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Ann [662]
3 years ago
8

Radon is the heaviest naturally radioactive

Physics
1 answer:
Burka [1]3 years ago
4 0

Answer:

Yes, it is the heaviest gas.

Explanation:

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PROJECTILE MOTION-Why isn't there any accelertion in the x direction while there is an acceleration of -g in the y direction? ..
Ahat [919]
There is no acceleration of g in the x direction because the gravitational acceleration points downward. Also, on most studies we ignore the tidal forces since we are dealing with small bodies compared to the size of the earth.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
5 0
4 years ago
All contraceptives prevent STIs true or false
NNADVOKAT [17]
False. Not all contraceptives prevent STIs.

Contraceptives are items used or taken to prevent conception. These are birth control pill, patch, ring, IUD, birth control shots, and condoms. Among these contraceptives, condoms are the only type of birth control that prevents STIs or Sexually Transmitted Infections.
7 0
4 years ago
An astronaut is in space with a baseball and a bowling ball. The astronaut pushes both objects in the same direction. If both ba
Crazy boy [7]

Answer:

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

Explanation:

This problem is an application of momentum and momentum. When the astronaut pushed balls, he needed more force to move the ball of greater mass (bowling). The expression for soul is

      p = m v

Besibol Blade

      p1 = m1 v

Bowling ball

      p2 = m2 v

As a mass greater than that of baseball, at the moment of the bowling wave the moment of the baseball ball is also greater

      p2 >> p1

3 0
4 years ago
A bat emitts a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away
Varvara68 [4.7K]

The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.

From the question given above, the following data were obtained:

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

Time (t) =?

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 ×  t = 16.84

Divide both side by 343

t = 16.84/343

t = 0.05 s

Thus, the time between  when the bat emits the sound and when it hears the echo is 0.05 s.

<h3>How does a bat know how far away something is?</h3>

A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.

Learn more about time elapses between when the bat emits the sound :

<u>brainly.com/question/16931690</u>

#SPJ4

Correction question:

A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)

8 0
2 years ago
Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
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