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andrey2020 [161]
4 years ago
8

I NEED HELP!!!!! PLEASE ANSWER! FIRST CORRECT ANSWER WILL GET BRAINLIEST AND I WILL FOLLOW YOU!

Physics
1 answer:
Naily [24]4 years ago
6 0

Average speed = (total distance) / (time to cover the distance)

We know:

    Average speed = 65 km/hr
    Total distance = 1,000 km 
    Time to cover it = (Driving Time) + 4 hours.

so we can write:

          65 km/hr  =  (1,000 km) / (Driving Time + 4hr)

          (I'm going to start calling the driving time 'DT'.
           Notice that DT is a number with the units of 'hours'.)

Multiply each side by    (DT + 4hr)

           (65 km/hr) (DT + 4hr)  =  1,000 km   

Eliminate parentheses on the left side:

           (65·DT km  +  260 km)  =  1,000 km

Subtract  260km  from each side:

              65·DT km          =    740 km

Divide each side by 65 :

                DT   =   11.38 hours .

DT (Driving Time) is the time you spent actually driving.
You had to cover the complete 1,000 km in that time.
So while you were driving, you had to do it at a speed of

                  1,000 km / 11.38 hrs  =  87.8 km/hr .
__________________________________________

As long as we're already totally bored by this question,
let's work on it some more, and check my answer:

... Driving for 11.38 hours at a speed of 87.8 km/hr, you cover

                     (11.38 hr) x (87.8 km/hr)  =  999.164 km  (close enough to 1,000) .

So far, so good.  The distance is taken care of.

With the 4-hour stop, the total trip takes 4 more hours = 15.38 hours.
So the average speed is

                     (1,000 km) / (15.38 hr)  =   65.02 km/hr

                                                                Close enough to 65 km/hr.  yay !

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8 0
3 years ago
When traveling through gases, _____________ also determines how fast a sound wave travels.
Tpy6a [65]

Answer:

density of the gas

Explanation:

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5 0
4 years ago
What must you do to calculate a meaningful value of distance from the following equation?
ss7ja [257]

Answer:

Convert all the times to either hours or seconds

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5 0
3 years ago
Read 2 more answers
One charge is decreased to one-third of its original value, and a second charge is decreased to one-half of its original value.
Nadya [2.5K]
The original Coulomb force between the charges is:

Fc=(k*Q₁*Q₂)/r², where k is the Coulomb constant and k=9*10⁹ N m² C⁻², Q₁ is the first charge, Q₂ is the second charge and r is the distance between the charges.

The magnitude of the force is independent of the sign of the charge so I can simply say they are both positive. 

Q₁ is decreased to Q₁₁=(1/3)*Q₁=Q₁/3 and
Q₂ is decreased to Q₂₂=(1/2)*Q₂=Q₂/2. 

New force:

Fc₁=(k*Q₁₁*Q₂₂)r², now we input the decreased values of the charge

Fc₁=(k*{Q₁/3}*{Q₂/2})/r², that is equal to:

Fc₁=(k*(1/3)*(1/2)*Q₁*Q₂)/r²,

Fc₁=(k*(1/6)*Q₁*Q₂)/r²

Fc₁=(1/6)*(k*Q₁*Q₂)/r², and since the original force is: Fc=(k*Q₁*Q₂)/r² we get:

Fc₁=(1/6)*Fc

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4 0
3 years ago
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(a) Calculate the electric field strength near a 10.0 cm diameter conducting sphere that has 1.00 C of excess charge on it. (b)
Romashka-Z-Leto [24]

Electric field strength near the surface of sphere is given by

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Now we have

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Part B)

Since charge on the sphere is very large to electric field near its surface is too high. So this is reasonable result near it.

Part c)

As per our assumption the charge is uniformly distributed on the surface and there is no effect of surrounding charge or surrounding electric field.

4 0
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