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Yuri [45]
3 years ago
10

Tips to get friends?

Physics
1 answer:
Mice21 [21]3 years ago
3 0

Answer:

be yourself

Explanation:

You might be interested in
Which is best supported by the data in the chart?
bija089 [108]

Answer:

Option C: Current X has a lower potential difference than Current Y.

Explanation:

The chart above only shows the potential difference of difference current.

A careful observation of the chart shows that Current X has a lower potential difference than Current Y.

7 0
3 years ago
A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a
mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required t=\frac{distance}{speed}=\frac{d}{38}hour

Speed when car return from hill = 66 km/h

So time required to return fro hill t=\frac{d}{66}h

Total time t_{total}=\frac{t}{38}+\frac{t}{66}

Total distance = d+d =2d

So average speed=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h

8 0
3 years ago
An object moves from the position +16m to the position +43m in 12s. What us the total displacement
NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

                                     =43m -16m

                                     =27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V =  (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12  ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

3 0
3 years ago
A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

a=\dfrac{v-u}{t}.............(1)

Now, the second equation of motion is :

s=ut+\dfrac{1}{2}at^2

Put the value of a in above equation as :

s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2

s=\dfrac{t(u+v)}{2}

u=\dfrac{2s}{t}-v

u=\dfrac{2\times 47}{8.6}-2.3

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0
3 years ago
A crime suspect fled the scene when police officers entered the building. How far could he run in 10 minutes if he can run 5 mil
emmainna [20.7K]
(5 mi/hr) x (1hr/60min) x (10min) = 5 x 10 / 60 = <em>5/6 mile</em>

(5/6 mile) x (1,760 yd/mile) = <em>1,466 and 2/3 yards</em>
3 0
3 years ago
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