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a_sh-v [17]
3 years ago
8

PLEASE HELP

Physics
2 answers:
Firlakuza [10]3 years ago
5 0
A bird flying near the mountain
siniylev [52]3 years ago
3 0
A person standing on the moon watching the earth rotate
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Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

  • V_r = resultant velocity of the plane
  • \theta = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

5 0
3 years ago
Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
What is the frequency of a wave with a wavelength of 15 m and a wavespeed of<br> 300 m/s?
STatiana [176]

Answer: f=20 (i think)

Explanation:

all I did was divide 300 and 15.

300/15= 20

6 0
2 years ago
cylindrical container is to be constructed to be open at the top with a volume of 27π cubic meters using the least amount of mat
Llana [10]

Answer:

radius comes out to be 3 m

height of the cylinder comes out to be 3m

Explanation:

given

volume of cylinder = 27π m³

π r² h = 27π

   r² h = 27.............(1)

surface area of cylinder open at the top

S = 2πrh + π r²

S = 2\pi \dfrac{27}{r} + \pi r^2

\frac{\mathrm{d} s}{\mathrm{d} r}=\frac{\mathrm{d}}{\mathrm{d} r} (2\pi \dfrac{27}{r} + \pi r^2)

\frac{\mathrm{d} s}{\mathrm{d} r}=54\pi \dfrac{-1}{r^2}+2\pi r

\frac{\mathrm{d} s}{\mathrm{d} r}=0

for least amount of material requirement.

\dfrac{54\pi }{r^2} = 2\pi r\\r=3m

hence radius comes out to be 3 m

for height put the value in the equation 1

so, height of the cylinder comes out to be 3m

3 0
3 years ago
Read 2 more answers
in the system to the right, the pulleys are frictionless and the system hangs in equilibrium. Determine the values of each of th
irga5000 [103]
Look up pulleys problem through Khan academy and a video should pop up with a problem similar and you should be able to walk through it .
4 0
2 years ago
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