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3 years ago

Name three situations in which force is created. Describe the cause the cause of the force in each situation.

1 answer:

6 0

There are four aerodynamic forces affecting an airplane while in flight, they are 1.Lift 2. Weight 3.Thrust 4. Drag. Lift is cause by different wind velocity under and over an airfoil shape of the wing of an airplane. According to Bernoulli's principle as velocity increases pressure decreases, as velocity decreases pressure increases. This pressure difference create a force known as "Lift". Weight is the downward force cause by gravitational force exerted by earth on an airplane. Thrust is a forward force created by combustion engine usually cause by the expansion of gases as it passes through the nozzle of an engine. Drag force is opposite of thrust, cause by friction air on the body and surface of an airplane

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Which best describes what happens to sunlight after passing through the keyhole of a door

Law Incorporation [45]

Given the answers I would say B, Is your best choice because when the light goes through a hole it spreads out, because most light rays (<) outward.

7 0

4 years ago

A resistor rated at 250 k Ω is connected across two D cell batteries (each 1.50 V) in series, with a total voltage of 3.00 V. Th

Andrej [43]

Given Information:

Resistance = R = 250 Ω

Voltage = V = 3 V

Tolerance = ±5%

Required Information:

Maximum current = Imax = ?

Minimum current = Imin = ?

Answer:

Imax = 12.6 uA

Imin = 11.4 uA

Explanation:

As we know from Ohm's law

I = V/R

The current will be maximum when R is minimum and the current will be minimum when R is maximum

Rmax = 250,000 + 0.05*250,000

Rmax = 250,000 + 12500

Rmax = 262500 Ω

Rmin = 250,000 - 0.05*250,000

Rmin = 250,000 - 12500

Rmin = 237500 Ω

Imax = V/Rmin

Imax = 3/237500

Imax = 12.6 uA

Imin = V/Rmax

Imin = 3/262500

Imin = 11.4 uA

6 0

3 years ago

Which statement is true according to Hooke's law?

Mkey [24]

Hooke's law:

F = -kx

F is the spring force, k is the spring constant, and x is the displacement of the spring. The negative sign is there to indicate that the force acts in the opposite direction of the displacement (e.g. if you stretch a spring towards the right, you will feel a force pulling towards the left).

Looking at Hooke's law, the magnitude of the force is directly proportional to its displacement; that is to say, if the displacement increases, so does the magnitude of the spring force, and if the magnitude of the force increases, this necessitates an increase in the displacement.

1. As the force on a spring increases, the distance it stretches decreases.

An increase in force necessitates an increase in displacement. This statement is false.

2. As the force on a spring increases, the distance it stretches increases.

An increase in force necessitates an increase in displacement. This statement is true.

3. As the force on a spring increases, the spring's length remains unchanged.

An increase in force necessitates an increase in displacement. This statement is false.

4. The distance a spring stretches has nothing to do with how much force is applied to it.

Hooke's law shows that the spring force and its displacement are very much dependent on each other. This statement is false.

Only statement 2 is true. As the force on a spring increases, the distance it stretches increases.

3 0

3 years ago

If earth did not rotate how would air at the equator move?

Dmitry_Shevchenko [17]

Heat rises, and it is warmer at the equator, so I think warm air would rise at the equator and move towards the cooler poles.

3 0

4 years ago

Read 2 more answers

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as

Dmitriy789 [7]

Answer:

a) $v=5.6725\,m.s^{-1}$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^{-2}$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^{-1}$

putting the values in eq. (1)

$v^2=0^2+2\times 3.62 \times 10$

$v=8.5088\,m.s^{-1}$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20\times 8.5088$

$p=170.1764\,kg.m.s^{-1}$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)\times v'=p$

$(20+10)\times v'=170.1764$

$v'=5.6725\,m.s^{-1}$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$\frac{1}{2} (M+m).v'^2=(M+m).g.h$

$\frac{1}{2} (20+10)\times 5.6725^2=(20+10)\times 9.8\times h$

$h\approx 1.6420\,m$

above the normal hanging position.

3 0

3 years ago