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arsen [322]
3 years ago
6

A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin

g another right turn to travel 1.40 km north to arrive at its destination. What is the magnitude and direction of the truck's displacement from the warehouse?
Physics
1 answer:
yulyashka [42]3 years ago
3 0

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19 km

tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2

\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees north or west or (180 - 72.65) = 107.35 degrees north of east

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acontainer is filled whith mercury to alevel of 10m whit water to alevel of 8m and whit oil to alevel of 5m the densities oil ,w
Alborosie

Answer:

1450.4 KNm^{2}

Explanation:

Pressure = ρhg

where: ρ is the density of the liquid, h is the height and g the force of gravity.

Total pressure exerted by the liquids at the base = Pressure of oil + Pressure of water + Pressure of mercury

So that,

i. Pressure of oil = ρhg

(ρ = 0.8 g/cm³ = 800 kg/m³)

                        = 800 x 5 x 9.8

                        = 39200

Pressure of oil = 39200 Nm^{2}

ii. Pressure of water = ρhg

(ρ = 1 g/cm³ = 1000 kg/m³)

                                      = 1000 x 8 x 9.8

                                     = 78400

Pressure of water = 78400 Nm^{2}

ii. Pressure of mercury = ρhg

(ρ = 13.6 g/cm³ = 13600 kg/m³)

                      = 13600 x 10 x 9.8

                      = 1332800

Pressure of mercury = 1332800 Nm^{2}

So that,

Total pressure exerted by the liquids at the base = 39200 + 78400 + 1332800

                                               = 1450400

                                               = 1450.4 KNm^{2}

Total pressure exerted by the liquids at the base is 1450.4 KNm^{2}.

8 0
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Keeping the applied voltage constant at approximately 0.70 volts, measure the electric field strength at different points betwee
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Answer: I observed that the electric field strength is the same at all points between the plates. The value of the field is 70 volts per meter. This is exactly 100 times the applied voltage. The electric field lines point from the positive plate to the negative plate, as the downward arrow on the detector shows.

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A) Is it possible for an object that, object with zero acceleration have velocity? If yes give an example if not give its proof.
sveticcg [70]

Answer:

Yes, in case of uniform velocity

Explanation:

This is the case of uniform velocity. If a body covers equal displacement in equal intervals of time, then the velocity of a body is said to be ‘Uniform Velocity’. It meas that the velocity of a body remains constant during the motion and it does not change.

Since, acceleration is defined as the rate of change of velocity.

Therefore, if there is no change in velocity or in other words the change in velocity is zero, then the acceleration is also zero.

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