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adelina 88 [10]
3 years ago
13

A car travels 5 km due east, then 10 km due north, and then 5

Physics
1 answer:
Rufina [12.5K]3 years ago
8 0

Answer:

idk what the answer plz help!!!

Explanation:

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Which country or region is coal-rich and oil-poor?
zimovet [89]
THE ANSWER FOR UR QUESTION IS 

WHICH COUNTRY OR REGION IS COAL RICH AND OIL POOR
ANS:- UNITED STATES IS THE COUNTRY WHICH IS COAL RICH AND OIL POOR

HOPE THIS HELPS!!!

^_^ HAPPY TO HELP U ^_^







5 0
3 years ago
7) A crazy cat (yes, this is redundant) is running along the roof of a 60 m tall building. The cat is moving at a constant veloc
gregori [183]

Answer:

The distance from the base of the building to the landing site is 154 m.

The total flight time is 3.5 s.

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

Explanation:

The equations for the position and velocity vectors of the cat are as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector of the cat at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive).

v = velocity vector of the cat at time t.

Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.

When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:

r1y = y0 + v0y · t + 1/2 · g · t²

-60 m = 0 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

- 60 m = -4.9 m/s² · t²

-60 m / - 4.9 m/s² = t²

t = 3.5 s

The cat reaches the ground in 3.5 s

Now, we can calculate the horizontal component of r1:

r1x = x0 + v0 · t

r1x = 0 m + 44 m/s · 3.5 s

r1x = 154 m

The distance from the base of the building to the landing site is 154 m.

The total flight time was already calculated and is 3.5 s.

The velocity vector of the cat when it reaches the ground will be:

v = (v0x, v0y + g · t)

v = (44 m/s, 0 m/s - 9.8 m/s² · 3.5 s)

v = (44 m/s, -34.3 m/s)

The magintude of the vector "v" is calculated as follows:

|v| = \sqrt{(44 m/s)^{2}+(-34.3 m/s)^{2}} = 55.8 m/s

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

6 0
3 years ago
A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp
Kisachek [45]

Answer:

The force is  F= 46.25kN

Explanation:

The diagram for this question is shown on the first uploaded image  

At Equilibrium the summation of the of force on the vertical axis is zero

         i.e   \sum F_y =0

=>            F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)

 v_2 is the is the speed of water at the nozzle which can be mathematically evaluated as

                      v_2 = \frac{R}{A_n}

substituting  0.5m^3/s for R and \frac{\pi}{4}(12*\frac{1m}{100} )^2 for A_n

                    v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }

                         = 44.23 m/s

 v_1 is the is the speed of water at the pipe which can be mathematically evaluated as

                       v_1 = \frac{R}{A_p}

substituting  0.5m^3/s for R and \frac{\pi}{4}(30*\frac{1m}{100} )^2 for A_p

                                v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }

                                    = 7.07 m/s

\rho is he density of water with value \rho =1000 kg /m^3

Substituting values into the equation above

                  F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)

                                 = 21.99kN

At Equilibrium the summation of the of force on the horizontal axis is zero

                  i.e   \sum F_x =0

=>            F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)

               Since The speed at both A and B nozzle are the same then v_2 remains the same

 Substituting values

               F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)

=>                        F_x = 40.69kN

   Hence the force acting on the flange bolts required to hold the nozzle in place is

                      F = \sqrt{F_x^2 + F_y^2}

                         = \sqrt{40.69 ^2 + 21.99^2}

                         F= 46.25kN

                 

6 0
3 years ago
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What's the name of the compound Kl?
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Answer: Potassium iodide

Explanation: their you go

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2 years ago
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A 5kg rock and a 10kg rock are dropped from a height of 10m?
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Look out below ! You should step nimbly to one side, to avoid being hit by one or the other of those hazardous weight objects when they arrive (at the same time).
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