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Y_Kistochka [10]
2 years ago
5

Brody and Ethel are arguing about waves. Brody says that waving his hand back and forth is a scientific wave. Ethel

Physics
2 answers:
butalik [34]2 years ago
6 0

Answer:

b

Explanation:

77julia77 [94]2 years ago
3 0

Answer:

B.)Ethel is correct because energy is not carried from one place to another.

Explanation:

stay safe 2021

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Which moment corresponds to the maximum potential energy of the system?
MissTica
6489 for the founding product
4 0
3 years ago
The Gibson family have bought tickets to the Christmas Market. There are 4 mothers, 2 grand-mothers and 4 daughters. What is the
koban [17]
The minimum number of tickets that could admit all of them is six (6).

This thing is impossible to explain in words, so I shall attempt it with a diagram:

Here are the six ladies:

       ( A )      ( B )
          |           |
          |           |
       ( C )      ( D )
          |           |
          |           |
       ( E )       ( F )  

--  'E'  and  'F'  are the daughters of  'C'  and  'D' .

--  'C'  and  'D'  are the daughters of  'A'  and  'B' .

So look what we have now:

--  'A'  and  'B'  are the mothers of  'C'  and  'D' .
     There's 2 of the mothers.

--  'C'  and  'D'  are the mothers of  'E'  and  'F' .
     There's the OTHER 2 mothers. 
 
--  'A'  and  'B'  are the grandmothers of  'E'  and  'F' .
    There's the 2 grandmothers.

--  'E'  and  'F'  are the daughters of  'C'  and  'D' .
     There's 2 of the daughters.

--  'C'  and  'D'  are the daughters of  'A'  and  'B' .
     There's the OTHER 2 daughters.

You want to know what ? !
The group is even bigger than THAT.
There are also 2 GRAND-daughters in the family ...  'E'  and  'F' .

So now you have a list of 12 people ! ... 4 mothers, 2 grandmothers,
4 daughters, and 2 grand-daughters ... and they all get in to the
Christmas Market with only six tickets.    Legally !

Such a deal !

Don't forget :  Christmas this year is also the first day of Chanukah !
                     All for the same price !

5 0
3 years ago
A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the
lana [24]
(a) Let's convert the final speed of the car in m/s:
v_f = 61 km/h = 16.9 m/s
The kinetic energy of the car at t=19 s is
K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
P= \frac{W}{\Delta t}
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W

(c) The instantaneous power is given by
P_i = Fv_f
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
a= \frac{v_f-v_i}{\Delta t}=  \frac{16.9 m/s}{19 s}=0.89 m/s^2
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W
5 0
3 years ago
I need help with 5 please and thank you
Likurg_2 [28]

Answer: A

Explanation:

7 0
3 years ago
How do you do this? Plz help answer
Contact [7]

Answer: Really

Explanation:

Just look it up for this page and maybe you will find an anwser sheet.

7 0
2 years ago
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