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masya89 [10]
3 years ago
15

How does the shape and size of the continents today help support the Theory of Continental Drift?

Physics
1 answer:
astraxan [27]3 years ago
4 0

The size and shape of the continents help support the continental drift because they all most fit together like puzzle pieces
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Infer why the doppler effect can or can not occur in all waves.
Darya [45]

Answer:

it can occur in all waves because all wave have a frequency

8 0
3 years ago
The current through two identical light bulbs connected in series is 0.25 A. The voltage across both bulbs is 110 V. The resista
Rus_ich [418]

The resistance of a single light bulb is 220 ohms per bulb.

<h3>What is Ohm's Law?</h3>

Ohm's Law is a formula used to determine how voltage, current, and resistance in an electrical circuit relate to one another.

Ohm's Law (E = IR) is as basic to students of electronics as Einstein's Relativity equation (E = mc2) is to physicists.

E = I x R

The formula reads voltage = current x resistance, or V = A xΩ., or volts = amps x ohms.

110volts divided by .25amps = 440 ohms. 440 divided by 2 =220 ohms per bulb.

R = 110/(2*0.25) = 220 ohms

to learn more about Ohms law go to - brainly.com/question/14296509

#SPJ4

4 0
2 years ago
Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air
frosja888 [35]

The frequency of the wave has not changed.

In fact, the frequency of a wave is given by:

f=\frac{v}{\lambda}

where v is the wave's speed and \lambda is the wavelength.

Applying the formula:

- In air, the frequency of the wave is:

f=\frac{400 m/s}{2 m}=200 Hz

- underwater, the frequency of the wave is:

f=\frac{1600 m/s}{8 m}=200 Hz

So, the frequency has not changed.

3 0
3 years ago
Read 2 more answers
Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
A particle with positive charge q = 9.61 10-19 C moves with a velocity v = (5î + 4ĵ − k) m/s through a region where both a unifo
user100 [1]

Answer:

\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

Explanation:

The total force on the particle is given by

\vec{F}=q\vec{v}\ X\ \vec{B}+q\vec{E}

Then, by replacing we have:

q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

where the cross product can be made with the determinant method.

Hope this helps!!

3 0
4 years ago
Read 2 more answers
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