 m/s

5 0

2 years ago

A proton is projected into a magnetic field that is directed along the positive x axis. Find the direction of the magnetic force

bulgar [2K]

Te direction of the magnetic force for the velocity of the proton in the

-ve y direction will be +ve z direction.

As we know that the right-hand rule is based on the relation of magnetic fields and the forces that they exert on moving charges.When a charged particle moves under a magnetic field, it exerts a force on the particle, which is not in the same direction but different than the direction of the magnetic field.Under the right-hand rule, if we point our pointer finger in the direction of the charged particle is moving and the middle finger is representing the direction of the magnetic field then our thumb depicts the direction of the magnetic force which is exerted on the charged particle.

So, we are given that the direction of the velocity of the proton is in the negative y direction and the direction of the magnetic field is in the positive x direction, so the magnetic force is acting in the positive z direction.

To know more about the right-hand rule refer to the link brainly.com/question/9750730?referrer=searchResults.

#SPJ4

4 0

2 years ago

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(c) $v_{1} = 2.04\times 10^{8}\ m/s$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

(c) To calculate the velocity of light in medium 1:

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

5 0

3 years ago

Read 2 more answers

7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee

wlad13 [49]

Answer:

a)

$N_{top}=9.8N\\N_{bottom}=33.4N$

b) $v_{min}=4.4m/s$

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is $a_{cp}=\frac{v^2}{R}$ . At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

$-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg$

Which means:

$N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N$

The limit for falling off would be $N_{top}=0$ , so the minimum speed would be:

$0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s$

3 0

3 years ago