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2 years ago

Which evidence did Alfred Wegener’s original theory of continental drift have access to?

Physics

1 answer:

Sophie [7]2 years ago

4 0

Answer:

Evidence for continental drift

Wegener knew that fossil plants and animals such as mesosaurs, a freshwater reptile found only South America and Africa during the Permian period, could be found on many continents. He also matched up rocks on either side of the Atlantic Ocean like puzzle pieces.

Explanation:

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A television of mass 12 kg sits on a table. The coefficient of static friction between the table and the television is 0.83. Wha

Zinaida [17]

The normal reaction between the television and the table is
N = 12 × 9.8 m/s² = 117.6 Newtons

But the static coefficient of friction is μ = 0.83

When the television is about to slide on the table, the applied force should overcome the force due to static friction;
Thus; the applied force should be at least
F = μN
= 0.83 × 117.6 N
= 97.608 Newtons
Therefore; the minimum applied force will be 97.6 Newtons.

4 0

3 years ago

Read 2 more answers

Ice has a density of 0.920 g/cm3. An ice cubes mass is 100 g. What is its volume?

Tasya [4]

Answer:

$\boxed {\tt 108.695652 \ cm^3}$

Explanation:

Volume can be found by dividing the mass by the density.

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

The mass is 100 grams and density is 0.920 grams per cubic centimeters.

Therefore,

$m=100 \ g \\d= 0.920 \ g/cm^3$

Substitute the values into the formula.

$v=\frac{100 \ g}{0.920 \ g/cm^3}$

Divide. Note that the grams, or "g" will cancel each other out.

$v- 108.695652 \ cm^3$

The volume of the ice cube is 108.697652 cubic centimeters.

3 0

3 years ago

A multimeter in an RL circuit records an rms current of 0.600 A and a 50.0-Hz rms generator voltage of 110 V. A wattmeter shows

jeka57 [31]

Answer:

(a) The impedance in the circuit is $Z=183.33\Omega$ .

(b)The resistance is $R=38.89\Omega$ .

(c) The inuctance is 0.57 H.

Explanation:

(a)

The expression for the impedance is as follows:

$Z=\frac{V_rms}{I_rms}$

Here, $V_rms$ is the rms voltage and $I_rms$ is the rms current.

Put $V_rms=110 V$ and $I_rms=0.600 A$ .

$Z=\frac{110}{0.600}$

$Z=183.33\Omega$

Therefore, the impedance in the circuit is $Z=183.33\Omega$ .

(b)

The expression for the average power is as follows;

$P_{a}=I_{rms}^{2}R$

Here, $P_{a}$ is the average power and R is the resistance.

Calculate the resistance by rearranging the above expression.

$R=\frac{P_{a}}{I_{rms}^{2}}$

Put $P_{a}=14W$ and

$R=\frac{14}{{0.600}^{2}}$

$R=38.89\Omega$

Therefore, the resistance is $R=38.89\Omega$ .

(c)

The expression for the impedance is as follows;

$Z^{2}=R^{2}+X_{L}^{2}$

Here, $X_{L}$ is the inductive reactance.

Put $Z=183.33\Omega$ and $R=38.89\Omega$ .

$(183.33)^{2}=(38.89)^{2}+X_{L}^{2}$

$X_{L}=179.16\Omega$

The expression for the inductive reactance in terms of frequency is as follows;

$X_{L}=2\pi fL$

Here, L is the inductance.

Calculate the inductance by rearranging the above expression.

$L=\frac{X_{L}}{2\pi f}$

Put $X_{L}=179.16\Omega$ and f=50Hz.

$L=\frac{179.16}{2\pi (50)}$

L=0.57 H

Therefore, the inuctance is 0.57 H.

4 0

3 years ago

During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapu

natali 33 [55]

Answer:

110.7 J

Explanation:

Hooke's law is represented by the formula:

F = ke where F is the force in Newton, K is force constant and e is extension in m

work done = 1/2ke² = 1/2 K ( e² - e₀²) and e₀ is the extension at relaxed length

e₀ =0

work done = 0.5 × 82N/m × (2.70 m)² = 110.7 J

4 0

3 years ago

Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det

Leokris [45]

Answer:

0.037 N/m

Explanation:

The web acts as a spring, so it obeys Hook's law:

$F=kx$ (1)

where

F is the force exerted on the web

k is the spring constant

x is the stretching/compression of the web

In this problem, we have:

- The mass of the fly is $m=15 mg=15\cdot 10^{-6} kg$

- The force exerted on the web is the weight of the fly, so:

$F=mg=(15\cdot 10^{-6}kg)(9.81 m/s^2)=1.47\cdot 10^{-4}N$

- The stretching of the web is

$x=4.0 mm=0.004 m$

So if we solve eq.(1) for k, we find the spring constant:

$k=\frac{F}{x}=\frac{1.47\cdot 10^{-4} N}{0.004 m}=0.037 N/m$

3 0

3 years ago