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Mice21 [21]
3 years ago
7

Do your relationships tend to have this quality?

Physics
2 answers:
dedylja [7]3 years ago
5 0
What quality? /////////
allochka39001 [22]3 years ago
3 0

Oh yes, yes ! Each and every one of my many, many relationships has been obviously and indelibly characterized by that exact quality ! Your question fills me with nothing so much as a portentous sense of deja vu ... all over again !

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Temperatures of gases inside the combustion chamber of a four‑stroke automobile engine can reach up to 1000°C. To remove this en
Alona [7]

Answer:

thats a lot, which one u want me to do?

Explanation:

8 0
2 years ago
Hiii please help i’ll give brainliest if you give a correct answer please thanks!
lakkis [162]

Answer: the first one

Explanation: good luck!

8 0
3 years ago
Read 2 more answers
What is the definition of input force?
andrezito [222]

Answer:

The input force (effort) is the amount of effort used to push down on a rod, or pull on a rope in order to move the weight. In this example, the force the little guy is using to pull the elephant is the input force.

Explanation:

5 0
2 years ago
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Sound travels through air at 343 m/s,
sasho [114]

The sound wave will have traveled 2565 m  farther in water than in air.

Answer:

Explanation:

It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.

Distance = Velocity × Time.

So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.

As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.

Distance = 343 × 2.25 =771.75 m

And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.

Distance = 1483×2.25=3337 m.

Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.

Difference in distance covered in water and air = 3337-772 m = 2565 m

So the sound wave will have traveled 2565 m  farther in water than in air.

5 0
3 years ago
Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre
castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:  

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)  

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619  

=

t

+

C

Here C =constant of integration.   

x

=

0  at  t

=

0

, we get:  C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619  =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x  is measured in miles and  t  in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),    

     to get:

      = −

4.7619  

      =  

1

−

4.7619

      = −

3.7619

  or  x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in   m

i

l

/

differentiate  

     equation (1) with respect to time.

     we have to eliminate x from the equation (1) using equation (2).  

     Eliminating x we get:

     v

=

7.5×

     Now differentiating above equation w.r.t time we get:

      a

=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

       a

=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

(c)  required time for the jogger to run 6 miles is obtained by setting  

        x

=

6  in equation (2).  We get:

        −

4.7619

(

1

−

(

0.04

×

6  )

)^

7

/

10=

t

−

4.7619

         or

         t

=

0.832

h

r

s

6 0
3 years ago
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