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Yuki888 [10]
3 years ago
5

Starting from your campsite you walk 3.0 km east, 6.0 km north, 1.0 km east, and then 4.0 km west. How far are you from your cam

psite
Physics
1 answer:
Hatshy [7]3 years ago
6 0
Think of it like a graph. You start at the origin which is (0,0).  go three to the east which now you are (3,0). Then, six to the north. Now, you are at (3,6).  1 to the east, ((4,6).  Then you go 4 to the west which is back tracking. So, you end at (0,6) which is saying you are now 6 km north from your campsite. 

Hope this helps!
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PLEASE HELP!
lana66690 [7]

Answer:

x = 4.32 [m]

Explanation:

We must divide this problem into three parts, in the first part we must use Newton's second law which tells us that the force is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force = 700 [N]

m = mass = 2030 [kg]

a = acceleration [m/s²]

Now replacing:

F=m*a\\700=2030*a\\a = 0.344[m/s^{2}]

Then we can determine the final speed using the principle of conservation of momentum and amount of movement.

(m_{1}*v_{1})+Imp_{1-2}=(m_{1}*v_{2})

where:

m₁ = mass of the car = 2030 [kg]

v₁ = velocity at the initial moment = 0 (the car starts from rest)

Imp₁₋₂ = The impulse or momentum (force by the time)

v₂ = final velocity after the impulse [m/s]

(2030*0) + (700*5)=(2030*v_{2})\\3500 = 2030*v_{2}\\v_{2}=1.72[m/s]

Now using the following equation of kinematics, we can determine the distance traveled.

v_{2}^{2} =v_{1}^{2}+2*a*x

where:

v₂ = final velocity = 1.72 [m/s]

v₁ = initial velocity = 0

a = acceleration = 0.344 [m/s²]

x = distance [m]

1.72^{2}=0^{2} +(2*0.344*x) \\2.97 = 0.688*x\\x = 4.32 [m]

8 0
3 years ago
A ball rolls at 3 meters per second. how much time does it take for the ball to roll 6 meters
natka813 [3]
2 seconds skins b the answer
5 0
3 years ago
Read 2 more answers
Two particles are fixed to an x axis: particle 1 of charge −8.00 ✕ 10⁻⁷ C at x = 6.00 cm, and particle 2 of charge +8.00 ✕ 10⁻⁷
victus00 [196]

Answer:

0 N/C

Explanation:

Parameters given:

q_1 = -8.00 * 10^{-7} C

x_1 = 6.00 cm

q_2 = +8.00 * 10^{-7} C

x_2 = 21 cm

The distance between q_1 and q_2 is

21 - 6 = 15cm

Electric field is given as

E = \frac{kq}{r^2}

r = 15/2 = 7.5cm = 0.075m

The electric field at their midpoint due to q_1 is:

E = \frac{9 * 10^9 * -8.0 * 10^{-7}}{0.075^2}

E_1 = -1.28 * 10^6 N/C

The electric field at the midpoint due to q_2 is:

E = \frac{9 * 10^9 * 8.0 * 10^{-7}}{0.075^2}

E_2 = 1.28 * 10^6 N/C

The net electric field will be:

E = E_1 + E_2

E = -1.28 * 10^6 + 1.28 * 10^6

E = 0 N/C

7 0
4 years ago
A 1.8-kg object moves in the x direction according to the following function:
DanielleElmas [232]
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then the speed is          x ' (t)  =  4t + 3    (first derivative of  'x'  wrt  't')

and the acceleration is     x ' ' (t)  =  4      (second derivative of  'x'  wrt  't')

Apparently, then, the acceleration is constant, and is not a function of time.

After 2.7 seconds or 2.7 years, the acceleration is  4 .

Force = (mass) x (acceleration)

Force =  (1.8)    x      (4)

<em>Force =            7.2 newtons </em>
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3 years ago
What does Neil Degrasse Tyson mean when he says "Wolves domesticated humans" 15000 years ago?
Aliun [14]

Answer:

Explanation:

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