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3 years ago

As an airplane is flying 20.0 degrees north of east at a speed of 750 km/h. How fast is it moving to the north

Physics

2 answers:

dedylja [7]3 years ago

7 0

<span>in this case velocity of airoplane will be 750*sin20'

</span>

Gwar [14]3 years ago

3 0

Answer:

$v_y=71.25\ m/s$

Explanation:

Given that,

Speed of an airplane, v = 750 km/hr = 208.33 m/s

An airplane is flying 20.0 degrees north of east, $\theta=20^{\circ}$

To find,

Velocity in north direction.

Solution,

Let $v_y$ is the speed of the airplane to the north. The y component of any motion is given by :

$v_y=v\ sin\theta$

$v_y=208.33\ m/s\ sin(20)$

$v_y=71.25\ m/s$

So, the airplane is moving with a speed of 71.25 m/s to the north direction.

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Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

Mariulka [41]

Answer:

a) 0.0288 grams

b) $2.6*10^{-10} J/kg$

Explanation:

Given that:

A typical human body contains about 3.0 grams of Potassium per kilogram of body mass

The abundance for the three isotopes are:

Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.

Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.

However, the amount of potassium that is present in such person is :

0.012% × 240 grams

= 0.012/100 × 240 grams

= 0.0288 grams

the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:

First the Dose in (Gy) = $\frac{energy \ absorbed }{mass \ of \ the \ body}$

= $\frac{1.10*10^6*1.6*10^{-14}}{80}$

= $2.2*10^{-10} \ J/kg$

Effective dose (Sv) = RBE × Dose in Gy

Effective dose (Sv) = $1.2 *2.2*10^{-10} \ J/kg$

Effective dose (Sv) = $2.6*10^{-10} J/kg$

5 0

3 years ago

A car drives around a curve with radius 400 m at a speed of 32 m/s. The road is banked at 7.0 degree. The mass of the car is 150

Ronch [10]

Answer:

The magnitude of the centripetal force to make the turn is 3,840 N.

Explanation:

Given;

radius of the cured road, r = 400 m

speed of the car, v = 32 m/s

mass of the car, m = 1500 kg

The magnitude of the centripetal force to make the turn is given as;

$F_c = \frac{mv^2}{r}$

where;

Fc is the centripetal force

$F_c = \frac{mv^2}{r} \\\\F_c = \frac{(1500)(32)^2}{400}\\\\F_c = 3,840 \ N$

Therefore, the magnitude of the centripetal force to make the turn is 3,840 N.

3 0

2 years ago

A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4

Svetllana [295]

The complete question is;

A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T.

a. Express the magnetic dipole moment μ in terms of the number of the turns N, the current I, and radius

b. Which direction does μ go?

Answer:

A) μ = 1738.87 A.m²

B) The direction of the magnetic moment will be in upward direction.

Explanation:

We are given;

The number of circular coils;

N = 410

The radius of the coil;R = 0.75m

The current in the coils; I = 2.4 A

The strength of magnetic field;

B =1.1T

The formula for magnetic dipole moment is given as;

μ = NIA

Where;

N is number of turns

I is current

A is area

Now, area; A = πr²

So, A = π(0.75)²

Thus,plugging in relevant values, the magnetic dipole moment is;

μ = 410 * 2.4 * π(0.75)²

μ = 1738.87 A.m²

B) According to Fleming's right hand rule, the direction of the magnetic moment comes out to be in upward direction.

7 0

3 years ago

The solids settled at the bottom is called a) sludge b) sewage c) waste d) litter

Anna71 [15]

Answer:

the solid particles that settle down at the bottom of the container are known as sediment.

Explanation:

5 0

2 years ago

If the resulting trajectory of the charged particle is a circle,what is , the angular frequency of the circular motion?Express i

Maru [420]

Answer:

ω = B₀q/m

Explanation:

The centripetal force on the charge equal the magnetic force on the charge. (Since the trajectory is a circle). So, mrω² = B₀qv.

v = rω The speed of the charge and r = radius of path

mrω² = B₀qrω

ω = B₀q/m

The angular frequency ω = B₀q/m

8 0

2 years ago