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Advocard [28]
2 years ago
11

Anjali's plane had been flying through calm skies (no wind) with a velocity (speed and direction) vector

-dn.net/?f=%5Cvec%7Bv_1%7D" id="TexFormula1" title="\vec{v_1}" alt="\vec{v_1}" align="absmiddle" class="latex-formula"> = (400,100). Now, however, the air mass surrounding the plane is moving quickly (i, it's windy). Although Anjali has not adjusted any of the controls in the cockpit, the plane's new velocity is \vec{v_2} = (350, 50). (Speeds are in kilometers per hour.) What is the speed of the wind?
Physics
1 answer:
grin007 [14]2 years ago
5 0

Answer:

The speed of the wind is 70.7 units.

Explanation:

The velocity of the wind is equal to plane's velocity before the wind minus its velocity after the wind:

\vec{v_{w}}=\vec{v_1}-\vec{v_2}= (400,100)- (350,50)

\boxed{\vec{v_{w}} = (50,50)}

Now, the speed of the wind is the magnitude of \vec{v_{w}}:

|\vec{v_{w}}| = \sqrt{50^2+50^2}

\boxed{|\vec{v_{w}}| = 70.7}

which is the speed of the wind.

<em>P.S: No units are given for speed because no units were provided in the question.</em>

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A circular loop of wire is rotated at constant angular speed about an axis whose direction can be varied. In a region where a un
katovenus [111]

Answer:

here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

Explanation:

As we know by Faraday's law of electromagnetic induction

Rate of change in magnetic flux will induce EMF in the coil

so here we will have

EMF = \frac{d\phi}{dt}

here we know that

\phi = NB.A

now if the magnetic flux will change with time then it will induce EMF in the coil

EMF = N\frac{d}{dt}(B.A)

so here induced EMF will be zero in the coil if the flux linked with the coil will remain constant

so here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

In such a way when coil will rotate then the flux linked with the coil will remains constant and there will be no induced EMF in it

5 0
2 years ago
A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its
Katyanochek1 [597]

Answer:

d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

d_{x}(t)=dcos(\theta (t)) (1)

  • d is the length of the minute hand (d=D/2)
  • D is the diameter of the clock
  • t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

\theta (t)=\omega *t (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}

Now, let's put this value on (2)

\theta (t)=\frac{\pi}{30}*t

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)

I hope it helps you!

6 0
2 years ago
Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to
Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate m=100 kg

Crate slows down in s=1.5 m

initial speed u=1.77 m/s

inclination \theta =30^{\circ}

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2

W_{gravity}=mg(0-h)=mgs\sin \theta

W_{gravity}=-mgs\sin \theta

W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N

change in kinetic energy=\frac{1}{2}\times 100\times 1.77^2=156.64 J

W_{friction}=156.64+735=891.645

(b)Coefficient of sliding friction

f_r\cdot s=W_{friciton}

891.645=f_r\times 1.5

f_r=594.43 N

and f_r=\mu mg\cos \theta

\mu 100\times 9.8\times \cos 30=594.43

\mu =0.7

5 0
2 years ago
A 5-kg concrete block is lowered with a downward acceleration of 2.8 m/s2 by means of a rope. The force of the block on the Eart
maksim [4K]

When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:

F = mg

Where,

m = mass

g = Gravitational acceleration

F = 5*9.8

F = 49N

Therefore the correct answer is E.

5 0
2 years ago
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