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Helga [31]
3 years ago
12

20 POINTS

Physics
1 answer:
bekas [8.4K]3 years ago
5 0
They can pretty much be by water i think
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An airplane flies at airspeed (relative to the air) of 280 km/h . The pilot wishes to fly due North (relative to the ground) but
MrRa [10]

Answer:

the pilot should head the plane 7.547^{\circ} towarrds south- west

Solution:

The airspeed of the airplane, v = 280 km/h

The velocity of the wind, v' = 52 km/h South-west

Angle, \theta = 225^{\circ}

Now, measured angle in the clockwise direction from North:

sin225 = sin(\pi + 45) =  - sin 45^{\circ}

Now,

vsinx - v'sin45 = 0

280sinx = 52sin45

x = sin^{- 1}(\frac{52}{280}\times \frac{1}{\sqrt{2}})

x = 7.547^{\circ} south- west

4 0
3 years ago
A student charges a balloon and then brings it near a metal sphere hanging from the
Hunter-Best [27]

Answer:

B

Explanation:

the balloon has a negative charge and the metal sphere has a positive charge

3 0
3 years ago
Read 2 more answers
A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

6 0
2 years ago
What happens when velocity and acceleration are at right angles to each other
photoshop1234 [79]
You get circular motion, where the acceleration is pointing towards the center of the circle, as long as they are constant, and not fluctuating.
7 0
3 years ago
In an incompressible three-dimensional flow field, the velocity components are given by u = ax + byz; υ = cy + dxz. Determine th
Ksivusya [100]

An incompressible flow field F in a 3D cartesian grid with components u,v,w:

F = u + v + w

where u,v,w are functions of x,y,z

Must satisfy:

∇·F = du/dx + dv/dy + dw/dz = 0

We have a field F defined:

F = u+v+w, u = ax+byz, v = cy+dxz

du/dx = a, dv/dy = c

Recall ∇·F = 0:

∇·F = du/dx + dv/dy + dw/dz = 0

a + c + dw/dz = 0

dw/dz = -a-c

Solve for w by separation of variables:

w = ∫(-a-c)dz

w = -az - cz + f(x,y)

f(x,y) is some undetermined function of x and y

The question states that w is not a function of x and y, therefore f(x,y) = 0...

w = -az - cz

8 0
3 years ago
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