What sort of fitness component would dance be<br> categorized under

A proton moves with a velocity of v with arrow = (3î − 5ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î

Allisa [31]

Answer:

The magnitude of the magnetic force this particle experiences is $2.6\times10^{-9}\ N$ .

Explanation:

Given that,

Velocity v= (3i-5j+k) m/s

Magnetic field B=(i+2j-k) T

We need to calculate the value $\vec{v}\times\vec{B}$

$(\vec{v}\times\vec{B})=3i+4j+11k$

We need to calculate the magnitude of the magnetic force this particle experiences

Using formula of magnetic force

$\vec{F}=q(\vec{v}\times\vec{B})$

Put the value into the formula

$\vec{F}=1.6\times10^{-19}\times(3i+4j+11k)$

$\vec{F}=(4.8i+6.4j+1.76k)\times10^{-19}$

$|F|=2.6\times10^{-9}\ N$

Hence, The magnitude of the magnetic force this particle experiences is $2.6\times10^{-9}\ N$ .

7 0

3 years ago

A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N,

Serggg [28]

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

$\sum F_x = ma_x\\$

$Fm - Ff = ma_x\\$ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

$Ff = \mu R\\Ff = \mu mg\\$

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

$Fm - Ff = ma_x\\Fm - \mu mg = ma\\$

Substitute the given parameters

$Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a = 1.736m/s^2\\$

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

7 0

3 years ago

Frank has a sample of steel that has a mass of 80 grams if the density is 8g/cm3 what is the volume

nikklg [1K]

Density is mass divided by volume. rho=m/v. So, v=m/rho. In frank's case this is 80/8 = 10 cm^3.

7 0

3 years ago

A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor

Mumz [18]

Answer:

The maximum induced emf in the rotating coil = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

To calculate the induced emf in the rotating cell,we can use the formula:

emf = NBAωsin(ωt)

For maximum induced emf, the value of sin(ωt) will be 1

$emf_max$ = NBAω ; if (A = l × w) , we have:

$emf_max$ = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

$emf_max$ = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e $emf_max$ = $N\frac{d∅}{dt}$

since $emf_max$ = 29.66 and N = 44; we have:

29.66 = $44\frac{d∅}{dt}$

$\frac{d∅}{dt}$ = $\frac{29.66}{44}$

= 0.674 Wb/s

5 0

3 years ago

A wave with a greater amplitude will transfer . . . . \

BabaBlast [244]

less mass is more mass but less energy in more mass. less mass has more energy

6 0

3 years ago

What sort of fitness component would dance be categorized under