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3 years ago

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An air craft heads. north at 400 km/hrs relative to the wind the air craft velocity is 300 km/hrs due north .find the velocity o

f the wind to the ground.

1 answer:

LUCKY_DIMON [66]3 years ago

8 0

Answer:

325

Explanation

seio na:

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A copper sphere was moving at 40 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f

USPshnik [31]

Answer:

Temperature increase = 2.1 [C]

Explanation:

We need to identify the initial data of the problem.

v = velocity of the copper sphere = 40 [m/s]

Cp = heat capacity = 387 [J/kg*C]

The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:

$E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]$

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3 years ago

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material o

zavuch27 [327]

Answer:

pp

Explanation:

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3 years ago

If the plane is flying in a horizontal path at an altitude of 98.0 m above the ground and with a speed of 73.0 m/s, at what hori

snow_lady [41]

Explanation:

The given data is as follows.

height (h) = 98.0 m, speed (v) = 73.0 m/s,

Formula of height in vertical direction is as follows.

h = $\frac{gt^{2}}{2}$ ,

or, t = $\sqrt{\frac{2h}{g}}$

Now, formula for the required distance (d) is as follows.

d = vt

= $v \sqrt{\frac{2h}{g}}$

= $73.0 m/s \sqrt{\frac{2 \times 98.0 m}{9.8 m/s^{2}}}$

= 326.5 m

Thus, we can conclude that 326.5 m is the horizontal distance from the target from where should the pilot release the canister.

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3 years ago

Ned tightens a bolt in his car engine by exerting 15 N of force on his wrench at a distance of 0.45m from the fulcrum. How much

____ [38]

Answer:

$T = 6.75\,N\cdot m$

Explanation:

Vectorially speaking, torque is the cross product between force and distance from fulcrum. Its magnitude is equal to the following expression:

$T = F\cdot r \cdot \sin \theta$

$T = F_{\bot} \cdot r$

Let assume that force is perpendicular to the distance from the fulcrum. So, the torque needed to turn the bolt is:

$T = (15\,N)\cdot (0.45\,m)$

$T = 6.75\,N\cdot m$

8 0

4 years ago

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f

Alborosie

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s$

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m$

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

4 0

3 years ago