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1 year ago

Explain how the forces need to change so the aeroplane can land

1 answer:

7 0

When an airplane is flying straight and level at a constant speed, the lift it produces balances its weight, and the thrust it produces balances its drag. However, this balance of forces changes as the airplane rises and descends, as it speeds up and slows down, and as it turns.

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Tyson throws a shot put ball weighing 7.26 kg. At a height of 2.1 m above the ground, the mechanical energy of the ball is 172.1

nika2105 [10]

Answer:

Velocity, v = 2.50 m/s

Explanation:

Given that,

Mass of the ball, m = 7.26 kg

Height above the ground, h = 2.1 m

Mechanical energy of the ball, T = 172.1 J

To find,

The velocity of the ball at the given point.

Solution,

The sum of potential energy and the kinetic energy is called the mechanical energy of an object. It is given by :

$T=K+P$

K is the kinetic energy

P is the potential energy

$T=\dfrac{1}{2}mv^2+mgh$

On rearranging the above equation,

$v^2=\dfrac{2(T-mgh)}{m}$

$v^2=\dfrac{2(172.1-7.26\times 9.8\times 2.1)}{7.26}$

v = 2.50 meters

Therefore, the velocity of the ball at the given point is 2.50 m/s.

7 0

1 year ago

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Magnetic fields within a sunspot can be as strong as 0.4T. (By comparison, the earth's magnetic field is about 1/10,000 as stron

WARRIOR [948]

Answer:

The speed of ejection is $2.06\times 10^{4}\ m/s$

Solution:

As per the question:

Magnetic field density, B = 0.4 T

Density of the material in the sunspot, $\rho = 3\times 10^{4}\ kg/m^{3}$

Now,

To calculate the speed of ejection of the material, v:

The magnetic field energy density is given by:

$U_{B} = \frac{B^{2}}{2\mu_{o}}$

This energy density equals the kinetic energy supplied by the field.

Thus

$KE = U_{B}$

$\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}$

where

m = mass of the sunspot in $1\ m^{3}$ = $3\times 10^{- 4}\ kg/m^{3}$

$v = \frac{B}{\sqrt{\mu_{o}m}}$

$v = \frac{0.4}{\sqrt{4\pi \times 10^{- 7}\times 3\times 10^{- 4}}} = 2.06\times 10^{4}\ m/s$

3 0

1 year ago

A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for

Lorico [155]

The force tending to lift the load (vertical force) is equal to 22.5N.

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

$VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N$

Hence, we can see that the force tending to lift the load off the ground (vertical force) is equal to 22.5N.

Have a nice day!

8 0

2 years ago

A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington

Roman55 [17]

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

The current has velocity vector (relative to the Earth)

$\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}$

$\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ$

which is approximately 41º west of north.

6 0

1 year ago

Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th

katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0

1 year ago

Read 2 more answers