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lianna [129]
3 years ago
10

Jenna flips a 10p coin four times. Here are her results:

Physics
1 answer:
ValentinkaMS [17]3 years ago
6 0

Answer:

Option B

Explanation:

The reliability of an experiment can be increased by conducting experiment sufficient number of times keeping other things constant. Hence, the 10 p coin needs to be flipped again to measure reliability for more than one times.

Hence, option B is correct

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Weight gain occurs when a person consumes too muchimany
MAVERICK [17]
The answer is calories !!
5 0
3 years ago
Read 2 more answers
A sailboat is heading directly north at a speed of 20 knots (1 knot 50.514 m/s). the wind is blowing toward the east with a spee
stellarik [79]
(a) The magnitude of the wind as it is measured on the boat will be the result of the two vectors. Since they are at 90°, the resultant can be determined by the Pythagorean theorem.
 
                   R = sqrt ((20 knots)² + (17 knots)²)
                    R = sqrt (400 + 289)
                      R = 26.24 knots

The direction of the wind will have to be angle between the boat and the resultant.
                   cos θ = (20 knots)/(26.24 knots)
                        θ = 40.36°

Hence, the direction is 40.36° east of north.

(b) As stated, the wind is blowing in the direction that is to the east. This means that it only has one direction. Parallel to the motion of the boat, the magnitude of the wind velocity will have to be zero. 
5 0
4 years ago
Why does earth have the youngest surface of all the terrestrial planets today?
kotykmax [81]

Answer:

Its the biggest terrestrial planet as the result of the interior being not able to cool down that much.

its rotates rapidly

Explanation:

the earth planet is not close to the sun,which puts it in the advantage of not loosing water.

8 0
3 years ago
A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depend
liq [111]
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

                     (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere. 
Only the initial and final y-coordinates matter.

We know that    F = - 2.85 y².  (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From  y=0  to  y=2.40  is a distance of  2.40  upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .


Now, integral of (y² dy)  =  1/3  y³ .

Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the  -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40.  Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
by some other route.  WHATEVER the route is, the work done by ' F ' 
is going to total up to be  -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then.  But that's my answer
and I'm stickin to it.  If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
3 0
3 years ago
A 4.50-kg wheel that is 34.5 cm in diametet rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1
lisabon 2012 [21]

Answer:

\alpha =10.93radian/sec^2

Explanation:

We have given given the final angular velocity \omega _{final}=13.5rad/sec

And \omega _{initial}=22rad/sec

Displacement \Theta =13.8radian

We have to find the angular acceleration \alpha

According to law of motion \omega _{final}^2=\omega _{initial}^2+2\alpha \Theta

So 13.5^2=22^2+2\times \alpha \times 13.8

\alpha =-10.93radian/sec^2

In question we have tell about magnitude only so \alpha =10.93radian/sec^2

4 0
3 years ago
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