Calculate the mass of the earth if the radius of the earth is 6.38 × 10^6

Communication with submerged submarines via radio waves is difﬁcult because seawater is conductive and absorbselectromagnetic wa

FrozenT [24]

Answer:

D. 4000 km

Explanation:

f = Frequency of wave that is being transmitted = 76 Hz

$\lambda$ = Wavelength of wave that is being transmitted

v = The velocity of electromagnetic waves through air = $3\times 10^8\ m/s$

Velocity of a wave is given by

$v=f\lambda\\\Rightarrow \lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{3\times 10^8}{76}\\\Rightarrow \lambda=3947368.42105\ m=3947.36842105 km\approx 4000\ km$

Hence, the approximate wavelength of the waves is 4000 km

8 0

3 years ago

A resistor with R = 300 Ω and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate

schepotkina [342]

Answer:

Explanation:

impedance z=(XL^2+R^2)^1/2

power across te resistor ==i^2r

286/300

I=.976

3 0

3 years ago

If an electron moves in a circle of radius 21 cm perpendicular to a B field of 0.4 T, what are the speed of the electron and the

kodGreya [7K]

Answer:

a)

$v = 4.048 *10^6$ m/s

b)

Angular frequency = $1.92 * 10^7$

Explanation:

As we know

$v = \frac{qBr}{m}$

q is the charge on the electron = $3.2 * 10^{-19}$ C

B is the magnetic field in Tesla = $0.4$ T

r is the radius of the circle = $0.21$ m

mass of the electrons = $6.64 * 10^{-27}$ Kg

a)

Substituting the given values in above equation, we get -

$v = \frac{3.2 * 10^{-19}*0.4*0.21}{6.64 * 10^{-27}} \\v = 4.048 *10^6$ m/s

b)

Angular frequency =

$\frac{4.048 * 10^6 }{0.21} \\1.92 * 10^7$

8 0

2 years ago

g If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of th

Reil [10]

Answer:

Speed; v = 17 m/s

Explanation:

We are given;

Radius; r = 110m

Angle; θ = 15°

Now, we know that in circular motion,

v² = rg•tanθ

Thus,

v = √(rg•tanθ)

Where,

v is velocity

r is radius

g is acceleration due to gravity

θ is the angle

Thus,

v = √(rg•tanθ) = √(110 x 9.8•tan15)

v = √(288.85)

v = 17 m/s

8 0

3 years ago

two horses are pulling a 325 kg wagon, initially at rest. The horses exert 250 N and 178 N forward forces, respectively. Ignorin

Dovator [93]

Answer:

AFter 3.5 s, the wagon is moving at: $4.62\,\,\frac{m}{s}$

Explanation:

Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):

Net force = 250 N + 178 N = 428 N

Therefore, the acceleration from Newton's 2nd Law is:

$F=m\,*\,a\\a = \frac{F}{m} \\a= \frac{428}{325}\, \frac{m}{s^2} \\a\approx 1.32 \,\,\frac{m}{s^2}$

So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:

$v_f=v_i+a\,*\,t\\v_f=0+1.32\,*\,3.5\,\,\frac{m}{s} \\v_f=4.62\,\,\frac{m}{s}$

7 0

3 years ago

Calculate the mass of the earth if the radius of the earth is 6.38 × 10^6​

Calculate the mass of the earth if the radius of the earth is 6.38 × 10^6