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NNADVOKAT [17]
3 years ago
13

At time=0s, the object is at the 21.0-meter position along the roadway. Where is the object at time = 10 s?

Physics
1 answer:
Jet001 [13]3 years ago
4 0
B.) 70m is your answer!
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A car starting at position 16m travels out to position 67m and turns around and speeds back to position 16m. What was the displa
enyata [817]

Answer: D. 102m

Explanation: Because if you subtract 67m-16m=51. 51 +51 is 102m

7 0
3 years ago
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A plane has an airspeed of 112 km km/h. It is flying on a bearing of 81 degrees° while there is a 28 km divided by h28 km/h wind
Darya [45]

Answer:

speed = 90.7934 km/hr

bearing = 88.55°​

Explanation:

given data

plane speed v = 112 km/hr

bearing = 81°  

θ  = 90 - 81 = 9°

wind speed w = 28 km/hr

bearing (θ2)  =  225°

to find out

ground speed and the bearing of the​ plane

solution

we get here resultant ground speed of plane that is

speed = [ v cos(θ) + w cos(θ2)] i + [ v sin(θ) + w sin(θ2)] j

speed = [112 cos(9) + 28 cos(225)] i + [112 sin(9) + 28 sin(225)] j

speed = 90.822 i - 2.278 j

| speed | = \sqrt{90.822^2-2.278^2}

| speed | = 90.7934 km/hr

and

tan(θ) =  \frac{2.278}{90.822}

θ = 1.4367°

bearing of the​ plan will be

bearing  = 90 - 1.4367

bearing = 88.55°​

8 0
3 years ago
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

6 0
3 years ago
If you see an emergency vehicle approaching either from the front or rear you should: Stop right where you are Speed up and out
Lorico [155]

You should pull over to the curb and stop

8 0
3 years ago
A reservoir dam has a vertical surface in contact with the water that is 125 m wide and 11.0 m high. Air pressure is one atmosph
pishuonlain [190]

Answer:

The answer to the question is;

The net force acting on the dam is 73,001,115 N.

Explanation:

To solve he question, we note that

Pressure = Force/Area and the formula for pressure is

Pressure = hρg

While force, F= Pressure × Area

For a small area at the bottom of the dam we have

= hρg×dh×width

Where

w = width of the dam = 125 m

ρ = Density of water = 1.0×10³ m³

g = Acceleration due to gravity = 9.81 m/s²

h = Height  of the dam = 11.0 m

Therefore force over the dam wall = \int\limits^{11}_0 {h\rho gw} \, dh  = [\frac{h^2}{2} \rho gw]^{11}_0  

                                   = [ 11²/2×1×10³×9.81×125] - 0

                                = 73,001,115 N

The net force acting on the dam = 73,001,115 N.

4 0
4 years ago
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