Which evidence did Wegener use to develop the theory of continental drift?

During an olympic bobsled run, the jamaican team makes a turn of radius 9.80 m at a speed of 96.6 km/h. what is the ratio of the

GREYUIT [131]

Answer:

Acceleration will 7.5 times g = 7.5 g

Explanation:

Centripetal acceleration is given by a = v²/r.

The speed in m/s will be; (96.6 Km/h )÷ 3.6

= 26.83 m/s.

Centripetal force = v²/r

= 26.83²/9.80

= 73.454 m/s²

But; g = 9.8 m/s²

Therefore the acceleration in terms of g will be;

= 73.454/9.8

= 7.5 g

Acceleration is 7.5 g

6 0

3 years ago

When Cesium metal is illuminated with light of wavelength 300 nm, the photoelectrons emitted have a maximum kinetic energy of 2.

Allisa [31]

Answer:

i have no clue im sorry

Explanation:

8 0

3 years ago

An object is moving with an initial velocity of 6.5m/s. It is then subject to a constant acceleration of 2.1m/s^2 for 14s. How f

alina1380 [7]

Answer:

Distance travel s = 296.8 m

Explanation:

Given:

Initial velocity u = 6.5 m/s

Constant acceleration a = 2.1 m/s²

Time T = 14 s

Find:

Distance travel s

Computation:

s = ut + (1/2)at²

s = (6.5)(14) + (1/2)(2.1)(14)²

s = 91 + 205.8

Distance travel s = 296.8 m

4 0

3 years ago

An air-filled capacitor consists of twoparallel plates, each with an area of 7.60 cm2,separated by a distance of 1.90 mm.A 16.0

Svetradugi [14.3K]

Answer:

(A)8,421.1 V/m

(B) $3.54 x 10^{-12} F$

(C) $53.1 x 10^{-12} C$

(D) $6.987 x 10^{-8} C/m^{2}$

Explanation:

area (A) = 7.6 cm^{2} = 0.00076 m^{2}

distance (d) = 1.9 mm = 0.0019 m

potential difference (v) = 16 V

(A) electric field (E) = Δv / d = 16 / 0.0019 = 8,421.1 V/m

(B) capacitance = (∈₀A)/d

where ∈₀ = permitivity of free space = $8.85 x 10^{-12}m^{-3}.kg^{-1} .s^{4}.A^{2}$

capacitance = $\frac{8.85 x 10^{-12}x0.00076}{0.0019}$

capacitance (C) = $3.54 x 10^{-12} F$

(C) charge (q) = C x v = $3.54 x 10^{-12} x 16$ = $56.64x 10^{-12} C$

(D) surface charge density = charge (q) / area = $56.64 x 10^{-12}$ / 0.00076 = $7.453 x 10^{-8} C/m^{2}$

3 0

3 years ago

A 100-cm long dipole is excited by a sinusoidally varying current with an amplitude i0=2 a. Determine the time average power rad

Mnenie [13.5K]

For A 100-cm long dipole is excited by a sinusoidally varying current with an amplitude i0=2 , the time average power radiated is mathematically given as

P=0.1577w

<h3>What is the time average power radiated by the dipole if the oscillating frequency is 150 mhz?</h3>

Generally, the equation for the is mathematically given as

$\lambda =\frac{c}{f}$

Therefore

$\lambda=\frac{3\times 10^{8}}{10^{6}}$

lambda=300m

In conclusion, for the power

$P=40\pi^{2}(I_{0})^{2}(\frac{l}{\lambda})^{2}\\\\P=40* (3.14)^{2}\times6^{2} (\frac{1}{300})^{2}$

P=0.1577w