All categories

2 years ago

Which activity would be the best choice foir a lifelong fitness program

Physics

2 answers:

Fantom [35]2 years ago

7 0

I think jogging/running/walking because you don't need any equipment and you can structure around it on your own time.

Xelga [282]2 years ago

6 0

Answer:

I would say walking or running

Explanation:

Because its simple and you can choose your own pace

You might be interested in

Provides most of the energy used in the world today.

lara [203]

fossil fuels is used the most often in the world.

6 0

2 years ago

A box has a length of 18.1 cm and a width of 19.2cm tall ?

ladessa [460]

Volume = l*w*h = (18.1 cm)(19.2 cm)(20.3 cm) = 7,054 cm^3.

3 0

2 years ago

PLEASE NEED AN ANSWER SO I CAN SUBMIT IT!!! THANK YOU IN ADVANCE (will give brainliest)

alexandr402 [8]

Answer:

Im only 12 and i need the points so ima try my best.

Explanation:

574.780616 m6 kg3 s-6 K-3 mol-3

6 0

2 years ago

A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle

Contact [7]

<h2>Answers: </h2>

1) 1.359, 1.403

2) $2.207(10)^{8}m/s$ , $2.138(10)^{8}m/s$

Explanation:

The described situation is known as Refraction.

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index $n$ is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum ( $c=3(10)^{8}m/s$ ) and the speed of light $v$ in the second medium:

$n=\frac{c}{v}$ (1)

On the other hand we have the Snell’s Law:

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

Where:

$n_{1}$ is the first medium refractive index . We are told is the air, hence $n_{1}\approx 1$

$n_{2}$ is the second medium refractive index

$\theta_{1}$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=38.1\º$ :

$(1)sin(57.0\º)=n_{2}sin(38.1\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}$

$n_{2}=1.359$ (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=36.7\º$ :

$(1)sin(57.0\º)=n_{2}sin(36.7\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}$

$n_{2}=1.403$ (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

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

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

Substituting (3) in this equation:

$v_{red}=\frac{3(10)^{8}m/s}{1.359}$

$v_{red}=2.207(10)^{8}m/s$ >>>>Speed of red light

<h2>1a) Violet light</h2>

Using again equation (1):

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

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

Substituting (4) in this equation:

$v_{violet}=\frac{3(10)^{8}m/s}{1.403}$

$v_{red}=2.138(10)^{8}m/s$ >>>>Speed of violet light

3 0

3 years ago

hich of the following statements are true for magnetic force acting on a current-carrying wire in a uniform magnetic field?Check

leonid [27]

Explanation:

The magnetic force acting on a current carrying wire in a uniform magnetic field is given by :

$F=I(L\times B)$

$F=ILB\ sin\theta$

Where

$\theta$ is the angle between length and the magnetic field

The magnetic force is perpendicular to both current and magnetic field. It is maximum when it is perpendicular to both current and magnetic field.

So, the correct options are :

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.
.The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.
The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

7 0

3 years ago